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7/17/2019 MAT 211 CourseGuide_Lecture Notes_Summer 2015 (2)
Office: Rm. 6004-A, SECSOffice hours: ST: 11:30 a.m.-13:00 p.mMW: 11:30 a.m – 13:00 p.m or by appointment
Pre-requisite: MAT 101 or equivalent. Instructional Format p/w: 2×1½ -hours lectures
Course objectives
An understanding of statistics is required in the implementation of uncertainty calculations in differentfields. It is understandable clearly by anyone, today information is everywhere and one will be bombarded with the numerical information. What is needed then? Skills are needed today to deal with allof numerical information. First, need to be critical consumers of information presented by others and
second, need to be able to reduce large amounts of data into a meaningful form so that one can makeeffective interpretations, judgments and decisions. The course ‘MAT 211 Probability and Statistics’ is animportant foundation course offered by IUB and suited for all undergraduate students who wish to majorunder the non-SECS, IUB. It covers all the usual topics in statistics and explains how theories can applyto solve real world problems. Topics include: Elementary Descriptive Techniques, Probability Theorywith Important Probability Distributions, Sampling Theory, Statistical Inference, Linear Correlation andRegression Theories and others. By the end of the course, students should have acquired sufficient skillsto be able to: follow statistical arguments in reports and presentations; understand how to apply thestatistical tools to make effective decisions and find that many of the topics and methods students learncan be used in other courses in their future education; finally, express statistical findings in non-technicallanguage.
Textbook: All students should collect:Anderson D.R., Sweeney, D.J. and Thomas A.W. (2011), Statistics for Business and Economics (11thEdition), South-Western, A Division of Thomson Learning.
Recommended ReferenceMurray R. Spigel and Larry J. Stephens (2008), Schaum’s Outline of Theory and Problems of Statistics(Fourth edition), Schaum’s Outline Series, McGraw-Hill.
Evaluation criteriaHomework will be assigned weekly. Students are not required to hand those back for grading butcompleting the given homework is essential for understanding the material and performing satisfactorilyon examinations.
The weighting scheme is as follows:Class Attendance – 5%, Two Class tests (CT) – 35%(20% + 15%), Mid-term test (MT) - 20% and Finaltest (FT) - 40%
7/17/2019 MAT 211 CourseGuide_Lecture Notes_Summer 2015 (2)
Rules and regulations Students are required to attend classes on time and to take well-organized notes. If a student misses or fails to attend a class, it is his/her sole responsibility to obtain missing
information (for examples, change of exam dates, omit/add some topics, lecture notes, new homeworks etc).
For a test, no extensions or alternative times are possible and also there is no word for make-up.
For any unavoidable circumstances, the test will be strictly held on the next lecture.
No extra work will be given to improve the grade. Students are required to show matured behaviour in class. For examples, cellular phones will be
shut off during class lectures and examinations. Eating, drinking, chewing gum, readingnewspapers, socialization and sleeping are not permitted in class.
Any kinds of cheating in class are strictly prohibited and may result in a failing grade forthe course.
Students are advised to obtain a scientific calculator for use in the class. It is noticeable that thetwo variables calculator is needed for all types of calculations.
51%-55%: C-, 46%-50%: D+, 40%-45%: D, below 40%: F
Incomplete (I-Grade)I-grade will be given only to a student who has completed the bulk of the course works and is unable tocomplete the course due to a serious disruption not caused by the student’s own slackness.
Mid-term and Final Test: All sections will have a common examination. Materials and date will beannounced later.
