Introductionwebpages.sou.edu/~ordonez/BA282Web/BA_282_COURS… · Web viewIdentify the critical value for the test (one tail). a. 0.0199. b. 1.96. c. -1.96. d. -1.645. e. 1.645.

Lecture Outlines for Applied Business StatisticsRene Leo E. Ordonez, Southern Oregon University

Summer 2010

Note: The problems below were based on the text Doing Statistics for Business with Excel, 2nd Edition, by Pelosi and Sandifer. The problems in each section are courtesy of same text.

Page Number

Reference to Excel and Minitab (Statistical functions) 3

Introduction 8

Probability Distributions 10

Sampling Distributions and Confidence Intervals 19

Hypothesis Testing: An Introduction 39

Inferences: One Population (Hypothesis Testing) 56

Comparing Two Populations 68

Improving and Managing Quality 83

Experimental Design and Analysis of Variance (ANOVA) 89

Analysis of Qualitative Data (Chi-square) 102

Regression and Correlation 113

Sample Midterm Exam 140

Sample Final Exam 148

Statistical Procedure EXCEL 2007 MINITAB

Descriptive Statistics(mean, median, etc.)

Data> Data Analysis > Descriptive Statistics Stat > Basic Statistics > Display Descriptive Statistics

Confidence Interval Estimates Mean Proportion

Data > Data Analysis > Descriptive StatisticsNONE

Stat > Basic Statistics > 1 Sample t (or 1 Sample z)Stat > Basic Statistics > 1 Proportion

One Population Hypothesis Test Mean Proportion

Data > Data Analysis > Descriptive StatisticsNONE

Stat > Basic Statistics > 1 Sample t (or 1 Sample z)Stat > Basic Statistics > 1 Proportion

Two Populations Hypothesis Test Means of 2 Dependent Samples Means of 2 Independent Samples (small samples, equal vars.) Means of 2 Independent Samples (small samples, unequal vars) Means of 2 Independent Samples (large samples)

Variances of 2 Populations

Proportion of 2 Populations

Data > Data Analysis > t-Test:Paired Two Sample for Means t-Test: Two Sample Assuming Equal Variances t-Test: Two Sample Assuming Unequal Variances z-Test: Two Sample for Means

F-Test Two-Sample for Variances

NONE

Stat > Basic Statistics > Paired tStat > Basic Statistics > 2 Sample tStat > Basic Statistics > 2 Sample tStat > Basic Statistics > 2 Sample z

Stat > Basic Statistics > 2 Variances

Stat > Basic Statistics > 2 Proportions

Analysis of Variance One Factor (use for comparing 2 or more population means) Two Factor With Replication Two Factor Without Replication Interaction Effect Plot

Data > Data Analysis > Anova: Single Factor Anova: Two-Factor Wtih Replication Anova: Two-Factor Wtihout ReplicationNONE

Stat > ANOVA > Oneway Unstacked (or Stacked)Stat > ANOVA > TwowayStat > ANOVA > TwowayStat > ANOVA > Interactions Plot

Chi-square Analysis Goodness of Fit Test Comparing Proportions of Two or More Groups Testing Independence of Two Nominal Variables Comparing Variances of Two Populations

NONENONENONE

Data > Data Analysis > F-Test Two-Sample for Variances

NONEStat > Tables > Cross Tabulation (for raw data)Stat > Tables > Chisquare Test (for summarized data)

Stat > Basic Statistics > 2 Variances

Regression and Correlation Analysis Data > Data Analysis > Regression Stat > Regression Fitted Line Plot Regression Residual Plots

STATISTICAL PROCESSING USING MINITAB

A free 30-day trial copy of the full commercial version of Minitab can be downloaded from www.minitab.com

Basic StatisticsConfidence Interval EstimationHypothesis Testing

One-way and Two-way ANOVA

Use for generating means, standard deviation, etc

Use for One-way ANOVA(procedure for comparing means of two or more independent samples)

Use for two-way ANOVA with replication

Use for testing a population mean(n ≥ 30 or known)

Use for testing a population mean (n < 30, unknown, and normal)

Use for testing a population proportion (approximation to the binomial)

Use for generating interactions plot for two-way ANOVA with replication

Use for testing variances (or standard deviation) of two groups

Use for comparing means of two INDEPENDENT sample

Use for comparing means of two DEPENDENT samples

Use for comparing proportions of two populations

http://www.minitab.com/

BA 282: Applied Business Statistics Course Outline

Chisquare Tests

Regression and Correlation Analysis

Lecture Notes to Accompany Page 4 of 151BA 282: Applied Business Statistics Ordonez, School of Business, SOU

Use for performing Chisquare test using RAW data (procedure for testing whether two qualitative variables are independent)

Use for performing Chisquare test using TABULATED data

Use for generating output for regression and correlation analysis(for simple and multiple models)


STATISTICAL PROCESSING USING EXCEL

Data > Data Analysis


Use for generating means, standard deviation, etc

Use for testing variances (or standard deviation) of two groups

Use for One-way ANOVA

Use for two-way ANOVA with replication

Use for regression and correlation analysis

Use for comparing means of two DEPENDENT sample

Use for comparing means of two INDEPENDENT samples(n < 30 and equal variances)

Use for comparing means of two INDEPENDENT samples(n < 30 and unequal variances)

Use for comparing means of two INDEPENDENT samples(n ≥ 30)


Important Note:

If you don’t see Data Analysis option under the DATA tab you have to add it in. Here are the steps:

1. Click on the Microsoft Icon (upper left corner), then select Excel Options

2. Click Add Ins, then click Analysis Tool Pak VBA, then Go







3. Select Analysis Toolpak and Analysis Toolpak VBA, then click OK



INTRODUCTION

1. What is statistics? A science that deals with rules and procedures that govern how to:

collect summarize describe interpret data

2. Why study statistics?

Decisions! Decisions! Decisions!

3. The Importance of understanding probabilitySome 'real life' examples (it’s just a game!)

Monty Hall DilemmaSuppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors? (Craig. F. Whitaker,Columbia, MD )

Three Shell Game

Operator: Step right up, folks. See if you can guess which shell the pea is under. Double your money if you win.

After playing the game a while, Mr. Mark decided he couldn't win more than once out of three.

Operator: Don't leave, Mac. I'll give you a break. Pick any shell. I'll turn over an empty one. Then the pea has to be under one of the other two, so your chances of winning go way up.

4. Ways of assigning (determining) probabilities

Subjective - describes an individual's personal judgement about how likely a particular event is to occur. It is not based on any precise computation but is often a reasonable assessment by a knowledgeable person.

Relative -- Relative probability is another term for proportion; it is the value calculated by dividing the number of times an event occurs by the total number of times an experiment is carried out.

Objective (classical) – is probability based on symmetry of games of chance or similar situations. For example:

Coin tossing experiment P(head)Die tossing experiment P(“one”)



Monty Hall Dilemma P(win)5. Important statistical terms and concepts

(KNOW THESE DEFINITIONS AND SYMBOLS!)

Population vs. sample

population -- any entire collection of people, animals, plants or things from which we may collect data. It is the entire group we are interested in, which we wish to describe or draw conclusions about

sample -- a group of items selected from a population. Conclusions about the population are drawn by studying the sample.

Parameter vs. statisticsparameter – a numeric characteristic of a population

statistic – a numeric characteristic of a sample. It is used to estimate and unknown population parameterParameters are often assigned Greek letters ( e.g. , , ), whereas statistics are assigned Roman letters (e.g. s, p).

Common measures of central tendencyMean, median, mode

Common measures of dispersionRange, variance, standard deviation

Common symbols used in statistics

Important Symbols: Must Know

POPULATION SAMPLESize N n


StatisticsParameters


One Population

Mean

Variance 2 s2

Standard deviation s Proportion pTwo Populations Comparing Means 1 vs 2 1 vs 2

Comparing Proportions 1 vs 2 p1 vs p2

Comparing Variances 21 vs 2

2 s21 vs s2

2

Comparing Standard Deviations 1 vs 2 s1 vs s2



PROBABILITY DISTRIBUTIONS

1. Definitions

probability – likelihood or chance of an event occurring

experiment -- any process or study which results in the collection of data, the outcome of which is unknown.

random variable -- an outcome of an experiment. It need not be a number, for example, the outcome when a coin is tossed can be 'heads' or 'tails'. However, we often want to represent outcomes as numbers. Usually denoted by the letter “X”Example:

toss a coin 5 times (experiment), observe the number of heads (variable)

Randomly select 20 students (sample), record each student’s GPA (variable)

2. Random variables

discrete (185) - usually involves counting (e.g. number of defectives, number of correct answers, etc.) If a random variable can take only a finite number of distinct values, then it must be discrete

in the coin tossing experiment above, the random variable is “number of heads” x = {0, 1, 2, 3, 4, 5}

continuous (186) – usually involves something that is measured A continuous random variable is one which takes an infinite number of possible values. Examples include height, weight, the amount of sugar in an orange, the time required to run a mile

in the student sampling above, the random variable is GPAx = {0 to 4.0}

3. Common Discrete Probability Distributions Uniform

Binomial (191)The trials must meet the following requirements:a) the total number of trials (n) is fixed in advance;b) there are just two outcomes of each trial; success and failure;c) the outcomes of all the trials are statistically independent;d) all the trials have the same probability of success


x


Example: coin tossing Hypergeometric (200)

a) each trial has just two outcomes; success and failure;b) the outcomes of all the trials are statistically dependent;c) the probability of success changes from trial to trial

Poisson (203)-- typically, a Poisson random variable is a count of the number of events that occur in a certain time interval or spatial area. For example, the number of cars passing a fixed point in a 5 minute interval; the number of calls received by a switchboard during a given period of time

4. Common Continuous Probability Distributions

Uniform (219)

Exponential (Not covered but will be introduced and covered in BA 380-Operations Management)

Normal (223 to 228)

5. The Normal Distribution (223 to 228) Characteristics

bell-shaped mean = median = mode area underneath the curve equals 1 symmetric about the mean (left side is mirror-image of right side)


BA

f(x)

x


area left of mean = 0.50 = area right of mean asymptotic



6. How to find areas (probabilities) using the Normal table (229 to 234)

A MUST UNDERSTAND CONCEPT!

Standard Normal Distribution

z 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.090 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359

0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.57530.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.61410.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.65170.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.68790.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.72240.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.75490.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.78520.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.81330.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.83891 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621

1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.88301.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.90151.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.91771.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.93191.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.94411.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.96331.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.97061.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.97672 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817

2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.98572.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.98902.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.99162.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.99362.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.99522.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.99642.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.99742.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.99812.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986



EXCEL FUNCTIONS FOR NORMAL DISTRIBUTION

IMPORTANT: LEARN HOW TO USE THESE FUNCTIONS!


0

= 1

z


Exercises in Using the Standard Normal Table

Use Excel’s =NORMSDIST(z) function or the Standard Normal Distribution to answer the problems below.

1) Draw the normal distribution, and shade and find the areas (probabilities) of the following expressions:

e.g. P ( Z > 0 ) = ?

a) P ( Z < 1.0 ) b) P ( Z > 1.0 )

c) P ( Z < 1.0 ) d) P ( 1.0 < Z < 1.0 )



e) P ( 1.0 < Z < 2.5) f) P ( Z > 2.65 )

2) Given probabilities and their respective probability expressions, draw the normal distribution, shade the areas and find the corresponding z values:

Use Excel’s =NORMSINV(area) function or the Standard Normal Distribution to answer the problems below.

a) P ( Z < z ) = 0.95 b) P ( Z > z ) = 0.95



c) P ( Z < z ) = 0.25 d) P ( Z > z ) = 0.25



Learning it! Exercises

The amount of money spent by students for textbooks in a semester is a normally distributed random variable with a mean of $235 and a standard deviation of $15

(a) Sketch the normal distribution that describes the amount of money spent on textbooks in a semester.

(b) What is the probability that a student spends between $220 and $250 in any semester?

(c) What percentage of students spend more than $270 on textbooks in any semester?

(d) What percentage of students spend less than $225 in a semester?



The actual amount of a certain brand of orange juice in a container marked half gallon is a normally distributed random variable with a mean of 65 oz. and a standard deviation of 0.35 oz.

(a) What percentage of the containers contain more than 64.5 oz?

(b) What percentage of the containers contain between 64 and 66 oz?

(c) If federal law says that 98% of all the containers must be or above the labeled weight, does this brand of orange juice meet the requirement?