Course Plan
Lecture # Topics Text/Reference
Lecture 1 Introduction: Definition: variable, scales ofmeasurement, raw data, qualitative data,quantitative data, cross-sectional data, time seriesdata, census survey, sample survey, target population, random sample, computer andstatistical packages
Course Guide, pp.7-8
HW: Text, Ex: 2,4,6,9-13, pp.21-23
Lecture 2 Summarizing qualitative data- Frequencydistribution, relative frequency distribution, barchart, pie chartApplications from real data
Summarizing quantitative data- Frequencydistribution, relative frequency distribution,cumulative frequency distribution, Applicationsfrom real data
Course Guide, pp.9-11HW: Text, Ex:4-10, pp.36-39
HW: Text, Ex:15-21, pp.46-48
Ex: 39, 41and 42, pp.65-67
Lecture 3 Histogram, ogive, line chart, stem and leaf displayApplications from real data
Course Guide, pp.12-14
HW: TextEx:15-21, pp.46-48Ex:25-28, pp.52-53
7/17/2019 MAT 211 CourseGuide_Lecture Notes_Summer 2015 (2)
Summarizing bi-variate data: Cross-tabulation,scatter diagram, Applications from real data
HW: Text, Ex:31, 33-36, pp.60-61
Lecture 4 Measures of average: simple mean, percentiles(median, quartiles), modeApplications from real data
Course Guide, pp.15-16
HW: Text, Ex: 5-10, pp.92-94
Lecture 5 Measures of variability: variance, standarddeviation, coefficient of variation, detectingoutliers (five number summary), Applicationsfrom real data
Lecture 17 Interval estimation: Parameter, statistic, margin oferror (ME), statistical tables (z-table, t-table, chi-square table, F-table), confidence interval of population mean, confidence interval of population SD, Applications from real data
Course Guide, pp.45-53
Lecture 18 Interval estimations about two population means,standards deviations
Applications from real data
Text, Chapter 10
Lecture 19 Test of hypothesis
Concept of hypothesis, null hypothesis,alternative hypothesis, one-tail tests, two-tail test,tests of population mean (large samples test, smallsamples test), test of population SD
Applications from real data
Course Guide, pp.55-67
Lecture 20 Lecture 19 continued Course Guide, pp.55-67
Lecture 21Test of hypothesis
Tests of two populations means, two standarddeviationsApplications from real data
Course Guide, pp.68-69Text, Chapter 11
Lecture 22 Correlation analysisConcepts of covariance and correlation(Numerical measures of bi-variate data),
Regression analysisLinear and multiple regression model, prediction,coefficient of determination
Data, elements, variable, observations, raw data, qualitative data, quantitative data, scales of measurement
population, random sample, census, sample survey, cross-sectional data, time series data, Computer andstatistical analysis, glossary.
Textbook: Anderson D.R., Sweeney, D.J. and Thomas A.W. (2011), Statistics for Business andEconomics (11th Edition), South-Western, A Division of Thomson Learning.
Data (or Variable) - Changing characteristics.Examples: Gender, Grade, Family size , Score, Age, and many others.
Gender, Grade- Qualitative data (letter)
Family size, Score, Age - Quantitative data (numeric value)
Family size – Whole number – Discrete data
Score, Age – Continuous data
Note: ID #, cell # are qualitative data
Observations- Data size
Variable denoted by X, Y, Z or denoted by first letter (e.g. Score – S, Age –A)
Elements – Variable (X), elements x1, x2, …., xn
Raw data – Data collected by survey, census etc. It is known as ungrouped data.
Note: Always we have raw data. We have to process or make data summary by various statisticaltechniques (we will learn all by Chapters 2-3).
Scales of measurement
Before analysis, scale of each of selected variables have to define. Specially, when we do our analysis bystatistical packages (e.g. SPSS, Minitab, Strata even in Excel also). We have to assign scale for each ofvariables those involve in our analysis.
There are four kinds of scale: nominal, ordinal, interval and ratio
Nominal, ordinal - Qualitative data
Nominal scale – The variables like Name, ID, Address, Cell # declare this scale. Not possible to doanalysis.Ordinal scale – Qualitative data like test performances (excellent, good, poor etc), quality of food (goodor bad) etc. possible to order. Some analysis is possible.
7/17/2019 MAT 211 CourseGuide_Lecture Notes_Summer 2015 (2)
Interval scale: Shows properties of ordinal data and interval between values are meaningful. ExampleScore for 5 students. Apply ordinal concept and differences of each of two students is meaningful.
Ratio scale – Have properties of Interval data. In addition ratio of the data values are meaningful. .
Example Score for 5 students. Apply interval concept and ratio of each of two students score ismeaningful.
Details see Textbook, p.6
Target population: The set of all elements in a particular study.
Random sample – A subset of target population. Set to set will vary for each of draws
Census - Method to collect data about target population .
Sample survey- Method to collect data about random sample.
For the purpose of statistical analysis, distinguishing time series data and cross-sectional data aremeaningful.
Time series data – Data collected over several time periods. For example, Exchange rate, interest rate,gross national product (GNP), grosses domestic product (GDP) and many others. These sorts of data w.r.ttime are meaningful.
Cross-sectional data – Data collected at same time. For example, company’s profit, students profile wecollect at the same time.
Note that in this course most of the data will be considered as cross-sectional data.
Computer and Statistical packages
Because statistical analysis generally involves large amount of data. That’s why analysis frequently usescomputer software for this work. Several very useful softwares are available in computing literature.These are: SPSS, Minitab, Matlab, Excel, Stara and many others.
HW: Text
Ex: 2,4,6,9-13, pp.21-23
7/17/2019 MAT 211 CourseGuide_Lecture Notes_Summer 2015 (2)
Aim of presentation of raw data; Tabular form of raw data (e.g. Summarizing qualitative and quantitative
Data).
The aim of presentation of raw data is to make a large and complicated set of raw data into a morecompact and meaningful form. Usually, one can summarize the raw data by
(a) The tabular form
(b) The graphical form and
(c) Finally numerically such as measures of central tendency, measures of dispersion and others.
Under the tabular and the graphical form, we will learn frequency distribution (grouping data), bar graphs,
histograms, stem-leaf display method and others.
Presentation of data can be found in annual reports newspaper articles and research studies. Everyone isexposed to those types of presentations. Hence, it is important to understand how they are prepared andhow they should be interpreted.
As indicated in the Lecture 1, data can be classified as either qualitative or quantitative.
The plan of this lecture is to introduce the tabular methods, which are commonly used to summarize boththe qualitative and the quantitative data.
Summarizing qualitative data
Recall raw data and find the following data:
Table 1: Test Performances of MAT 211
Make a tabular and graphical summary of the above data.
Solution: Define T - Test performances and n =15. It is a qualitative data.
Good Good Excellent
Excellent Poor Excellent
Poor Excellent Good
Excellent Excellent Poor
Poor Good Good
7/17/2019 MAT 211 CourseGuide_Lecture Notes_Summer 2015 (2)
So far we have focused on tabular and graphical methods for one variable at a time. Often we need tabularand graphical summaries for two variables at a time.
Tabular Method-Cross-tabulation and Graphical method- scatter diagram are such two methods tomake decision from two qualitative and/or quantitative variables.
Tabular Method-Cross-tabulation:
Problem-1:Consider the following two variables: Quality rating and meal price($) for 10 restaurants. Data are asfollows:Quality rating: good, very good, good, excellent, very good, good, very good, very good, very good,goodMeal price($): 18,22,28,38,33,28,19,11,23,13.Make a tabular summary (or cross-table and make a data summary).
7/17/2019 MAT 211 CourseGuide_Lecture Notes_Summer 2015 (2)
Solution: Define X - Quality rating and Y - Meal price. Here n =10
Table: Crosstabulation of X and Y for 10 restaurants
Y
X 10-20 20-30 30-40 Total
Good || (2) || (2) (0) 4Very good || (2) || (2) |(1) 5
Excellent (0) (0) |(1) 1
Total 4 4 2 n=10
Data summary:We see that there are 2 restaurants their quality of food is very good and meal prices are ranging 20$ to30$, 1 restaurant quality of food is excellent, 4 restaurants meal prices are ranging 10$ to 20$ and so on.
Graphical method-scatter diagramScatter diagram provide the following information about the relationship between two variables.
• strength• shape – linear, curved etc.
• Direction – positive or negative
• Presence of outliers
Problem -2: Now consider the following two variables: # of commercials and total sales for 5 sound equipment stores.Data are as follows:
# of commercials: 2, 5, 1, 3, 4 and total sales: 50, 57, 41, 54, 54
Data summary: There is a positiverelationship exists between # ofcommercials and total sales for 5sound equipment stores.
Figure: Scatter diagram of Sales and # of commercials for 5 sound equipment stores
HW: TextEx: 31, 33-36, pp.60-61
0
10
20
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0 1 2 3 4 5 6
S a l e s
Comm
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Definition of average: It is a single central value that represents the whole set of data. Differentmeasures of averages are: simple mean, weighted mean, median, mode, quartiles, percentiles.
We will learn the above measures for the raw data and grouped data.
Mean: Denoted by and calculated by ∑ / .
For example, for a set of monthly starting salaries of 5 graduates: 3450, 3550, 3550, 3480, 3355.
Define X - monthly starting salaries of 5 graduates. Here ∑ / = 3477.
Median, Percentiles, Quartiles
It is denoted by pi , i =1, 2, …, 99 that means there are 99 percentiles.50th percentile is known as median and it is denoted by p50.
25th percentile is known as first quartile and it is denoted by p25.
75th percentile is known as 3rd quartile and it is denoted by p75. p50 is also known as 2nd quartile (Q2).Thus, there are 3 quartiles: These are p25 (Q1), p50 (Q2) and p75(Q3).
Calculation of percentiles: Need to sort the data
3355 3450 3480 3550 3550
For Q2: i = (pn)/100 = (50*5)/100= 2.50. The next integer 3. Thus, Q2 is 3480.