The size of a gift/specialty store in a regional super mall is a normally distributed random variable with a mean of 8,500 sq ft and a standard deviation of 260 sq ft. What is the probability that a randomly selected gift/specialty store in a regional super mall is:

a) more than 8000 sq ft?

b) between 8300 and 9000 sq ft?

c) less than 9,500 square feet



SAMPLING DISTRIBUTIONS AND CONFIDENCE INTERVALS

1. The Distribution of the Sample Mean ( ) and the Central Limit Theorem (266 to 269)

Central Limit Theorem Definition (271):

When randomly sampling from a population, the distribution of the sample mean( ) is:

approximately normal regardless of the original population distribution so as the sample is large (at least 30. But this sample size restriction is not required if the population is normal to begin with) with

a mean equal to and

a standard deviation equal to


x x

x x

=

X X

X

X X


2. Confidence Intervals

Use: For estimating unknown population parameters

Definition of confidence interval (290) A probability that the interval contains the true population parameter

e.g. P ( U ≤ ≤ L ) = 1

Components of confidence interval – point estimate and margin of error

a. Point estimate (290)

A single number that is calculated from sample data

Is used to estimate a population parameter

e.g. sample mean is a point estimate for population mean, sample proportion is a point estimate of population proportion

POPULATION SAMPLESize N nOne Population

Mean Variance 2 s2

Standard deviation sProportion pTwo Populations

Comparing Means 1 vs 2 1 vs 2

Comparing Proportions 1 vs 2 p1 vs p2

Comparing Variances 21 vs 2

2 s21 vs s2

2

Comparing Standard Deviations 1 vs 2 s1 vs s2


Point estimators a.k.a. statistics

Parameters


b. Margin of error (e)

when added to and subtracted from the point estimator gives the upper and lower limit for the range of values where the population parameter could be found

is affected (or determined) by: confidence level sample size population variability

Interpreting the confidence interval (294 - 295)

Say that you computed a 95% confidence interval estimate for the mean of a certain population as 3.2 and 3.5

correct interpretation : “We are 95% confident that the interval 3.2 and 3.5 contains the true population mean”

incorrect interpretation: “There is a 95% chance that the population mean is in the interval 3.2 and 3.5”

3. Computing a confidence interval for the population mean ()

z-dist for Large samples or known (290-297)

t-dist for small samples and unknown (298-304)



4. Computing a confidence interval for qualitative data (the population proportion ()) (305-310)

5. Sample Size Calculations (311-313)

For estimating a population mean

For estimating a population proportion(Using the Normal distribution as an approximation to the Binomial distribution)

Factors affecting sample size requirement(1) confidence level(2) variability of the population(3) acceptable level of margin of error





CONFIDENCE INTERVAL ESTIMATION

Confidence IntervaI for a. Known

b. Unknown

Confidence Interval for

SAMPLE SIZE (n) DETERMINATION

For estimating

For estimating



STUDENT’S t-DISTRIBUTION TABLE



Learning it! Exercises (by hand and using Excel)

7.6 A manufacturer of pain reliever claims that it takes an average of 12.75 minutes for a person to be relieved of headache pain after taking its pain reliever. The time it takes to relief is normally distributed with a standard deviation of 0.5 minutes. A sample of 12 people is taken and the data are shown here:

12.9 13.2 12.7 13.1 13.0 13.1 13.0 12.6 13.1 13.0 13.1 12.8

a. Find the sample mean

b. Find the standard error of the sample mean

c. If the manufacturer claims that the mean is 12.75 minutes, find the z-score of the sample mean

d. What do you think of the manufacturer’s claim based on the z-score? (translation – if the claim made by the manufacturer is correct, how likely is it to observe a sample at least as large as your answer to (a))



7.8 The U-Male-It Hardware chain has 10 different stores within a certain geographical region. The dollar value of customer sales is normally distributed with an average sale of $35.25 and a standard deviation of $2.50. You have recently been hired to manage one these stores and under your leadership the average sales based on a sample of 100 customers is $36.50. You are very proud of the increased average sale and point this out to senior management.

a. Find the z-score for $36.50

b. Based on this z-score, is your pride justified?(translation – is the difference observed between the original average sale of $35.25 and the sample average sales of $36.50 significant, or is the difference mainly attributable to chance?)



7.16A national grocery chain is considering opening a store at a particular location. To be sure that enough traffic goes by that location, the grocery chain took a sample of vehicles crossing the intersection on 40 days. The results are shown in the table below:

Number of Cars CrossingLocation per Day

1431 1540 1293 13401302 1700 1533 14021255 1840 1272 14671377 1642 1572 12201450 1139 1520 14771483 1227 1227 15151529 1684 1257 12421588 1782 1238 13501535 1491 1276 13671533 1513 1420 1375

a. Find a 95% confidence interval for the average number of cars that pass this location on a daily basis. The standard deviation is assumed to be 165 cars.

b. The company has decided to open a store at this location only if there is a daily average of at least 1400 cars passing this location. Based on your confidence interval, would you advise the company to open a store at this location? Explain why or why not.



7.20

The police department is concerned about the ability of officers to identify drunk drivers on the road. Before instituting a new training program they take a sample of 28 arrests and record the level of alcohol in the blood at the time of the arrest. Assume that the level of alcohol in the blood is normally distributed. The data are shown below:

92 93 108 173 194 133 207127 256 184 253 159 101 133204 182 173 105 153 150 180209 141 151 133 147 209 252

(a) Find a 90% confidence interval for the average alcohol level in the blood at the time of arrest

(b) Find a 95% confidence interval for the average alcohol level in the blood at the time of arrest



Excel Solution

Data > Data Analysis > Descriptive Statistics


Margin of Error

Descriptive statistics


Minitab Solution

Stat > Basic Statistics > 1-Sample t > Options



7.21

A large amusement park has recently added 5 new rides, including a large roller coaster called Mind Eraser. Management is concerned about the waiting times on the new roller coaster. A random sample of 10 people is selected and the time (in minutes) that each person waits to ride the Mind Eraser is recorded and shown below:

43 80 48 61 7466 54 72 58 68

(a) Find a 95% confidence interval for the average waiting time for the Mind Eraser, assuming that the waiting time is normally distributed.

(b) The park management thinks that if customers have to wait more than 60 minutes for a ride, then the park should increase the staff to reduce the waiting time. Based on your confidence interval, does the park need to increase the staff? Explain why or why not.



7.25

I asked 100 imaginary friends (only to avoid the time and cost of data collection) the following question: “Do you regularly watch MTV’s Beavis and Butthead?” Of the 100 friends, 35 of them answered yes.

(a) Calculate a 95% confidence interval for the viewership of this show.

(b) MTV is considering canceling the show if less than one-third of the population regularly watches the show. Based on this information, what will MTV do?



Minitab Solution

Stat > Basic Statistics > 1-Proportion

(No Excel procedure)



7.31How many stores must be sampled for the woman who wants to buy a ranch to be 95% confident that the error in estimating the average fat content per pound in steaks sold in the Portland, Maine area is at most 0.05 oz? The standard deviation of fat content is known to be 0.30 oz.

7.32How many months must be sample for analysts to be 99% confident that the error in estimating the average monthly price of peanut butter is at most $0.02? Assume the standard deviation is $0.035



7.42In an effort to improve the quality of the CD players that your company makes, you have started to sample the component parts that you purchase from an outside supplier. You will accept the shipment of parts only if there is less than 1% defectives in the shipment. Recognizing that you cannot test the entire shipment (or population), you select a sample of 25 components to test. You find 3 defectives in the sample.

(a) Find a 90 percent confidence interval for the proportion of components in the population that are defective.

(b) Based on your confidence interval, should you accept the shipment? Why or why not?



7.44A hotel is studying the proportions of rooms that are not ready when customers check in to the hotel.

(a) How many rooms must be in the sample for the hotel to be 95% confident that the margin of error is at most 1%?

(c) How many rooms must be in the sample for the hotel to be 95% confident that the margin of error is at most 3%?



Lecture Notes to Accompany Page 42 of 151Applied Business Statistics School of Business, SOU


HYPOTHESIS TESTING: AN INTRODUCTION

1. What is a hypothesis test? (327)

a hypothesis is an idea, an assumption, or a theory about the behavior of one or more variables in one or more populations

a hypothesis test is a statistical procedure that involves formulating a hypothesis and using sample data (n) to decide on the validity of the hypothesis i.e. is the sample consistent with the hypothesis (in which case you believe the hypothesis) or whether the sample is inconsistent with the hypothesis (in which case you choose not to believe it or to reject it)

important! in statistical testing, regardless of the specific hypothesis that you are testing, the basic procedure is the same! Your understanding of the concepts introduced in this chapter is crucial for the remaining chapters!

2. Steps in performing hypothesis test: (328-332)Step 1: Set up the null and alternative hypothesesStep 2: Identify the significance level () for determining the critical valueStep 3: Identify the appropriate distributionStep 4: Collect the sample data (for determining the computed value)Step 5: Compare the computed value to the critical value (or the p-value to the significance level)Step 6: Make a statistical conclusion (reject the null or fail to reject the null)Step 7: Make a managerial conclusion (usually a statistical test is conducted to assist in a decision-making process)

3. Null vs. Alternative Hypotheses and decision rule (329)

Important things to remember about H0 and H1

H0: null hypothesis and H1: alternate hypothesis H0 and H1 are mutually exclusive and collectively exhaustive H0 is always presumed to be true H1 has the burden of proof

a random sample (n) is used to “reject H0” or to “fail to reject H0 “

If we conclude 'do not reject H0', this does not necessarily mean that the null hypothesis is true, it only suggests that there is not sufficient evidence against H0 in favor of H1; rejecting the null hypothesis then, suggests that the alternative hypothesis may be true.

equality is always part of H0 (e.g. “=” , “≥” , “≤”). “≠” “<” and “>” always part of H1



Structure:

4. Setting up the null and alternative hypotheses (Is the test two-tailed (non-directional) or one-tailed (directional)?) and establishing the Rejection Region (333)

identify the parameter being tested (, , 2) determine how many populations are included in the test Is “the claim” the null hypothesis or the alternate hypothesis?

In actual practice, the status quo is set up as H0

If the claim is “boastful” the claim is set up as H1 (we apply the Missouri rule – “show me”). Remember, H1 has the burden of proof

In problem solving, look for key words and convert them into symbols (see examples below)

Some Examples

Keywords InequalitySymbol

Part of:

Larger (or more) than > H1

Smaller (or less) < H1

No more than H0

At least ≥ H0

Has increased > H1

Is there difference? ≠ H1

Has not changed = H0

Has “improved”, “is better than”. “is more effective”

See note below

H1

The direction of the test involving claims that use the words “has improved”, “is better than”, and the like will depend upon the variable being measured.

For instance, if the variable involves time for a certain medication to take effect, the words “better” “improve” or more effective” are translated as “<” (less than, i.e. faster relief).

On the other hand, if the variable refers to a test score, then the words “better” “improve” or more effective” are translated as “>” (greater than, i.e. higher test scores)

The equality (, ≥, =) is always part of the null hypothesis. 5. Two types of error in hypothesis testing: Type 1 () vs.Type 2 () (330-333)


z0


Statistical definitions:Type 1 () – the probability of rejecting a TRUE H0

Type 2 () – the probability of failing to reject (or “accepting”) a FALSE H0

True ConditionStatistical Conclusion H0 TRUE H0 FALSEReject H0 Type 1 () Correct

Fail to reject H0 Correct Type 2 ()

More on Type 1 (): in addition to its definition as “the probability of rejecting a TRUE H0 it is also:

known as the significance level of a test (or simply, the significance level) usually ranges between 0.01 and 0.10 (which level is ‘best’? see next subsection) used to generate the critical value for a test an area at the tail end of a distribution, and this area is known as the reject H0 region (or the rejection region) The critical value marks the boundary between the reject H0 and fail to reject H0 regions

Which should be avoided - Type 1 or 2 error?

For a given sample size (n), there is a trade-off between Type 1 and Type 2 errors, that is, decreasing one will increase the other To decrease both types at the same time, a larger sample size must be taken However, because of cost, time, and practicality of sampling concerns, oftentimes we need to choose between type 1 and type 2 errors. Which should we decrease? Depends on the cost associated with each type of error

EXAMPLES:



In each of the example below, the Type 1 and Type 2 errors are defined in non-statistical terms. Can you identify the ‘cost’ associated with each type of error? For instance, in criminal cases, the cost associated with a Type 1 error (that is, a jury convicting an innocent person) is the freedom, or worse yet, the life of the accused. Now compare this to the cost of a Type 2 error. As a society, which do we consider as worse?

Justice system - criminal and civil cases

H0: InnocentH1: Guilty

True ConditionStatistical Conclusion Innocent Guilty

Reject H0

(Guilty)

Type 1 () – conclude that

accused is guilty when in fact is

innocent

Correct

Fail to reject H0 (not Guilty) Correct

Type 2 () – conclude that accused is not

guilty when in fact is

Business - quality control situations – process monitoring

H0: Process is in controlH1: Process is not in control

True ConditionStatistical Conclusion Process OK Process Not OK

Reject H0

(process not OK)


process is not in control when in

fact is

Correct

Fail to reject H0 (process OK) Correct


process is OK when in fact is not



Business - quality control situations- quality assurance

H0: Lot of shipment is goodH1: Lot of shipment is not good

True ConditionStatistical Conclusion Good Lot Not Good Lot

Reject H0

(shipment is not good)

Type 1 () – conclude that lot is not good when in

fact is(producer’s risk)

Correct

Fail to reject H0 (shipment is good)

Correct


shipment is good when in fact is not(consumer’s risk)



6. P-values (339)

The probability value (p-value) of a statistical hypothesis test is the probability of getting a value of the test statistic as extreme as or more extreme than that observed by chance alone, if the null hypothesis H0, is true. (see example below)

It is the probability of wrongly rejecting the null hypothesis if it is in fact true.

When used as a decision rule in hypothesis testing, the p-value is compared to the significance level (α). If the r-value is smaller, the conclusion is to reject the null hypothesis (or, we say that the result “is significant.”

Here’s a decision rule using the p-value as a decision rule – this applies to ALL forms of hypothesis tests!

Remember this very important rule!

Important interpretation! Small p-values suggest that the null hypothesis is unlikely to be true. The smaller it is, the more convincing is the rejection of the null hypothesis. It indicates the strength of evidence for say, rejecting the null hypothesis H0, rather than simply concluding 'reject H0' or 'do not reject H0'.



P-value Example:

example:

Hypothesis Major Concepts, Hypothesis One Population Determining Appropriate Test, Hypothesis Testing Major Concepts_Pvalues



TESTING A POPULATION MEAN ()

H0: = value H0: value H0: value H1: value H1: < value H1: > value

Reject H0 if: Reject H0 if: Reject H0 if:Z > Z/2 Z < Z Z > Z

t > t/2, n 1 t < t, n 1 t > t, n 1

TESTING A POPULATION PROPORTION ()



TESTING A POPULATION VARIANCE (2)

H0: 2 = value H0: 2 value H0: 2 value H1: 2 value H1: 2 < value H1: 2 > value

Reject H0 if: Reject H0 if: Reject H0 if:2 < 2 1-/2 2 < 2 1- 2 > 2 2 > 2 /2




Setting up the Hypotheses and Determining Type I and II Levels

For items 8.7 to 8.27 below, do the following:(a) State the Null and Alternative hypotheses(b) State the consequence of a Type I error(c) State the consequence of a Type II error(d) Suggest a value for , and justify your choice

8.7 Administrators at a small college are concerned that part-time evening students may not be familiar with all the services of the College. They wish to offer an orientation program to these students but recognize that most of the part-time students work during the day and are generally very busy. The administrators do not want to prepare an elaborate presentation if only a handful of part-time students will attend. Hence, they will conduct the orientation if more than 25% of the part-time students are interested in attending.

8.8 A company CEO is thinking about setting up an on-site day-care program for its employees. The CEO has stated that she will do so only if more than 80% of the employees favor such a decision. Set up the null and alternative hypotheses.

8.9 In an attempt to improve quality many manufacturers are developing partnerships with their suppliers. A local fast-food burger outfit has partnered with its supplier of potatoes. The burger outfit buys potatoes in bags that weigh 20 lbs. It wishes to set up the null and alternative hypotheses to test if the bags do weigh on average 20 lbs.

8.10 You are a connoiseur of chocolate chip cookies and you do not think that Nabisco’s claim that every bag of Chips Ahoy cookies has 1000 chocolate morsels is correct. Set up the null and alternative hypotheses to test this claim.



8.11 Antilock brake systems (ABS) have been hailed as a revolutionary safety feature. A study by the National Traffic Safety Administration looked at fatal accidents. The claim is that cars with ABS are in fewer fatal crashes than those without.

8.12 A college placement office wonders whether there is a difference between the average salary of engineering graduates and business school graduates.

8.13 Your new television has a 1-year warranty. You are given the option to buy a 3-year warranty and wonder if it is worth it. You wish to test the hypothesis that the average time before a problem occurs is more than 3 years

8.14 M&M/Mars claims that at least 20% of the M&M’s in each package are the new blue color

8.15 A computer center is arguing for more computers in the lab for students at a midsize college. The computer center at a university claims that the average amount of time that students spend on-line has increased from last year’s average of 1 hour per day.



8.16 It seems like you spend more money on groceries during the summer months when you eat more ice cream and drink more fluids. You know that you spend on average of $25 per week on groceries during the winter months. Set up the null and alternative hypotheses to decide if, on average, you spend more than this amount per week during the summer months.

8.17 M&M Mars claims that at least 20% of the M&M’s in each package are the new blue color. Set up the null and alternative hypotheses to test this claim.

8.18 The computer center at a university claims that the average amount of time that students spend on-line has increased from last year’s average of 1 hour per day. Set up the null and alternative hypotheses to test this claim.



(note: for the following problems use Excel’s Data Analysis Tools to generate the descriptive statistics to minimize hand calculation)

Problem 8.1

The School Committee members of a midsize New England city agreed that a strict discipline code had caused an increase in the number of student suspensions. The number of suspensions for a sample of schools in this city for the periods September 1992 to February 1993 is shown below:

CITY Number of SuspensionsCentral 245MCDI 1Chestnut 65Duggan 133Kennedy 97Forest Park 149Puttnam 1024Kiley 56Central Academy 254Commerce 114Bridge 7

The average number of suspensions for the previous year was 130.5 with a population standard deviation of 158.2

(a) Set up the null and alternative hypotheses to test if the average number of suspensions has changed

(b) Test your hypothesis using significance level of 0.05

(c) Find the p-value

(d) Display the data to see if it is reasonable to assume that the underlying population distribution is normal.

(e) Based on the p-value, what can you conclude about the average number of suspensions.



Minitab Solution

Stat > Basic Statistics > 1-Sample Z


Population standard deviation

Hypothesized mean

Direction of test (alternative hypothesis)

Raw data


Minitab Output


Confidence interval for the true mean

Computed statistic (compare to the critical statistic)

p-value of the test(compare to the significance level)


Problem 8.2

The Educational Testing Service (ETS) designs and administers the SAT exams. Recently the format of the exam changed and the claim has been made that the new exam can be completed in an average of 120 minutes. A sample of 50 new exam times yielded an average of 115 minutes. The standard deviation is assumed to be 2 minutes.

(a) Set up the null and the alternative hypotheses to test if the average time to complete the exam is has changed from 120 minutes.



(d) Based on the p-value, what can you conclude about the average time to complete the new exam?



Problem 8.28

A major manufacturer of glue products thinks it has found a way to make glue adhere longer than the current average of 90 days. The manufacturer wishes to see whether the glue products made this way have an average time to failure greater than 90 days. A sample of 30 tubes of new glue yield an average of 93 days before failing. The failure time is normally distributed with a standard deviation of 3 days.

(a) Set up the null and the alternative hypotheses to test whether average time to failure is greater than 90 days.



(d) Based on the p-value, what can you conclude about the average time to failure for the new product?



INFERENCES: ONE POPULATION (HYPOTHESIS TESTING)

1. Testing the Mean () z-dist for Large samples or known (334) t-dist for small samples and unknown (341)

2. Testing the Population Variance(2) (Not covered) 2 distribution

3. Testing the Population Proportion() (349) z-dist. (approx. to the Binomial dist.)

4. Hypothesis Testing using Minitab and Excel



TESTING A POPULATION MEAN ()



t > t/2, n 1 t < t, n 1 t > t, n 1

TESTING A POPULATION PROPORTION ()



TESTING A POPULATION VARIANCE (2)

H0: 2 = value H0: 2 value H0: 2 value H1: 2 value H1: 2 < value H1: 2 > value

Reject H0 if: Reject H0 if: Reject H0 if:2 < 2 1-/2, n - 1 2 < 2 1-, n-1 2 > 2 , n - 1

2 > 2 /2, n - 1




Problem 9.1

The cost of common goods and service in 5 cities is shown in the table below (USA Today):

City Aspirin(100)

Fast Food(hamburger,fries, soft drink)

Woman’sHaircut/Blow Dry

Toothpaste(6.4 oz)

Los Angeles $7.69 $4.15 $20.11 $2.42Tokyo $35.93 $7.62 $76.24 $4.24London $9.69 $5.80 $44.35 $3.63Sydney $7.43 $4.53 $29.93 $2.08Mexico City $1.16 $3.63 $17.94 $1.08

a. You have just returned from a business trip and you lost your receipt for the aspirin you purchased but would like to be reimbursed by your company (since you had taken the aspirin after a stressful business meeting!). You guesstimate a cost of $10.00. Your boss claims that the average cost of aspirin is less than $10.00. Using these data, can you “prove” your boss wrong? Conduct the necessary hypothesis test. Assume that all costs are normally distributed.

b. Based on these data, is there enough evidence to support your submitting a cost of $10.00 for the fast-food meal on your trip?

c. If you remove Tokyo from the data set do your answers to parts (a) and (b) change? What does this tell you about the effect of outliers on the hypothesis test of µ when you have a small sample?



Problem 9.2

The marketing material for a New England Ski resort advertises that they can make snow whenever the temperature is 32°F or below. To demonstrate how often this happens their material includes the following line graph of the weekly average temperatures (See graph in text).

The data that generated the graph are shown below:

Week 1 2 3 4 5 6 7 8 9 10 11 12 13 14Temperature 18 19 24 35 33 14 22 20 23 33 27 23 30 35

Is there enough evidence for the ski resort to claim that the average weekly temperature is less than 32°F?

Assume that the average weekly temperature is normally distributed.



Problem 9.4

If you like shopping for the best deal on long-distance phone services, then you’ll enjoy sorting through offers from 10 different marketers vying to be your energy supplier. Residents of 16 communities will be the first in Massachusetts to wade into the coming nationwide experiment in deregulation of the natural gas industry. The average consumer uses 1232 therms of natural gas, for which the average cost has been $520.24. The table shows proposed costs to deliver 1232 therms of gas from 10 competitors:

Company Cost ($)All Energy Marketing Co 478.66Broad Street/ Energy One 450.24Global Energy Services 468.16Green Mountain Energy Partners 471.24KBC Energy Services 435.53Louis Dreyfus Energy Services 472.24National Fuel Resources 468.22NorAm Energy 442.20WEPCO Gas 443.52Western Gas Resources 457.81



Problem 9.5Computer centers at universities and colleges are certainly aware of the increased number of Web surfers. To begin to understand the demands that will be made on the computer center resources, one school studied 25 children in grades 7 to 12. The number of hours that these children spend on the Internet in 1 week is shown here:

5.0 4.4 5.7 5.6 5.5 5.2 5.0 4.8 3.6 4.14.6 4.9 4.0 6.7 5.5 5.4 6.7 5.8 5.4 4.85.9 5.1 3.8 4.1 6.7

Is there enough evidence to indicate that children spend more than a average of 5 hours per week Web surfing? Assume that the time spent Web surfing is normally distributed.



9.6

A company that sells mail-order computer systems has been planning inventory and staffing based on an assumption that the variance of their weekly sales is 180 ($10002). The weekly sales are normally distributed. The company selects 15 weeks at random from the past year and obtains the data (in thousands of dollars) shown below:

Weekly Sales: 191 222 222 223 223 225 227 228 229 232 234 234 236 244 253

a) What is the sample variance for these data?

b) Set up the hypotheses to test whether the population variance is different from 180.

c) At the 0.05 level of significance, what can you conclude about the company’s assumption?



9.7

In manufacturing, the amount of material that is wasted or lost during a process is very important. In preparing financial estimates, a company assumes that the percent material lost for its new process has a variance 10%2 . After the new process has been running for a month and appears to be stable, the cost analyst looks at the percent material lost and finds the following data:

Daily Loss: 10 12 12 13 14 14 18 19 19 20

a) What is the sample variance for these data?

b) Set up the hypotheses to test whether the actual variance is greater than the value the company has been assuming. Assume that the daily loss is normally distributed.

c) At the 0.05 level of significance, what can you conclude?



9.10

Companies are increasingly concerned about employees playing video games at work. In addition to reducing productivity, this habit shows down networks and uses valuable storage space. A recent article stated at 80% of all employees play video games at work at least once a week. A large company that employs many engineers wonders if its employees are as bad as the article claims. If they are, the company will install software that detects and removes video games from the network. The company surveys (anonymously) 100 employees and finds that 85 of the employees surveyed have played video games at work in the past week.

a. Set up the null and alternative hypotheses to test whether the proportion of the company’s employees that play video games is greater than the proportion stated in the article.

b. At the 0.05 level of significance, test the hypotheses.

c. What do you recommend that the company do?