For Q1: i = (pn)/100 =(25*5)/100 = 1.25. The next integer 2. Thus, Q1 is 3450.
For Q3: i = (pn)/100 =(75*5)/100 = 3.75. The next integer 4. Thus, Q3 is 3550.
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Now consider the following data: 3450, 3550, 3550, 3480, 3355, 3490
Here ∑ / = 3.4792e+003 = 3479.2
Sort the data to calculate percentiles: 3355 3450 3480 3490 3550 3550
For Q2: i = (pn)/100 = (50*6)/100= 3. It is an average value of 3rd and 4th observations of the sorted data.Thus, Q2 = (3480+3490)/2 = 3485.
For Q1: i = (pn)/100 =(25*6)/100 = 1.50. The next integer 2. Thus, Q1 is 3450.
For Q3: i = (pn)/100 =(75*6)/100 = 4.5. The next integer 5. Thus, Q3 is 3550.
Mode: It is the value that occurs with greatest frequency. Denoted by M0.
Consider the following observations
(1)
3450, 3550, 3550, 3480, 3355 - M0 is 3550.
(2) 3450, 3550, 3550, 3480, 3450 - M0 are 3450 and 3550.
(3) 3450, 3550, 3550, 3450, 3450 - M0 is 3450
(4)
3450, 3650, 3550, 3480, 3355 – no Mode.
Data Summary:
Mean = 3477 it means that most of graduates monthly starting salaries is about 3477$.
Median = 3485 it means that 50% graduates monthly starting salaries are observed below 3485$ and theremaining (50%) graduates monthly starting salaries are observed over 3485$.
First quartile = 3450 it means that 25% graduates monthly starting salaries are observed below 3450$ andthe remaining (75%) graduates monthly starting salaries are observed over 3450$.
Third quartile = 3550 it means that 75% graduates monthly starting salaries are observed below 3550$and the remaining (25%) graduates monthly starting salaries are observed over 3550$.
Mode = 3450 it means that the most common graduates monthly starting salaries is 3450$.
HW: Text
Ex: 5-10, pp.92-94
7/17/2019 MAT 211 CourseGuide_Lecture Notes_Summer 2015 (2)
Recall the concept of average (Ref. Lecture 4). Follow the following: Say for example, suppose we have
the following 2 sets of raw data:
1) 15, 15,15,15,15 – Average 15 and variation 0.2) 15, 16,19, 13, 12– Average 15 and variation 2.73.
Statistical meaning of variation
Make a question - is there any difference exist between each of observations from the average value?Suppose X – score of CT1 (class test 1) and for example, suppose it is calculated average score 15.
Next investigation will be to see differences between each of student’s marks to average marks.
If difference is 0, very easy to say student score and average score is same.
If differences give us a positive (negative) sign (+(-)), we can say that student score is greater(lower) thanthe average score.
How we can measure variation of a data set. Various measures (or formulas) are available to detectvariation. These are:
1. Range, R = H-L, H-highest value of a data set and L – Lowest value of a data set
2. Inter-quartile range, IR = p75 – p25, p75- 75th percentile and p25- 25th percentile
3.
Variance (denoted by
) and is calculated by
∑
.
4.
*Standard Deviation (denoted by and is calculated by ∑ ). That means SD = sqrt(variance).
Note: Measures of variation cannot be negative. At least can be 0, recall which indicates all students gotsame scores.
Calculation for variance and SDRecall monthly starting salaries of 5 graduates: 3450, 3550, 3550, 3480, 3355, where we found ∑ / = 3477 (see L3).
Calculation Table for variance and SD
X 3450 729
3550 5329
3550 5329
3480 9
3355 14884
Here variance, ∑ = 26280/4= 6570 and SD = sqrt(variance) = 81.05$.
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Data summary: SD = 81.054 indicates that graduates salary varies from the average salary 3477$.
Note: Variance cannot be interpreted because its unit comes as a square. For example if mean = 3477$then variance = 6570$2. Taking square root of variance removes this problem (going back to the originalunit of data), which is standard deviation (SD).
So, no interpretation for variance and talk always on SD measure.