9.11

An alumni office is interested in serving their alumni better in order to encourage more donations to the college. A survey of 200 alumni was conducted to determine whether half-day training sessions offered on the campus were of interest. If more than 75% of the alumni were interested, the college would start a program. The survey showed that 160 of the alumni surveyed were interested in such a program.

a. Set up the null and alternative hypotheses to test whether the college should implement the program.


c. What do you recommend that the college do?



9.12

A company that makes computer keyboards has specifications that allow it produce a product that has a maximum of 3% defective. The company has been receiving more customer complaints than usual. A sample of 50 keyboards has 2 defectives.

a. Set up the null and alternative hypotheses to test whether the proportion defective keyboards has exceeded the amount allowed by the specifications.


c. What do you recommend that the company do?



COMPARING TWO POPULATIONS

1. Comparing Means of Two Populations (1 vs. 2)

Dependent vs. Independent Samples

Comparing Means using Two Independent Samples - Large samples (z-distribution) (365)- Small samples (t-distribution) (375)

Comparing Means using Two Dependent Samples (384)- t-distribution

2. Comparing Proportions of Two Populations (1 vs. 2) (371) Using the z-distribution as approximation to the Binomial

3. Comparing Variances of Two Populations (21 vs. 2

2) (404) Using the F-distribution

are available





COMPARING TWO POPULATION MEANS (1 vs. 2)

H0: 1 = 2 H0: 1 2 H0: 1 2 H1: 1 2 H1: 1 < 2 H1: 1 > 2


t > t/2, n1+ n2 2 t < t, n1+ n2 2 t > t, n1+ n2 2

COMPARING TWO POPULATION PROPORTIONS (1 vs. 2)

H0: 1 = 2 H0: 1 2 H0: 1 2 H1: 1 2 H1: 1 < 2 H1: 1 > 2


COMPARING TWO POPULATION VARIANCES (2

1 vs. 22)

H0: 21 = 2

2

H1: 21 2

2

Reject H0 if:F > F(/2,v1,v2)






10.1

Many studies have been done comparing consumer behavior of men and women. One such on-going study concerns take-out food. In particular, the study focuses on whether there is a difference in the mean number of times per month that men and women buy take-out food for dinner. The most recent results of the study are shown below:

PopulationMen Women

Sample Size 34 28Sample Mean 25.6 21.2Population Standard Deviation 4.2 3.7

Because the study has so much historical data, information is known about the population standard deviations.

a. Set up the hypotheses to test whether there is a difference in the mean number of times per month that a person buys take-out food for dinner for men and women.

b. Use the Z test with known population variances to set up and perform the test. Use a level of significance of 0.05.

c. Find the p value for the test.

d. Do the data provide evidence that the mean number of times per month for men differs from that for women?

e. Does the choice of α in this case affect the decision?



10.2

Professional employees who work for large corporations often contend that the mean salary paid by a company differs by location in the United States. To test that claim, data were collected on financial analysts working for a large corporation at locations in New England and in the upper Midwest. Because there is an extensive history of salary data, the population standard deviations are available. The study found the following results:

PopulationNew England Upper Midwest

Sample Size 25 20Sample Mean 22.3 18.5Population Standard Deviation 1.5 2.2

a) Set up the appropriate hypotheses to test whether the company’s analysts in New England were paid more, on the average, than those working in the upper Midwest.

b) Use the Z test with known population variances to set up and perform the test. Use a level of significance of 0.05.

c) Find the p value for the test.

d) Do the data support the contention that the mean pay for analysts in New England is higher than that of analysts in the upper Midwest?



10.11

Having learned about the paired t test you realize that you really should have used the test for the data on software price comparison. The data are repeated below:

Top Ten BusinessSoftware Packages

ComputabilityPrice ($)

PC ConnectionPrice ($)

Windows 95 Upgrade 88 95Norton Anti-Virus 59 70McAfee ViruScan 49 60First Aid 97 Deluxe 54 58Clean Sweep III 37 37Norton Utilities 68 75Netscape Navigator 45 40MS Office Pro 97 Upgrade 300 310First Aid 97 32 35Win Fax Pro 95 95

a) Calculate the differences between the prices for each type of software package. Just looking at the differences, do you think that one company charges more than the other? Why or why not?

b) Calculate the average difference and the standard deviation of the differences.

c) Set up the hypotheses to test whether the mean difference in price between the two companies is zero.

d) Assuming that the data are normally distributed, at the 0.05 level of significance, is there a difference in the mean price of software for the two companies?

e) Did these results differ from the last time you analyzed the data? Why do you think this happened?



Minitab Solution

Stat > Basic Statistics > Paired t



Excel Solution

Data > Data Analysis > t-test: Pair Two Sample for Means



10.12

A hospital administrator is concerned about the length of time that the nursing staff washes their hands. A recent study in health care showed that longer washing greatly reduces the spread of germs. The hospital observed the amount of time that a sample of nine nurses in the Cardiac Care Unit (CCU) washed their hands. The data were colleted in such a way that the employees did not know that they were being observed. The hospital then showed the nurses an educational video on the negative effects of shortened time spent hand washing. After the video, the hospital again watched and timed the group of nurses washing their hands. The data are shown below:

Observation Unit Time 1 (s) Time 2 (s)1 CCU 3 162 CCU 2 73 CCU 0 54 CCU 5 85 CCU 2 156 CCU 0 157 CCU 2 208 CCU 3 169 CCU 0 18

a) Calculate the differences between the times for each nurse. Just looking at the differences, do you think that, on the average, they washed their hands longer the second time? Why or why not?

b) Calculate the average difference and the standard deviation of the differences.

c) Set up the hypotheses to test whether there was an increase in the average amount of time spent washing hands.

d) Assuming that the data are normally distributed, at the 0.05 level of significance, what can you conclude?

e) Can you conclude that the video caused the nurses to wash their hands longer? Why or why not?



Problem 10.7

Women who smoke suffer an increased risk of dying of breast cancer, according to a recently published study. In the study about, out of 319,000 women who never smoked there were 468 deaths from breast cancer, whereas out of 120,000 smokers, there were 187 deaths.

(a) Calculate the sample proportion of women who died of breast cancer for smokers and non-smokers.

(b) Ste up the hypotheses to test whether the proportion of women who die of breast cancer is higher for smokers than non-smokers.

(c) At the 0.05 significance level, can you conclude that smoking causes breast cancer? If not, what can you conclude?



Problem 10.18

Selling personal computers is big business and consumers are becoming increasingly aware of vendor reputation. A recent study of two vendors of desktop personal computers reports on the units that need repair for Dell Computers and Gateway 2000. Of 1584 computers manufactured by Dell Computer 427 needed repair, whereas for Gateway 2000, 825 of 2662 computers needed repair.

(a) Calculate the sample proportion of computers needing repair for each company.

(b) Set up the hypotheses to test whether the proportion of computers needing repairs is different for the two companies.

(c) At the 0.05 level of significance, what can you conclude?



Minitab Solution

(No Excel Solution)



10.20

Consider the problem in which the Board of Realtors for Greater Bridgeport, CT, was looking at the average selling prices of homes. The data are given again below:

Population1995 1996

Sample Size 25 25Sample Mean $151,116 $160,669Sample Standard Deviation $5,332 $6,468

a) Assuming that the populations are normally distributed, set up the hypotheses to test whether the population variances are equal at the 0.10 level of significance.

b) Was the decision to test using the pooled variance justified?



10.21

In your quest for the perfect golf clubs you made an assumption about the population variances when you tested your hypotheses. The data you collected are given below:

PopulationBrand X Brand Z

Sample Size 15 15Sample Mean 255 271Sample Standard Deviation 8.7 9.1

a) Set up the appropriate hypotheses to test whether the variance of Brand Z clubs is the same as the variance for Brand X.

b) Assuming that the populations are normally distributed, at the 0.10 level of significance was your decision to pool the variances a good one?

c) In general, would a difference in variation between the clubs be a factor in your purchase decision?



Procedures for Testing Independence



EXPERIMENTAL DESIGN AND ANOVA (ANALYSIS OF VARIANCE)

1. Definition of termso Factor and response variable (428)o ANOVA and treatment (410)

2. Sources of Variance (411) Treatment or Between Groups Variation (a.k.a. explained, factor, treatment) Random or Within Groups Variation (a.k.a. unexplained, random, error)

3. One-way ANOVA (410) Review of variables

- quantitative vs. qualitative- dependent vs. independent

Using ANOVA as procedure for comparing means of two or more groups Using ANOVA as procedure for determining whether a qualitative independent variable

and quantitative dependent variable are related

4. Two-Way ANOVA with Replication – a.k.a. Two-way ANOVA with Interaction (427) Using ANOVA as procedure for comparing means of two or more groups (Factor A and

Factor B) Using ANOVA as procedure for determining whether a qualitative independent variable

and quantitative dependent variable are related (Factor A and Factor B) Testing the presence of interaction between Factors A and B



ANALYSIS OF VARIANCE (ANOVA)

A.ONEWAY ANOVA

H0: 1 = 2 =3 = … = t

H1: The population means are not all the sameReject H0 if: F > F,v1,v2

Where: v1 = (t-1) v2 = (N - t)

B.TWOWAY ANOVA(with replication)

1. Testing for Main Effects (Factor A)

H0: 1 = 2 =3 = … = t (No level of factor A has an effect)H1: The population means are not all the same (at least 1 level has an effect)

Reject H0 if: MSA/MSE > F,v1,v2

Where: v1 = (a -1) v2 = ab(r -1)

2. Testing for Main Effects (Factor B)

H0: 1 = 2 =3 = … = t (No level of factor B has an effect)H1: The population means are not all the same (at least 1 level has an effect)

Reject H0 if: MSB/MSE > F,v1,v2

Where: v1 = (b -1) v2 = ab(r -1)

3. Testing for INTERACTION EFFECTS (AB)

H0: There are NO interaction effectsH1: At least 1 combination of factor A and B levels has an effect

Reject H0 if: MSAB/MSE > F,v1,v2

Where: v1 = (a -1)(b - 1) v2 = ab(r -1)




14.1A diaper company is considering 3 different filler materials for their disposable diapers. Eight diapers were tested with each of the 3 filler materials, and 24 toddlers were randomly given a diaper to wear. As the child played, fluid was injected into the diaper every 10 minutes until the product failed (leaked). The amount of fluid (in grams) at the time of failure was recorded for each diaper. The data are shown below:

Material 1 Material 2 Material 3791 809 828789 818 814796 803 855802 781 844810 813 847790 808 848800 805 836790 811 873

(a) What is the response variable and what is the factor?

(b) How many levels of the factor are being studied?

(c) Is there any difference in the average amount of fluid the diaper can hold using the three different filler materials? If so, which ones are different?

(d) What is your recommendation to the company and why?



MINITAB OUTPUT

Stat > ANOVA > One-way (or One-way (Unstacked))

Results for: Problem 14_1.MTW

One-way ANOVA: Grams versus Material

Analysis of Variance for Grams Source DF SS MS F PMaterial 2 9808 4904 29.83 0.000Error 21 3452 164Total 23 13260 Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev -------+---------+---------+---------1 8 796.00 7.50 (----*----) 2 8 806.00 11.12 (----*----) 3 8 843.00 17.70 (----*---) -------+---------+---------+---------Pooled StDev = 12.82 800 820 840

EXCEL OUTPUT

Tools > Data Analysis > Oneway ANOVA

Anova: Single Factor

SUMMARYGroups Count Sum Average VarianceMat1 8 6368 796 56.28571Mat2 8 6448 806 123.7143Mat3 8 6745 843.125 314.4107

ANOVASource of Variation SS df MS F P-value F critBetween Groups 9864.083 2 4932.042 29.92679 0.000000711968 3.466795Within Groups 3460.875 21 164.8036

Total 13324.96 23



14.3Grading homework is a real problem. It takes an enormous amount of time and many students do not do a very good job or copy answers from other students or the back of the book. A teacher of elementary statistics decided to conduct a study to determine what effect grading homework had on her students’ exam scores. She taught 3 sections of Elementary Statistics and randomly assigned each class one of three conditions: (1) no homework given, (2) homework given, but not collected, and (3) homework give, collected, and graded. After the first exam, she collected the data (exam scores). They are shown in the Excel data file Homework.xls

(a) What is the response variable and what is the factor?

(b) How many levels of the factor are being studied?

(c) Is there any difference in the average amount of fluid the diaper can hold using the three different filler materials? If so, which ones are different?

(d) What is your recommendation to the company and why?

MINITAB OUTPUT:

Results for: Problem 14_3.MTW

One-way ANOVA: C2 versus C1

Analysis of Variance for C2 Source DF SS MS F PC1 2 1700.4 850.2 8.91 0.001Error 45 4295.4 95.5Total 47 5995.8 Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev -------+---------+---------+---------1 16 74.500 11.051 (------*------) 2 16 70.313 9.016 (------*------) 3 16 84.500 9.107 (------*------) -------+---------+---------+---------Pooled StDev = 9.770 70.0 77.0 84.0



14.16The manufacturer of batteries is designing a battery to be used in a device that will be subjected to extremes in temperature. The company has a choice of 3 materials to use in the manufacturing process. An experiment is designed to study the life of the battery when it is made from materials A, B, C and is exposed to temperatures of 15, 70, and 125 degree Fahrenheit. For each combination of material and temperature, 4 batteries are tested. The lifetimes in hours of the batteries are shown below:

Temperature15F 70F 125F

Material A 130 34 20155 40 7074 80 82

180 75 58Material B 150 126 25

188 122 70159 106 58126 115 45

Material C 138 174 96110 120 104168 150 82160 139 83

(a) Calculate the average life for each of the material types.

(b) Calculate the average life for each of the 3 temperatures.

(c) Calculate the average life for each of the 9 treatment groups.

(d) Plot the 9 treatment means on a graph with temperature factor on the x axis, and the life of the battery in hours on the y axis. Use different color for each of the 3 materials and connect the averages for those of the same material. What do you speculate about the interaction effect based on the graph?

(e) Confirm your suspicions by doing a two-way ANOVA and testing to see if there is a significant interaction effect.

(f) What materials do you recommend to this manufacturer and why?



MINITAB OUTPUT

Stat > ANOVA > Two-way



Interaction Plot

Stat > ANOVA > Interaction Plot



Excel Solution

Data >Data Analysis > Two-way ANOVA (with Replication)



14.20A manufacturer of adhesive products designed an experiment to compare a new adhesive product to a competitor’s product. The adhesive product, or glue, is used by automobile manufacturers. The response variable was the strength of the glue measured by tensile strength in pounds per square inch (psi). The ability to adhere to oil-contaminated surfaces under different humidity conditions was studied. There were 2 levels for factor A: no oil or oil. Oil contamination was applied by hand dipping the samples in an oil solution and allowing them to air dry at room temperature for 2 hours. There were two levels for factor B: 50% humidity and 90% humidity. Three samples were tested for each of the combinations of factor A and factor B. The tensile values (psi) for the new product are shown below:

Humidity No Oil Oil50% 175 43

100 42175 44

90% 95 95115 10585 116

(a) Does the product behave significantly differently if the surface is oil contaminated?

(b) Does the product behave significantly differently at different humidity levels?

(c) Is there any significant interaction effect present?



Two-way ANOVA: PSI versus Surface, Humidity

Analysis of Variance for PSI Source DF SS MS F PSurface 1 7500 7500 13.52 0.006Humidity 1 85 85 0.15 0.705Interaction 1 9747 9747 17.56 0.003Error 8 4439 555Total 11 21772

Individual 95% CISurface Mean ----------+---------+---------+---------+-No oil 124.2 (--------*--------)Oil 74.2 (--------*--------) ----------+---------+---------+---------+- 75.0 100.0 125.0 150.0

Individual 95% CIHumidity Mean ---------+---------+---------+---------+--50% 96.5 (-----------------*------------------)90% 101.8 (------------------*-----------------) ---------+---------+---------+---------+-- 84.0 96.0 108.0 120.0



Analysis of Qualitative Data (Chi-square)

1. The Chi-square test Explained (640)

2. Four Uses of the Chi-square Distribution

(a) Testing for goodness-of-fit (641)

is a test for comparing a theoretical distribution, such as a Normal, Poisson etc, with the observed data from a sample

answers the question: “does the sample come from a specified distribution?”

(b) Testing (comparing) proportions of two or more groups (651)

(c) Testing whether two categorical (a.k.a. nominal, qualitative, classification) variables are independent (651)

(d) Testing the variance of a population (covered in earlier chapter)

Chi-square Concepts and Solved Problems are Available



CHI-SQUARE (2) DISTRIBUTION

1. Goodness-of-fit Test

H0: The sample comes from a specified distributionH1: The sample does not come from a specified distribution

Reject H0 if: 2 > 2 ,(k - 1)

2. Test of Independence of 2 Categorical Variables(Also used for comparing proportions of 2 or more groups)

H0: Variables 1 and 2 are not dependentH1: Variables 1 and 2 are dependent

Reject H0 if: 2 > 2 ,(r - 1)(c - 1)

3. Testing A Population Variance (2)

H0: 2 = value H0: 2 value H0: 2 value H1: 2 value H1: 2 < value H1: 2 >value

Reject H0 if: Reject H0 if: Reject H0 if:2 > 2

/2,n-1 2 < 2 1-,n-1 2 > 2

,n-1

2 < 2 1/2,n-1




15.1The administration of a university has been using the following distribution to classify ages of their students:

Age Group

Estimated % ofStudent Population

Less than 18 2.718 – 19 29.920 – 24 53.4Older than 24 14

A recent student survey provided the following data on age of students:

Age Group FrequencyLess than 18 618 – 19 11820 – 24 102Older than 24 26

Set up a table that compares the expected and observed frequencies for each group.

Based on the table, do you think that the data represent the established distribution?

Set up the hypothesis for the Chi-square goodness of fit test.

Perform the goodness of fit test at the 0.05 significance level.

Based on the chi-square test, is the estimated age distribution that the university is correct?



15.2As part of a survey on the use of Office Suites Software, the company doing the polling wanted to know whether its population was uniformly distributed over the following age distribution: under 25, 25 to 4, 44 and up. The company looked at the data it had collected so far had found the following distribution:

Age Group

Number of Respondents

Under 25 7325 to 44 6145 and up 66

200

Based on the data, do you think that the respondents are uniformly distributed over the age categories?

Set up the hypothesis to test whether the data are uniformly distributed over the age categories.

Find the expected frequency distribution and perform the chi-square goodness of fit test.

At the 0.05 level of significance, would you say that the respondents were uniformly distributed over the age groups?



15.6

In an experiment to study the attitude of voters concerning term limitations in Congress, voters in Indiana, Ohio, and Kentucky were polled with the following results:

Opinion Indiana Kentucky OhioSupport 82 107 93Do Not Support 97 66 74

(a) Set up the hypothesis to test whether the proportion of voters who support congressional term limits is the same for all three states.

(b) Calculate the proportion of voters that support congressional term limits for each state individually. Based on these values, do you think there is a difference in the proportions?

(c) Calculate the overall proportion of voters who support term limits for Congress.

(d) Calculate the expected frequencies for each cell and find the value of the chi-square test statistic.

(e) At the 0.05 level of significance, is there a difference in the proportion of voters who support congressional terms limits among the three states?



Minitab

Stat > Table > Chi-square Test

(No Excel Solution)



15.7In a survey about satisfaction with local phone service, those respondents who rated their current service as excellent and those who rated Poor to Very Poor were asked to classify their current local service provider. The results are given in the table below:

Type of CompanyCurrent Service

SourceLong

DistanceLocal Phone Power

CableTV

Cellular Phone

Excellent 264 444 131 215 198Poor – Very Poor 1394 1318 485 431 572

(a) Set up the hypothesis to test whether the proportion of people who rated their company as excellent is the same for each type of company.

(b) Calculate the proportion of people who rate their current phone service as excellent.

(c) Calculate the expected frequencies for each cell and find the value of the chi-square test statistic.

(d) If you wanted to perform the test at the 0.05 significance level, what would be the critical value of the test?

(e) At the 0.05 level of significance, is there a difference in the proportion of people who rate their local phone service as excellent among the different types of companies?



Chi-Square Test: Long Distance, Local Phone, Power, CableTV, Cellular Phone

Expected counts are printed below observed counts

Long Dis Local Ph Power CableTV Cellular Total 1 264 444 131 215 198 1252 380.74 404.63 141.46 148.35 176.82

2 1394 1318 485 431 572 4200 1277.26 1357.37 474.54 497.65 593.18

Total 1658 1762 616 646 770 5452

Chi-Sq = 35.796 + 3.831 + 0.773 + 29.947 + 2.536 + 10.671 + 1.142 + 0.230 + 8.927 + 0.756 = 94.610DF = 4, P-Value = 0.000



15.10

A report by the Department of Justice on rape victims reports on interviews with 3721 victims. The attacks ere classified by age of the victim and the relationship of the victim to the rapist. The results of the study are given here:

Relationship of RapistAge of Victim Family

Acquaintance or Friend Stranger

Under 12 153 167 1312 to 17 230 746 172Over 17 269 1232 739

(a) Set up the hypotheses to test whether age of victim and relationship of rapist are independent.

(b) Calculate the expected frequencies for each cell.

(c) How many degrees of freedom will the chi-square test for independence have? Using a level of significance of 0.01, what is the critical value for the test?

(d) Calculate the value of the chi-square test statistic.

(e) Is the age of the victim independent of the relationship to the rapist?



MINITAB

Stat > Table > Chisquare Test

Chi-Square Test: C1, C2, C3


C1 C2 C3 Total 1 153 167 13 333 58.35 191.96 82.69

2 230 746 172 1148 201.15 661.77 285.07

3 269 1232 739 2240 392.50 1291.27 556.24

Total 652 2145 924 3721

Chi-Sq =153.539 + 3.246 + 58.734 + 4.136 + 10.720 + 44.849 + 38.857 + 2.720 + 60.050 = 376.852DF = 4, P-Value = 0.000



15.11A company that manufactures cardboard boxes for packaging cereals wants to determine whether type of defect that a particular box has is related to the shift on which it was produced. It compiles the following data. In each case, if a box had multiple defects the most serious defect was recorded.

Type of DefectShift Printing Rips/Tears Size1 55 60 852 58 63 793 89 63 48

(a) Set up the appropriate hypotheses for the test.

(b) Calculate the expected frequencies for each cell and calculate the value of the chi-square test statistic.

(c) How many degrees of freedom will the chi-square test for independence have?

(d) Using a level of significance of 0.01, are defect type and shift related?



Chi-Square Test: Printing, Rips/Tears, Size


Printing Rips/Tea Size Total 1 55 60 85 200 67.33 62.00 70.67

2 58 63 79 200 67.33 62.00 70.67

3 89 63 48 200 67.33 62.00 70.67

Total 202 186 212 600

Chi-Sq = 2.259 + 0.065 + 2.907 + 1.294 + 0.016 + 0.983 + 6.972 + 0.016 + 7.270 = 21.782DF = 4, P-Value = 0.000



Simple Linear Regression and Correlation(454-491)

Important Definition of TermsTest of IndependenceVariables

Quantitative (measured)Qualitative (category, classification, nominal)

Scatter plotRegression and correlationLinear vs. Curvilinear modelsSimple vs. Multiple Linear ModelsCorrelation coefficientCoefficient of determinationResidual (error) termObserved y vs. expected y

Important SymbolsYXSy/x

R2

Rab

Problems for Simple Linear Regression:

11.2 (p. 553)

Problems for Multiple Linear Regression:

Problem 12.1

Problem 12.5Problem 12.9



Steps in Regression/Correlation Analysis

1. Identify the response (y) and candidate predictor variables (x’s)

2. Collect y,x set of data

3. Plot each x versus y

4. From the plots in #3, select the most promising x

5. Perform Regression and Correlation Analysis

a. Select a model (linear or nonlinear) that fits the plot and then generate the regression equation using Excel or Minitab

b. Test the resulting model for significance using the slope (), correlation (), or the ANOVA tests

(If resulting model is NOT significant, go back to Step 1)

c. Test the model for appropriateness using the analysis of residuals. This tests whether the assumptions on the residual are met. These assumptions are:

Normal distribution Homoscedastic Indepedent

(If selected model is not appropriate, go back to Step 5a, else proceed to Step 7)

6. If model generated in Step 5 is significant but not appropriate, choose a different model (perhaps use curvilinear model) and repeat Step 5 until an appropriate model is found.

7. Use model for estimating:

(1) the response variable (y)

Point Estimate – substitute the value of x Into the regression equation

Interval Estimates:

1. Prediction Interval Estimate

2. Confidence Interval Estimate

(2) the actual slope (B) of the line

CI = b ( t /2, n-2 ) sb



Definitions of Relevant Terms

Types of Variables:y – response variable (a.k.a. dependent,

predicted, explained)

x – independent variable (a.k.a. predictor, explanatory)

Regression – provides a ‘best-fit’ mathematical equation for the values of y,x variables

-- expresses the relationship of y and x inequation form

mathematical equation may belinear or curvilinear

linear: Y = a + bX (Direct, linear)

Y = a – bX (Inverse, linear)

curvilinear: Y = a + bX + cX2

(quadratic)

Y = e-X (negative expon.)