Coefficient of variation
See Text, p.99
HW: Text, Ex: 16-24, pp.100-102
Detecting outliers (Five number summary)
See, Text, pp.109-111
HW: Text, Ex: 40-41, pp. 112-113
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So far we focused calculation of all measures of average and variation for ungrouped (raw data).Sometimes grouped data (frequency table) is available. In this situation, formula for ungrouped (raw data)is invalid. Follow the following:
Recall Tabular summary, where X- Test Score and n =15 (Lecture 2)
X Frequency (# of students) – f i, , i=1,2,3
46-56 2
56-66 2
66-76 2
76-86 3
86-96 6
Total n =15
X Frequency (# ofstudents) –f i, , i=1,2,3
Midpoints(mi)
f imi
46-56 2 51 102 1352
56-66 2 61 122 512
66-76 2 71 142 72
76-86 3 81 243 48
86-96 6 91 546 1176
Total n =15 1155 3160
Grouped mean (weighted mean)
∑ /
= 1155/15=77
Here variance, ∑ =3160/14=225.71 and SD = sqrt(variance) = sqrt(225.71) =
15.02
Data Summary: SD = 15.02 indicates that students score varies from the average score 77.
HW: Text, Ex: 54-55, pp.128-129
Text: Case problems 1, 2, 3, 4, pp.137-141
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We can get a general impression of skewness by drawing a histogram. To understand the concept ofskewness, consider the following 3 histograms:
Figure -1 Figure-2 Figure-3
Figure-1 is known as positively skewed or skewed to the right.
Figure -2 is known as normal/symmetric frequency curve.Figure-3 is known as negatively skewed or skewed to the left.
There are two types of skewness. These are (1) positively skewed or skewed to the right (2) negativelyskewed or skewed to the left.
Note that the normal/symmetric frequency curve is known as non-skewed curve (skewness is absent).
Definition: It gives us idea about the direction of variation of a raw data set.
Figure -1 - direction of variation is observed in left (most of frequencies).Figure -2 - direction of variation is observed in middle (most of frequencies).
Figure -3 - direction of variation is observed in right (most of frequencies).
Recall X – test score.
Figure 1 says us most of students have poor performances. It means that most of students score below theaverage value.Figure 2 says us most of students have average performances. It means that most of students score near to(more/less) the average value.Figure 3 says us most of students have good performances. It means that most of students score over theaverage value.
Measure of skewness
To detect whether skewness is present or not in a set of raw data, we will use the most commonly usedformula, known as Karl Pearson’s (known as Father of Statistics) coefficient of skewness. It is defined as
SK = 3(mean-median)/SD
Note that this formula will work for ungrouped/grouped data.
6
2 23
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0
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4
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8
2 2
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32
0
2
4
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2 2 23
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4
6
8
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Suppose X – test score. Let mean = 15, median(50th percentile or 2nd quartile) = 17 and SD =3.Here SK = -2.00.Data summary: SK = -2.00 it means that the test score is negatively skewed. It means that most ofstudents score over 15.
Let mean = 18, median= 14 and SD =5. Here SK = 2.40.Data summary: SK = 2.40 it means that the test score is positively skewed. It means that most of studentsscore below 18.
Let mean = 16, median= 16 and SD =5. Here SK = 0.Data summary: SK = 0 it means that the test score is symmetric. It means that few student’s score belowand over 16.
KurtosisSuppose if a distribution is symmetric, the next question is about the central peak: Is it high or sharp or
short or broad.Pearson (1905) described kurtosis in comparison with the normal distribution and used phases leptokurtic,
platykurtic and mesokurtic to describe different distributions.
If the distribution has more values in the tails of the distribution and a peak, it is leptokurtic. It is a curve
like two heaping kangaroos has long tails and is peaked up in the center.
If there are fewer values in the tails, more in the shoulders and less in the peak, it is platykurtic.
A platykurtic curve, like a platypus, has a short tail and is flat-topped.
HW: Text bookEx: 5 and 6, pp.92-93 (Calculate skewness and interpret)
7/17/2019 MAT 211 CourseGuide_Lecture Notes_Summer 2015 (2)
Interpretation: Table 2 shows that 9 programs need times 4.1 to 5.4 seconds, Table 3 shows that 30 percent programs need times 4.1 to 5.4 seconds and Table 4 shows that 18 programs need at most 5.4seconds and so on.
c) Descriptive Statistics: X
Variable n Minimum Maximum
X 30 0.20 8.1
Variable Mean Median(Q2) StDev Q1 Q3
X 5.0 4.75 1.859 4.02 6.12
Interpretation:
Mean = 5.0 seconds means that most of times to run a program need approx. 5 seconds.
Median = 4.75 seconds means that 50% programs to run need less than 4.75 seconds and rest of 50% need
more than 4.75 seconds.
Standard deviation = 1.859 seconds means all the times a program did not take 5 seconds to run.
7/17/2019 MAT 211 CourseGuide_Lecture Notes_Summer 2015 (2)
Random experiment, random variable, sample space, events (simple event, compound event), counting
rules, combinations, permutations, tree diagram, probability defined on events
Introduction
We finished our first important part of the course (known as data summary). Even we sat for CT1. Nowwe are moving in the 2nd very important part of the course, namely “Chance Theory”. It is also known as“Probability Theory”. The word “Chance or Probability” frequently we are using in our real life. Forexamples:
(i) What is the chance of getting grade-A for the course MAT 211?(ii) What is the chance that sales will decease if we increase prices of a commodity?(iii)
What is the chance that a new investment will be profitable?
and so many other situations not possible to record all.
It is true word “Chance or Probability” using frequently.
To understand consider a situation. For example if we ask the following question to 3 students:
What is the chance of getting grade-A for the course MAT 211?
Say for example, answered the following:
Student-1: Chance is 95%
Student-2 : Chance is 100%Student-3 : Chance is 10%
Let’s explain their predicted values under the chance theory. What we can observe:
Student 1 is 95% confident he/she is getting grade A. That means past experience tells us out of 100students, 95 students had grade A.
Student 2 is 100% confident he/she is getting grade A. All the students got grade A.
Student 3 is only 10% confident (less confident) he/she is getting grade A. Only 10 students out of 100students got grade A.
How calculated?Recall the relative frequency method, where relative frequency = frequency/n and apply this. We will getthe answer. Suppose n =100, # of students got grade A is 95 (frequency), here probability is 0.95.
This formula we will use to calculate probability. Follow the following
To calculate probability of an event (recall in the previous example, one possible event grade A), we haveto very familiar with the following terms:
7/17/2019 MAT 211 CourseGuide_Lecture Notes_Summer 2015 (2)
Random experiment, random variable, sample space, events
Random experiment – It is the process of getting all possible events. Events are also known asoutcomes.
Random variable- It is denoted by r.v. It is the event which one we will be interested from the all possible outcomes. In the previous example, grade A is the random variable.
Note that r.v. will vary from experiment to experiment.
Sample Space: It is very very important. Without it, will not be possible to calculate chance of an event.
It is denoted by S. It is all the possible outcomes of a random experiment. It is just the set (recall set –collection of all objects). Sometimes it will be not possible to calculate easily (note that to get the ideaabout S, we have to practice a lot!).
Several methods will be used to find S. These are:
Our knowledge, tree diagram (a wonderful method) and counting rules (permutation, combination). Wewill use combination approach most of times, however permutation approach also will be used.
Events – It is denoted by E. It is a possible outcome of our random experiment.
Formula to compute probability of an event: It is denoted by P(E) and calculated as
P(E) = (#of E)/S, 0≤P(E)≤1If
P(E) = 0, no chance to occur (improbable event).
P(E) = 0.5, 50% chance that the E will occur.P(E) – 1.0, 100% chance, that the E will occur (sure event)
Recall grade exampleP(grade A) =(#of E)/S = 95/100 = 0.95, where S = {all possible grades}, E = grade A.Summary: The randomly selected student will get grade A, chance is 95%.
Some random experiments and S (Text, p.143)
Random experiment 1:Toss a fair coin. S = {H,T}, H-head and T-tail. If E – head, then P(H) = ½ = 0.5 and P(T) = ½ = 0.5.
Random experiment 2:Select a part for inspection, S = {defective, non-defective}.
Random experiment 3:Conduct a sales call, S = {purchase, no purchase}.
Random experiment 4:Roll a fair die, S={1,2,3,4,5,6}.
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Random experiment 5:Play a football game, S = {win, lose, tie}.
Note that in the sample space, S all possible events read as “or”. Be careful not “and”. It is impossible toget H or T in one experiment. Win and lose in one game is also impossible (realize it!)
Important concepts
mutually exclusive events, equally like events, Tree diagram, combination
Mutually exclusive events – It is the event where two possible events cannot occur simultaneously. Toss a
coin, H and T cannot occur in a single random experiment. It is written as P(H∩T) =0.
If P(H∩T) ≠0, events are mutually inclusive. Toss two coins (or one coin two times), H and T can occur
in this random experiment. For example, P(H∩T) =0.50, where S = {HH, HT, TH, TT}.
Equally like events – Two events has equal chance of being occur. Toss a coin, P(H) = P(T) = 0.5.
Tree diagram – It is a technique to make a summary of all possible events of a random experimentgraphically.
Combination - It is a formula to make a summary of all possible events of a random experiment.
Counting rules – Two rules: Combination and Permutation
Combination – It allows one to count the number of experimental outcomes when the experimentinvolves selecting n objects from a set of N objects.