Y =1/X

Simple Linear Regression – regression model is linear with only ONE predictor variable

y = b0 + b1X

Multipe Linear Regression -- regression model is linear with only TWO OR MORE predictor variables

y = b0 + b1x1+ b2x2+ b3x3 + ...+ bKxk

Correlation Analysis – measures the strength of the relationship between Y,X

coefficient of correlation (r) – number that measures both the direction and the strength of the linear relationship between y and x

1 r 1

coefficient of determination (r2) – the percent of the variation in y that is explained by the regression model

0% r2 100%



Simple Linear Model and Assumptions

Models

Actual Population Model

Estimated (sample) Model

Assumptions on the residuals

1) Normally distributed2) Homoscedastic (constant variance across all x values)3) Statistically independent of each other



Testing the Model for Significance

a. Using the Slope ()Test

H0: = 0H1: 0

Reject H0 if t> t /2, n-k-1

b. Using the coefficient of correlation ()Test

H0: = 0H1: 0

Reject H0 if t> t /2, n-k-1

c. Using the ANOVA F -TestH0: the Model is not significantH1: the Model is significant

Reject H0 if F > F, v1, v2



Using the Model for Estimation/Prediction

A. Estimating the actual slope () of the model

b point estimate of the actual slope () of the model

Computing a confidence interval for the actual slope of the model

C.I. for = b t /2, n-k-1 (sb)

B. Using the model to estimate the actual value of y for a given value of x

point estimate of the actual value of y for a given value of x

computed by substituting the value of x into the regression equation

Confidence Interval (C.I.) the interval that contains the actual average value of the response variable (y/x) for a specific value of x

Prediction Interval (P.I.) the interval that contains the actual value of the response variable (Y) for a specificvalue of x



SIMPLE LINEAR REGRESSION:A Solved Example

EXAMPLE:

A manufacturer of small electric motors uses an automatic milling machine to produce the slots in the shaft of the motors. A batch of shafts is run and then checked. All shafts in the batch that do not meet required dimensional tolerances are discarded.

At the beginning of each new batch, the milling machine is readjusted since its cutter head wears slightly during the production of the batch. The manufacturer is trying to pick an optimal batch size, but in order to do this (s)he must know how the size of the batch affects the number of defective shafts in the batch. Thirty (30) batches were inspected, and the number of defectives in each batch was counted. The results are shown below:

Batchsize Defects100 5125 10125 6125 7150 6150 7175 17175 15200 24200 21200 22225 26225 29225 25250 34250 37250 41250 34275 49300 53300 54325 69350 82350 81350 84375 92375 96375 97400 109400 112



INITIAL MODEL (LINEAR)

MINITAB SOLUTION

A. Plot Batchsize and Number of Defects

GRAPH > PLOTGraph Variables:X: BatchsizY: Defects

STAT > REGRESSION > FITTED LINE PLOT

Response (Y): DefectsPredictor (X): BatchsizType of regression model:

Linear

B. Generate the Regression Equation

STAT > REGRESSION > REGRESSIONResponse: DefectsPredictors (X): BatchsizClick on Results:

Select In addition, sequential sums of…Click OK

Click on StorageSelect FitsSelect Residuals



Select Standardized ResidualsClick OK

Click OK


Least squares regression equation

Coefficient of determination


Generate the Residual Plots

STAT > REGRESSION > Residual PlotsResiduals: RESI1Fits: FITS1Click OK



EXCEL SOLUTION

Data > Data Analysis > RegressionInput:

Input Y range: DefectsInput X range: Batchsiz

Labels: <select>Output Range: <type in a cell address here>Residuals:

Residuals: <do not select>Standard Residuals: <do not select>Residual Plots: < select >Line Fit Plots: <select>

Normal Probability:Normal Probability Plots: < select>



Analysis of Residual Plots



REVISED MODEL (NONLINEAR - Quadratic)

Minitab

Delete all non-empty columns except Defects and Batchsiz

Compute C3 = Batchsiz * Batchsiz

Calc > Calculator Store result in variable: C3Expression: Batchsiz*BatchsizClick OK

Name C3 as "Batchsiz^2"

STAT > REGRESSION > REGRESSIONResponse: DefectsPredictors (X): Batchsiz Batchsiz^2Click on Results:

Select In addition, sequential sums of…Click OK

Click on StorageSelect FitsSelect ResidualsSelect Standardized ResidualsClick OK

Click OKRegression Analysis

The regression equation isdefects = 6.90 - 0.120 batchsiz +0.000950 batchsiz^2

Predictor Coef StDev T PConstant 6.898 3.737 1.85 0.076batchsiz -0.12010 0.03148 -3.82 0.001batchsiz 0.00094954 0.00006059 15.67 0.000

S = 2.423 R-Sq = 99.5% R-Sq(adj) = 99.5%

Analysis of Variance

Source DF SS MS F PRegression 2 34186 17093 2911.35 0.000Residual Error 27 159 6Total 29 34345

Source DF Seq SSbatchsiz 1 32744batchsiz 1 1442





EXCEL SOLUTION

Create Batchsiz^2 column (must be adjacent to Batchsiz), where Batchsiz^2 = Batchsiz * Batchsiz

Data > Data Analysis > RegressionInput:

Input Y range: DefectsInput X range: highlight Batchsiz Batchsiz^2 range of data

Labels: <select>Output Range: <type in a cell address here>Residuals:

Residuals: <do not select>Standard Residuals: <do not select>Residual Plots: < select >Line Fit Plots: <select>

Normal Probability:Normal Probability Plots: < select>



GENERATING PREDICTION AND CONFIDENCE INTERVALS FOR Y(Minitab)

Stat > Regression > Regression > Option


Values for batch and batch^2 to predict defect rates

Columns where the new values for the predictors variables can be found


(Note: Excel does not have this capability)


Predicted values for number of defects


Problem 11.2

In trying to look at the effects of shopping center expansion, the Commerce Department decided to look at the relationship between the number of shopping centers and the retail sales for different states in the same region,. It collected the data for the North Central states in the US and found the following:

State Number of Shopping Centers

Retail Sales($ billion)

Illinois 2096 41.8Indiana 905 21.4Michigan 1018 25.3Minnesota 471 13.9Ohio 1704 41.6Iowa 308 7.5Missouri 887 22.7Wisconsin 625 14.6South Dakota 58 1.3North Dakota 87 2.1Nebraska 264 5.7Kansas 481 11.6

(a) Create a scatter plot of the data.

(b) Find the regression equation relating retail sales and number of shopping centers.

(c) Plot the regression line on the same plot as the data. Do you think the line fits the data well? Why or why not?

(d) Use the regression line to predict retail sales for each state.

(e) Calculate the residuals for each state. Which state has the largest residual? Which state has the smallest? Do the residuals support your answer to part (d)?

(f) Find the standard error of the estimate.



0

5

10

15

20

25

30

35

40

45

0 500 1000 1500 2000 2500

Number of Shopping Centers

Reta

il Sa

les

($bi

llion

)

Series1

SUMMARY OUTPUT

Regression StatisticsMultiple R 0.9866955R Square 0.9735681Adjusted R Square 0.9709249Standard Error 2.3387601Observations 12

ANOVA

df SS MS FSignificance

FRegression 1 2014.691 2014.691 368.330 0.000Residual 10 54.698 5.470Total 11 2069.389

CoefficientsStandard

Error t Stat P-value Lower 95% Upper 95%Intercept 1.492612 1.071387 1.393158 0.193764 -0.894588 3.879812X Variable 1 0.021517 0.001121 19.191926 0.000000 0.019019 0.024015



Problem 11.3

As part of an international study on energy consumption, data were collected on the number of cars in a country and the total travel in kilometers. The data for 12 of the countries are shown here:

Country

Total CarsTravel

(million)Travel

(billion km)US 142.35 3140.29Finland 1.82 34.66Denmark 1.66 30.76Britain 21.32 352.76Australia 8.53 138.22Sweden 3.32 53.21Netherlands 5.53 83.69France 23.27 348.2Norway 1.59 23.54Italy 26.12 367.85Germany 43.75 608.52Japan 40.25 439.30

(a) Create a scatterplot of the data. Do you think that there is a relationship between the number of kilometers traveled and the number of cars?

(b) Find the least-squares regression line for the data. Interpret the value of the slope.

(c) Does the intercept make sense for these data? Why or why not?

(d) Plot the regression line on the same plot with the data. Does the line make you feel confident about predicting travel as a function of the number of cars?

(e) Use the regression line to predict the number of kilometers for Sweden and Japan. How well do the predictions agree with the original data?



-500

0

500

1000

1500

2000

2500

3000

3500

0 20 40 60 80 100 120 140 160

Total Cars (in million)

Trav

eled

(in

bilii

on K

ilom

eter

)

SUMMARY OUTPUT

Regression StatisticsMultiple R 0.98503096R Square 0.97028599Adjusted R Square 0.96731458Standard Error 156.136088Observations 12

ANOVA

df SS MS FSignificance

FRegression 1 7960585.694 7960586 326.5415 0.0000Residual 10 243784.7804 24378.48Total 11 8204370.475

Coefficients Standard Error t Stat P-value Lower 95%Upper 95%

Intercept -106.2068 55.1609 -1.9254 0.0831 -229.1129 16.6992X Variable 1 21.5814 1.1943 18.0705 0.0000 18.9204 24.2425



Problem 11.23

How much does advertising impact market penetration? To assess the impact of advertising in the tobacco industry, a study looked at the amount of money spent on advertising a particular brand of cigarettes and brand preference among adolescents and adults. The data are shown here:

BrandAdvertising($ million)

Brand PreferencesAdolescent

(%)Adult(%)

Marlboro 75 60 23.5Camel 43 13.3 6.7Newport 35 12.7 4.8Kool 21 1.2 3.9Winston 17 1.2 3.9Benson & Hedges 4 1 3.0Salem 3 0.3 2.5

(a) Look at the data for brand preference for adolescents and amount spent on advertising. Which variable is the dependent variable? Which is the independent variable?

(b) Create a scatter plot of advertising and adolescent brand preference. Do you think that there is a linear relationship between the two variables? Why or why not?

(c) Now create another scatter plot using adult brand preference instead. How does this plot compare to the one for adolescent brand preference? From the plots, do you think that adolescent or adult brand preference is more strongly related to advertising expenditures? Why?

(d) Find the least squares line for adolescent brand and advertising expenditures

(e) Interpret the meaning of the slope and intercept of the model. Do they make sense?

(f) Use the model to predict adolescent brand preference for each brand studied. How well do the predicted values agree with the actual data?

(g) Using a 0.05 significance level, is the model significant?



Adolescent Market

010203040506070

0 10 20 30 40 50 60 70 80

Adevertising ($million)

Bran

d Pr

efer

ence

(%)

Adult Market

0

5

10

15

20

25

0 10 20 30 40 50 60 70 80

Adevertising ($million)

Bran

d Pr

efer

ence

(%)



ADOLESCENT MARKET

SUMMARY OUTPUT

Regression StatisticsMultiple R 0.923547R Square 0.852939Adjusted R Square 0.823527Standard Error 9.063086

Observations 7

ANOVA df SS MS F Significance F

Regression 1 2382.011 2382.011 28.99957 0.002978Residual 5 410.6976 82.13953

Total 6 2792.709


Intercept -9.42472 5.365513 -1.75654 0.139344 -23.2172 4.367747

Advertising ($m) 0.786227 0.146 5.385125 0.002978 0.410923 1.161531

ADULT MARKET

SUMMARY OUTPUT

Regression StatisticsMultiple R 0.901096R Square 0.811974Adjusted R Square 0.774369Standard Error 3.536488

Observations 7

ANOVA df SS MS F Significance F

Regression 1 270.0463 270.0463 21.59205 0.005599Residual 5 62.53373 12.50675

Total 6 332.58


Intercept -0.58794 2.093665 -0.28082 0.790098 -5.96987 4.793986

Advertising ($m) 0.264725 0.05697 4.646724 0.005599 0.118279 0.411172

Multiple Linear Regression



(560 to 595)

Problem 12.1

A group of legislators wanted to look at factors that affect the number of traffic fatalities. They collected some data for 1994 from the NTSB on the number of fatalities for 50 states and the District of Columbia, the number of licensed drivers, the number of registered vehicles, and the number of vehicle miles traveled. A portion of the data is shown on page 584. Full dataset is in traffat.xls

SUMMARY OUTPUT

Regression Statistics Multiple R 0.982548538R Square 0.96540163Adjusted R Square 0.963193224Standard Error 154.5407481Observations 51

ANOVA df SS MS F Significance FRegression 3 31321046.9 10440349 437.1485 2.54274E-34Residual 47 1122493.613 23882.84Total 50 32443540.51


Intercept 51.7481659 30.43306219 1.700393 0.095666 -9.475200509 112.9715Licensed Drivers 0.06294764 0.048829545 1.28913 0.203662 -0.035284642 0.16118Registered Vehicles -0.211896991 0.055989427 -3.78459 0.000436 -0.324533083 -0.09926Vehicle Miles Travelled 0.029349954 0.003525079 8.326041 8.34E-11 0.022258416 0.036441

(a) How many independent variables are there in the model proposed? What are they?