For example, if we want to select 5 students from a group of 10 students, then
!!! !!! 252 possible ways students can be selected, here S = 252.
Permutation – It allows one to count the number of experimental outcomes when n objects are to beselected from a set of N objects, where the order of selection is important.
For example, if we want to select 5 students from a group of 10 students ( where order is important , then
!! !! 30240 possible ways students can be selected, here S = 30240.
Ex: 1. How many ways can three items can be selected from a group of six items? Use the letters A, B, C,
D, E and F to identify the items and list each of the different combinations of three items.Solution: S = 6 !!! 20 possible ways letters can be selected. Some examples, ABC, ABD,
ABE, ABF, ……. DEF.
Ex: 2. How many permutations of three items can be selected from a group of six items? Use the lettersA, B, C, D, E and F to identify the items and list each of the different permutations of items B, D and F.
Solution: S = 6 !! 120 possible ways letters can be selected.
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Different permutations of items B, D and F: BDF, BFD, DBF, DFB, FDB, FBD, 6 outcomes.
Ex:3: An experiment with three outcomes has been repeated 50 times and it was learned that E1 occurred20 times, E2 occurred 13 times and E3 occurred 17 times. Assign probabilities to the outcomes.
Solution: S = {E1, E2, E3}. Here P(E1) = 20/50=0.40, P(E2) = 13/50=0.26, P(E3) = 17/50=0.34
P(S) = P(E1)+ P(E2)+ P(E3)= 0.40+0.26+0.34 = 1.0
Ex:4: A decision maker subjectively assigned the following probabilities to the four outcomes of anexperiment: P(E1) = 0.10, P(E2) = 0.15, P(E3) = 0.40 and P(E4) = 0.20. Are these probabilityassignments valid? Explain
Solution: S = {E1, E2, E3, E4}. Here P(E1) = 0.10, P(E2) = 0.15, P(E3) = 0.40, P(E4) = 0.20.
P(S)=P(E1)+P(E2)+P(E3)+P(E4)=0.10+0.15+0.40+0.20=0.85<1.0. Thus, probability assignments invalid
because P(S) ≠1.
The above two problems tell us for any random experiment, P(S) = 1.
Ex:5: Suppose that a manager of a large apartment complex provide the following probability estimatesabout the number of vacancies that will exist next month
Provide the probability of each of the following events:
a. No vacancies b. At least four vacancies
c.
Two or fewer vacancies
Solution: S={0,1,2,3,4,5}
a. P(0)=0.05
b.
P(At least four vacancies) = P(4)+P(5)=0.20.
c. P(Two or fewer vacancies)= P(0)+P(1)+P(2)=0.05+0.15+ 0.35=0.55.
Ex:6: The National Sporting Goods Association conducted a survey of persons 7 years of age or olderabout participation in sports activities. The total population in this age group was reported at 248.5
million, with 120.9 million male and 127.6 million female. The number of participation for the top fivesports activities appears here
Participants
Activity Male Female
Bicycle riding 22.2 21.0
Camping 25.6 24.3
Exercise walking 28.7 57.7
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a. For a randomly selected female, estimate the probability of participation in each of the sportsactivities
b. For a randomly selected male, estimate the probability of participation in each of the sports
activitiesc.
For a randomly selected person, what is the probability the person participates to exercisewalking?
d. Suppose you just happen to see an exercise walker going by. What is the probability the walker isa woman? What is the probability the walker is a man?
Solution: S = {Br, C, EW, EE,S}, where Br - Bicycle riding, C- Camping, EW- Exercise walking, EE-Exercising with equipment and S – Swimming.
a. Female can come from any sports activities. Thus P(F) = (21/248.5) +(24.3/248.5) +… +(34.4/248.5).
b.
Male can come from any sports activities. Thus P(M) = (22.2/248.5) +(25.6/248.5) +… +
(26.4/248.5).c. Person can be male or female. Thus, P(EW) = P(Male EW) +P(Female EW) = (28.7/248.5)
+(57.7/248.5)=86.4/248.5=0.34 = 34%.d. We have to consider exercise walker population. Thus, P(woman/EW) = 57.7/ (28.7+57.7) =
Basic relationships of probability (addition law, complement law, conditional law, multiplication law)
Addition Law
Suppose we have two events A and B (A, B ∈S). The chance of occurring A or B is written as
•
P(A∪B) = P(A) + P(B) - P(AB), if two events are not mutually exclusive.
• P(A∪B) = P(A) + P(B), if two events are mutually exclusive.