(b) Use the computer output to write won the regression model.

(c) Interpret the coefficients of the model.

(d) Use the model to predict the number of traffic fatalities for the states shown in the data table.

(e) Compare the predicted values from the model to the actual values. Based on the plot, does the model do a good job of predicting the number of traffic fatalities?



Problem 12.9

In the problem about number of traffic fatalities the model was rerun, dropping the data on number of licensed drivers that had the lowest t statistic. The output is shown below:

Regression Analysis: Traffic Fata versus Registered V, Vehicle Mile

The regression equation isTraffic Fatalities = 46.0 - 0.163 Registered Vehicles + 0.0300 Vehicle Miles Travelled

Predictor Coef SE Coef T PConstant 46.04 30.32 1.52 0.135Register -0.16280 0.04132 -3.94 0.000Vehicle 0.029996 0.003513 8.54 0.000

S = 155.6 R-Sq = 96.4% R-Sq(adj) = 96.3%



(a) Write down the equation of the new two-variable model.

(b) Compare the new model to the model with three variables. How much does the model change when number of licensed drivers is dropped?

(c) Compare the value of R2 for both models. What does this make you think about the decision to drop number of licensed drivers from the model?

(d) Would you consider a two-variable model a good model? Why or why not?

(e) Based on the value of the R2 would you be satisfied with this model or would you want to consider other variables?



BA 282: APPLIED BUSINESS STATISTICSFall 1999 Name____________________Midterm Exam

Part 1: Multiple Choice

1. Using the Standard Normal distribution, the area between –1.5 and –2.4 is:a. 0.9250b. 0.0586c. -0.0568d. -0.9250e. None of the above

For questions 2 to 7, select the most appropriate pair of hypotheses for each statement.

2. The average age of SOU students is more than 21.5 years. a. H0: 21.5 H1: > 21.5b. H0: 21.5 H1: < 21.5c. H0: = 21.5 H1: 21.5d. H0: 21.5 H1: > 21.5e. H0: 21.5 H1: < 21.5

3. A new medication for headache is touted to relieve pain in less than 5 minutes. a. H0: 5 H1: > 5b. H0: 5 H1: < 5c. H0: = 5 H1: 5d. H0: 5 H1: > 5e. H0: 5 H1: < 5

4. A CPA review program is advertised as “guaranteed to improve your CPA test scores.” Fifty graduating accounting students from a business school were randomly assigned to two groups – group in which students didn’t participate (D) in the review program and the other group participated (P) in the review program. a. H0: P D H1: P > D

b. H0: P D H1: P < D

c. H0: P = D H1: P D d. H0: P D H1: P > D

e. H0: P D H1: P < D

5. A councilperson claims that there is no difference in the level of support to Measure 51 among Republican (R) and Democratic (D) voters in the Rogue Valley. a. H0: R D H1: R > D

b. H0: R D H1: R < D

c. H0: R = D H1: R D

d. H0: R D H1: R < D

e. H0: R = D H1: R D

6. A filling machine is supposed to fill an average of 12 ounces when operating properly. a. H0: 12 H1: > 12b. H0: 12 H1: < 12c. H0: = 12 H1: 12d. H0: 12 H1: > 12e. H0: 12 H1: < 12



7. A right-tailed test of a population mean results in a p-value that is practically zero. This means that the sample represents: a. A weak evidence supporting the null hypothesis b. A weak evidence supporting the alternative hypothesisc. A strong evidence supporting the null hypothesis d. A strong evidence supporting the alternative hypothesis

8. Of the following variations, which does not belong?a. Common cause variationb. Special cause variationc. Explained variationd. Nonrandom variation

9. A confidence interval estimate has two components – a point estimate and a margin of error. Which of these two components is affected by the confidence level?a. Point estimateb. Margin of Errorc. None of the above

10. A sample statistic (e.g. sample mean, sample proportion, or sample variance) is a random variable. Which type of random variable is a sample statistic?a. Continuousb. Discrete

11. "The distribution of the sample means of any type of distribution will approximate the normal distribution, as the sample size increases." This sounds like the definition of thea. Standard Normal Distributionb. Z-distributionc. Central Limit Theoremd. Binomial distribution approximated by the Normal distributione. None of the above

12. Which of the following does not belong?a. sb. pc. x-bard. e. All of the above belong to the same group

13. Which of the following is NOT true of a sample mean?a. It is a point estimateb. It is a statisticc. It is a continuous random variabled. All of the above (a-c) are true of a sample meane. None of the above (a-c) are true of a sample mean

14. The two components of a confidence interval estimate of a population parameter are:a. Confidence Level and Margin of Errorb. Sample Size and Statisticc. Point Estimate and Margin of Error



d. Sample Mean and Sample Proportione. Margin of Error and Sample Size

15. The conditions in using the Normal distribution as an approximation to the binomial distribution are that np and n(1-p) be both at least 5.a. Trueb. False

16. Which of the following will be the benefit derived from using a larger sample size in estimating an unknown population parameter?a. A larger margin of errorb. A smaller margin of errorc. A lower confidence leveld. A higher confidence levele. (b) and/or (d) f. (a) and/or (c)

17. Using the Standard Normal distribution table, find the area below the z-score of –2.50. a. -0.4938b. 0.4938c. 0.9938d. 0.0062e. None of the above

18. Which of the following pairs of hypotheses is NOT correct?a. H0: 3.5 H1: > 3.5b. H0: 3.5 H1: < 3.5c. H0: p < 0.035 H1: p > 0.035d. All of the above are correcte. None of the above are correct

19. For a one-tailed test of a population mean the significance level has been set at 0.01. Assume that the population standard deviation is not known, sample size is 10, and the population is normally distributed. What distribution is appropriate for performing the hypothesis test? a. z-testb. t-testc. F-testd. Binomiale. None of the above

20. In testing the mean of a population, which of the following is a necessary condition for using a t distribution?a. n is smallb. is not knownc. The population is infinited. All of thesee. (a) and (b) but not (c)

21. Assume that you took a sample and calculated the sample mean as 100. You then calculated the lower and upper limit of a 90 percent confidence interval for to be 90 and 110, respectively. What is the margin of error of the estimate? a. 0.10b. 90 percentc. 20



d. 10e. 100

22. A single value used to estimate an unknown population parameter is known as a(n) a. Point estimateb. Interval estimatec. Statisticd. Parametere. (a) and (c)

23. In hypothesis testing, we conclude to reject or fail to reject the null hypothesis by comparing the computed statistic to a critical statistic. Another form of decision rule is by comparing the p-value to a significance level. Which of the following is a correct decision rule?a. Reject H0 if z > p-valueb. Reject H0 if z > c. Reject H0 if p-value > d. Reject H0 if p-value < e. All of the above are correct forms of decision rule

24. Which of the following variations cannot be removed but only can be reduced by redesigning or improving the process?a. common cause variationb. special cause variation

25. If n = 45 and = 0.05, then the critical value of z for testing the hypotheses H0: 3.5 and H1: > 3.5 isa. 0.0199b. 1.96c. -1.96d. -1.645e. 1.645

26. When a null hypothesis is rejected, it is possible that:a. A correct conclusion has been madeb. A Type II error has been madec. A Type I error has been maded. (a) or (b) is correcte. (a) or (c) is correct

27. Which of the following actions will reduce the Type I and II errors simultaneously? a. Decreasing the significance level of a testb. Increasing the confidence level of a testc. Decreasing Beta errord. Increasing the sample sizee. Decreasing the sample size

28. One concludes whether to “reject” or “fail to reject” the null hypothesis based on a decision rule. The decision rule is nothing more than a comparison of a calculated value and a critical value. Which of the two is based on the significance level of a test?a. Calculated valueb. Critical value



29. Which of the following is NOT true of a critical valuea. It marks the boundary between the “reject H0“ and the “fail to reject to reject H0“ regionsb. It is based on the significance level of a testc. It is determined from the statistic derived from a sampled. All of the above are truee. None of the above are true

30. If one were to perform a hypothesis test using the following hypotheses:H0: shipment is GOOD and H1: shipment is BAD Which of the two types of errors is called the Producer’s risk? a. Type I (alpha)b. Type II (beta)c. Both (a) and (b)d. Neither (a) nor (b)

31. When the sample size as a proportion of the population size (n/N) gets larger, the value of the finite correction multiplier approaches which value?a. 0b. 1c. None of the above

32. In statistical process control charts are used as tools to monitor processes. All processes exhibit variability. When NOT in control, the ___________ variability is said to be present:a. common cause variabilityb. special cause variabilityc. none of the above

33. A 5-week diet program is claimed be effective in reducing the weights of participants in the program. Skeptical about the claim, you randomly select 10 applicants and weigh each one before and after the 5-week period. This problem involves:a. Comparison of two population proportions b. Comparison of means of two independent samples c. Comparison of means two dependent samples d. Comparison of variations of two independent samples

34. Which of the following sampling distributions would be used in comparing means of two populations, with n1 = 32, n2

= 40?a. Z testb. pooled t test (or equal variances t test)c. unpooled t test (or unequal variances t test)d. Binomial

Use for questions 35-37You work for a market research agency and you were asked to estimate the proportion of people with personal computers who are using Windows 97 as an operating system. How many people will you need to survey for your estimate to be within 2 percentage points of the actual value and be 90 percent confident with this estimate?

35. The problem described above involves:a. Testing a hypothesis about a population meanb. Computing a confidence interval estimate of a population averagec. Computing a sample size to estimate a true population proportiond. Estimating a confidence interval estimate of a true population mean



e. None of the above

36. How much is the stated margin of error? a. 90 percentb. 10 percentc. 0.10d. 2 percentage pointse. 1.645

37. Give the z-value that will be used for computing the 90 percent confidence interval estimate of the true population parameter. a. 1.96b. 1.32c. 0.10d. 2 percentage pointse. 1.645

Use for questions 38- 43C. Garr Smoke claims that no more than 5 percent of the 40-60 male group smoke cigars. Of 2500 males of this age group you recently sampled, 200 said they smoke cigars. At 0.05 significance level, do the refute C. Garr Smoke’s belief?

38. The sample statistic in this problem is:a. Population proportionb. Population meanc. Sample proportiond. Sample meane. None of the above

39. In this problem the statement “no more than 5 percent” is:a. The hypothesized population proportionb. The hypothesized population meanc. The sample proportiond. The sample meane. None of the above

40. State the null and alternative hypotheses of this problem a. H0: 5 H1: > 5b. H0: 5 H1: < 5c. H0: = 5 H1: 5d. H0: 0.05 H1: > 0.05e. H0: 0.05 H1: < 0.05

41. Identify the critical value for the test (one tail). a. 0.0199b. 1.96c. -1.96d. -1.645e. 1.645

42. If the computed value for the test is 6.88, the p-value is



a. almost 1b. almost 0c. close to the significance level

43. If the resulting p-value of the test is less than the significance level, then you would conclude toa. Reject the null hypothesisb. Fail to reject the null hypothesis

Use for questions 44- 47A torque wrench used in the final assembly of cylinder heads requires a process average of 135 lbs-ft. The process is known to have a standard deviation of 5.0 lbs-ft. For a simple random sample of 45 sample nuts that the machine has recently tightened, the sample average is 137 lbs-ft. Using a 0.05 significance level, determine whether the machine is operating at the desired level.

44. The appropriate hypotheses for the problem are:a. H0: 135 H1: > 135b. H0: 135 H1: < 135c. H0: = 135 H1: 135c. None of the above are correct

45. The appropriate distribution for the test above isa. t-distributionb. z-distributionc. F-distribution

46. The computed value isa. -0.40b. 0.40c. -2.68d. 2.68e. None of the above

47. Assuming that the critical value for this problem is 1.645, and another sample produced a computed value of 1.55. For this sample your statistical conclusion is to:a. Reject the null hypothesis and conclude that the process is operating at the desired levelb. Accept the null hypothesis and conclude that the process is operating at the desired levelc. Reject the null hypothesis and conclude that the process is not operating at the desired leveld. Accept the null hypothesis and conclude that the process is not operating at the desired level

Use for questions 48- 50A pharmaceutical company is testing two new compounds intended to reduce blood-pressure levels. The compounds are administered to two different sets of lab animals. In Group 1, 71 of 100 animals tested respond to drug 1 with lower blood-pressure levels. In Group 2, 58 of 90 animals tested respond to drug 2 with lower blood-pressure levels. The company wants to test at the .05 level whether drug 1 is more effective in reducing blood pressure levels than drug 2.

48. The problem involves which of the following procedures?a. Comparison of two population proportionsb. Comparison of means of two populations using dependent samples



c. Comparison of means of two populations using independent samplesd. Comparison of variances of two populations49. Using the following subscripts for the two groups: 1- drug 1; 2- drug 2, which of the following is the most appropriate pair of hypotheses? a. H0: 1 2 H1: 1 > 2

b. H0: 1 2 H1: 1 < 2

c. H0: 1 2 H1: 1 > 2

d. H0: 1 2 H1: 1 < 2

e. H0: 1 = 2 H1: 1 2

50. (Bonus) On which day is Thanksgiving celebrated?a. Mondayb. Tuesdayc. Wednesdayd. Thursdaye. Friday



BA 282: Applied Business StatisticsFinal Exam -- Spring 1999 Name ______________________

1. Find the value of 2 .05,20

2. Find the value of F.05,2,10

3. In ANOVA, we will tend to not reject the null hypothesis of equal population means whenever the calculated F is:a. smallb. largec. equal to the critical Fd. none of the above is correct

4. In a two-way ANOVA, in the xijk = + i + j + ()ijk+ ijk model, the term ()ij representsa. random error in the sampling processb. the effect that is due to factor Ac. the effect that is due to factor Bd. the interaction effect between level i of factor A and level j of factor Be. the level of significance at which the null hypothesis is rejected

5. Which of the following is a typical source of internal secondary data for business research?a. accounting or financial reportsb. sales information

c. production datad. all of the above

For questions 6 to 10, refer to the plot on the right.

6. The equation for the line going through the points would take the form of:a. Y = a + b+ cb. Y = a - bXc. Y = x -1d. Y = a + bX2



e. None of these is correct

7. In this particular problem, the researcher is trying to predict:a. Quantity demanded based on priceb. Price based on quantity demandedc. Both price and quantity demandedd. None of these is correct

8. If computed, the sign of b in the equation would be:a. Either positive or negativeb. Positivec. Negatived. None of the above

9. The correlation coefficient of the problem, if computed, could be:a. 1.00b. 0c. -1.0d. None of the above

10. Which of the following won't be true for the regression resulting from the data in the plot above?a. r2 = 100%b. r = 1c. sy.x = 0d. ANOVA p-value = 0e. All of the above are true

11. In multiple regression analysis, multicollinearity means: a. High correlation between the dependent variable and the independent variablesb. A high correlation between the response variable and some independent variablesc. A condition where 2 or more of the independent variables are highly correlated with each otherd. None of the above

Use the following regression output to answer questions 12 to 15.

The regression equation issales = 46.5 + 52.6 ad

Predictor Coef Stdev t-ratio pConstant 46.486 9.885 4.70 0.000ad 52.57 10.26 5.12 0.000

s = 6.837 R-sq = 76.6% R-sq(adj) = 73.7%


SOURCE DF SS MS F pRegression 1 1226.9 1226.9 26.25 0.000Error 8 374.0 46.7Total 9 1600.9

12. Identify the coefficient of determination __________



13. Write the standard deviation of the slope (sb )____________

14. Write the standard error of the estimate (sy.x)_________

15. Identify the slope ______________

16. List ONE of the 3 assumptions that underlie the simple regression model ___________

17. Suppose you wished to investigate the effect of consumption of alcohol (Y/N) and a common over-the-counter cold medicine (Y/N) on a person's reaction time. The appropriate statistical procedure for this experiment is:a. Chi-square test of independenceb. Analysis of Variance (Two factor)c. Analysis of Variance (One factor)d. Regression analysise. Discriminant analysis

18. Which of the following is not a linear function?a. Y = a + bXb. Distance = Rate of speed Travel timec. Total Profit = profit per unit Number units soldd. Total Cost = Fixed Cost + Unit Variable Cost Quantity Producede. All of the above are linear functions

19. There are two main uses of a multiple linear model: 1) for slope analysis, or 2) for estimating the value of Y given a value of X's. For which of the two uses is multicollinearity not a problem? _____________

20. The _____ interval is the interval that contains the actual average value of the response variable for a specific value of Xa. Confidenceb. Prediction

21. The _____ interval is the interval that contains the actual value of the response variable for a specific value of Xa. Confidenceb. Prediction

22. Residual is also known as:a. Errorb. Actual Y Estimated Yc. Observed Y Fitted Yd. None of the abovee. All of the above

PROBLEM 1:Ryerson Coil Pickling manager wishes to know how the level of pickling operation (measured in tons) affects the monthly overtime expense of the plant. He collects data for the last 17 months on actual tonnage processed and overtime cost. He then performs regression analysis on the data. Using the attached output, answer questions 23 to 37.

Partial Data:

Month 1 2 3 . . . 17Production (Tons) 29,668 23,577 27,117 . . . 19,365



Overtime Expense $11,000 $8,000 $9,000 . . . $7,000

23. What is the response variable in this problem? __________

24. What is(are) the independent variable(s)? ______________

25. Give the linear regression equation generated by the 17 observations. __________

26. Using the regression equation, estimate the plant’s overtime expense for a month where 30,000 tons of steel is planned to be processed. __________

27. The correlation between the response variable and the predictor variables could be best described as:a. Perfectly positively linearb. Perfectly negatively linearc. Positively correlatedd. Negatively correlated

28. For the planned production described in #26, give the 95% interval estimate for the plant’s actual overtime expense. ____________________

29. How much of the variation in overtime expense is explained by production level? _____________

30. For every ton increase in processing level, plant overtime expense salary is expected to:a. increase by $0.587b. increase by $5.87c. decrease by $0.587d. decrease by $5.87e. None of the above

31. If for the year 2000 the plant plans to process 30,000 tons of coils each month, give the 95% interval estimate of the actual average monthly overtime expense. ____________

32. What is the coefficient of correlation of this model?_____________

33. Using the slope test set up the null and alternative hypotheses to determine whether the model you identified in #25 is significant.

34. Write the decision rule for the hypotheses in #33 ______________

35. Identify the computed and critical values corresponding to the decision rule in #34. ________, ________

36. Based on #35, is the model significant? __________

37. Compute the 95% confidence interval for the actual change in overtime expense for every ton increase in production level. _____________

PROBLEM 2:In a recent survey of winter 1999 BA 282 students, 41 students from the Medford section and the Ashland sections responded. One of the objectives of the survey was to investigate what factors could possibly affect students’ success in the course's midterm exam. A regression analysis was performed in which 4 explanatory variables were included in the model. The variables were the following:

GPA – student’s overall GPA to date



243GRADE – student’s final grade in the prerequisite course, MA 243243WHEN – the number of terms ago the student took MA 243WHERE – where the student is currently taking BA 282 0 code for Ashland section, 1 for Medford

Use the attached regression output in answering questions 38 to 49

38. What is the dependent variable in this problem? __________

39. What are the predictor variables? ______________

40. Give the linear regression equation generated by the 41 observations. __________

41. Give an estimated midterm grade for a student in the Medford section who earned 3.5 in MA 243 grade a term ago, and currently holds a 3.25 overall GPA. __________

42. How much of the variation in BA 282 midterm exam can be explained by the regression equation? _____________

43. For every unit increase in MA 243 grade, BA 282 midterm grade is estimated toa. increase by .330 unitsb. decrease by .330 units c. increase by .781 unitsd. decrease by .781 unitse. None of the above

44. Compute the 80% confidence interval for the actual slope of the variable MA 243. ___________

45. What is the multiple coefficient of determination of this model? __________

46. Using the ANOVA test set up the null and alternative hypotheses to determine whether the model is significant.

47. Write the decision rule for the hypotheses in #46 ______________

48. Identify the computed and critical values (use a 0.05 significance level) corresponding to the decision rule in #47 ________, ________

49. Based on #48, is the model significant? __________

PROBLEM 3:Given a significance level of 0.05, is there a significant difference in the average midterm grades of the students in the 3 sections of BA 282? (output attached)

50. The most appropriate testing procedure for the problem stated above is:a. Chi-square test of independenceb. Regression and correlationc. Discriminant analysisd. Oneway ANOVAe. Twoway ANOVA

51. Write the appropriate hypotheses and decision rule for comparing the average test scores of the three sections.



52. Give the computed and critical F values for carrying the test in #51____________, ____________

53. At 0.05 significance level, which of the three sections has the largest average test scores (note: your answer here should be consistent with your answer to #51 and 52)? _______

PROBLEM 4:A test was conducted to determine if grade in MA243, or when MA 243 was taken, has any effect on the midterm grades in BA 282. Also of interest in the survey was to determine whether grade in MA 243 and the time it was taken have some interaction effect on the BA 282 midterm grades. But before the ANOVA test was conducted, the raw data were recoded MA 243 grades were re-classified into two groups – A’s and Non A’s. The term it was taken was also reclassified into two groups – one term ago and two or more terms ago. Grades in BA 282 midterm (not changed) are in 4 to 0 scale, representing A to F letter grades. Also, since twoway ANOVA with replication requires that each cell contain equal samples, six students from combination of MA 243 grade and term group were randomly selected to fill the cells – the resulting crosstabulation of the observations is shown below:

MA 243 GradeWhen MA 243 was Taken

One Term Ago Two or More Ago

Use 0.05 significance level for the following tests.

54. Write the appropriate hypotheses for testing whether MA 243 grade has an effect on BA 282 midterm exam. ________________

55. Does MA 243 grade have a significant effect on BA 282 midterm grades? If yes, which grade category performs better? _____________

56. Write the appropriate hypotheses for testing whether when MA 243 was taken has an effect on BA 282 midterm exam. ________________

57. Does the time when MA 243 was taken have a significant effect on BA 282 midterm grade? If yes, which time has higher average midterm grades? _____________

58. Is there a significant interaction between MA 234 grade and the time when it was taken on BA 282 midterm scores?

59 and 60 BONUS



PROBLEM 1:

MTB > Regress 'P_overt' 1 'Ton_prod';SUBC> Predict 30000.

Regression Analysis

The regression equation isP_overt = - 6776 + 0.587 Ton_prod

Predictor Coef StDev T PConstant -6776 4178 -1.62 0.126Ton_prod 0.5868 0.1760 3.33 0.005

S = 2566 R-Sq = 42.6% R-Sq(adj) = 38.7%



Predicted Values

Fit StDev Fit 95.0% CI 95.0% PI 10828 1306 ( 8044, 13611) ( 4691, 16965)

PROBLEM 2:

MTB > Regress 'MIDTERM' 4 'GPA' '243GRADE' '243WHEN' 'WHERE';SUBC> Constant;SUBC> Brief 2.

The regression equation isMIDTERM = - 0.92 + 0.781 GPA + 0.330 243GRADE - 0.242 243WHEN + 0.601 WHERE

41 cases used 5 cases contain missing values

Predictor Coef StDev T PConstant -0.923 1.020 -0.90 0.371GPA 0.7813 0.3569 2.19 0.035243GRADE 0.3302 0.1977 1.67 0.104243WHEN -0.2421 0.1339 -1.81 0.079WHERE 0.6011 0.3957 1.52 0.137

S = 0.9111 R-Sq = 40.7% R-Sq(adj) = 34.1%




Source DF SS MS F PRegression 4 20.5079 5.1270 6.18 0.001Residual Error 36 29.8823 0.8301Total 40 50.3902Problem 3

Analysis of Variance for MIDTERM Source DF SS MS F PSECTION 2 4.48 2.24 1.91 0.160Error 42 49.17 1.17Total 44 53.64 Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev --+---------+---------+---------+----MW 12 1.750 1.055 (---------*----------) TR-A 21 2.000 1.225 (-------*-------) TR-M 12 2.583 0.793 (---------*----------) --+---------+---------+---------+----Pooled StDev = 1.082 1.20 1.80 2.40 3.00

Problem 4

Two-way Analysis of Variance

Analysis of Variance for MIDTERM Source DF SS MS F Pma243 1 13.500 13.500 20.25 0.000ma243whe 1 1.500 1.500 2.25 0.149Interaction 1 0.167 0.167 0.25 0.623Error 20 13.333 0.667Total 23 28.500

Individual 95% CIma243 Mean ----+---------+---------+---------+-------A 3.00 (-------*-------)B,C,D 1.50 (-------*-------) ----+---------+---------+---------+------- 1.20 1.80 2.40 3.00

Individual 95% CIma243whe Mean ---+---------+---------+---------+--------One Term 2.50 (------------*-----------)Two and 2.00 (-----------*-----------) ---+---------+---------+---------+-------- 1.60 2.00 2.40 2.80




Introductionwebpages.sou.edu/~ordonez/BA282Web/BA_282_COURS… · Web viewIdentify the critical value for the test (one tail). a. 0.0199. b. 1.96. c. -1.96. d. -1.645. e. 1.645.

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