Keywords: Or, at least
Problem 1
Consider a case of a small assembly plant with 50 employees. Suppose on occasion, some of the workersfail to meet the performances standards by completing work late or assembly a defective product. At theend of a performance evaluation period, the production manager found that 5 of the 50 workers completed
work late, 6 of the 50 workers assembled a defective product and 2 of the 50 workers both completedwork late and assembled a defective product. Suppose one employee if selected randomly what is the probability that the worker completed work as late or will assembled a defective product?
Solution: Let L- work is completed late, D - assembled product as defective. Total employees S = 50.
We have to find P(L∪D). We know P(L∪D) = P(L) + P(D) - P(LD) = (5/50)+(6/50) – (2/50) =0.10+0.12-0.04 = 0.18 = 18%.
The chance is 18% the worker completed work as late or will assembled a defective product.
Problem 2
A telephone survey to determine viewer response to a new television show obtained the following data
Rating: Poor Below average Average Above average ExcellentFrequency: 4 8 11 14 13Suppose a viewer is selected randomly
(i) What is the chance that the viewer will rate the new show as average or better?(ii) What is the chance that the viewer will rate the new show as average or worse?
Solution: Total possible viewers S = 50.
(i)
P(average or better) = (11/50) + (14/50)+(13/50)
The viewer will rate the new show as average or better, chance is 76%.
(ii) P(average or worse) = (11/50) + (8/50) + (4/50) = 0.46 = 46%
The viewer will rate the new show as average or worse, chance is 46%.
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Suppose we have one event A, then the chance of not getting A event is defined as
P(Ac) = 1-P(A), A∈S,
Keyword: not
Recall Problem 1
(i) What is the chance that the randomly selected worker completed work will not be late?(ii) Suppose one employee if selected randomly what is the probability that the worker completed
work as late nor will assembled a defective product?
Solution: (i) P(Lc) = 1-P(L) = 1-0.10 =0.90 = 90%
The chance is 90% that the randomly selected worker completed work will not be late.
(ii) P(L∪D)c= 1- P(L∪D)= 1-0.18 = 0.82 = 82%
The chance is 82% that the randomly selected worker completed work as late nor will assembled adefective product.
Conditional law - Keyword: If, given, known, conditional
Suppose we have two events A and B (A, B ∈S), the chance of getting A when B is known (or B when Ais known) is defined as
• P(A/B) = P(AB)/P(B), P(B) ≠ 0
• P(B/A) = P(AB)/P(A), P(A) ≠0
To understand the concept, consider the following situation:
Roll a die. What is the chance of getting the die will show
(i) 2(ii) Even number(iii) 2 or even number(iv)
Not 2(v) 2 given that die will show even number(vi) 2 given that die will show odd number
Solution: S = 6. (i) P(2) = 1/6 (ii) P(Even number) =3/6 (iii) P(2∪even number) = (1/6)+(3/6)-(1/6)
Observe carefully (i) to (iv) are unconditional probabilities, but (v) to (vi) are conditional probabilities.Here to calculate (i) to (iv) we used unconditional sample space, whether to calculate (v) to (vi) we usedconditional sample space, where has given condition from the roll we need even or odd numbers.
Multiplication law
Suppose we have two events A and B (A, B ∈S), the chance of getting A and B is defined as
• P(AB) = P(A/B)P(B) if A and B events are dependent
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• P(AB) = P(A) P(B) if A and B events are independent
Keyword: both, joint, altogether, and
Problem
Consider the situation of the promotion status of male and female officers of a major metropolitan policeforce in the eastern United States. The force consists of 1200 officers, 960 men and 240 women. Over the past two years 324 officers on the public force received promotions. The specific breakdown of promotions for male and female officers is shown in the following Table
Table: Promotion status of police officers over the past two years
Men Women Total
Promoted 288 36 324
Not Promoted 672 204 876
Total 960 240 1200
a) Find a Joint probability table.
b)
Find marginal probabilities.c)
Suppose a male officer is selected randomly, what is the chance that the officer will be promoted?d) Suppose a female officer is selected randomly, what is the chance that the officer will not be
promoted?e)
Suppose an officer is selected randomly who got promotion, what is the chance that the officerwill be male?
f)
Suppose an officer is selected randomly who did not get promotion, what is the chance that theofficer will be female?
Solution: Here S = 1200 officers
a) Joint probability table for promotion status
Men Women Total
Promoted 0.24 0.03 0.27
Not Promoted 0.56 0.17 0.73
Total 0.80 0.20 1.00
b) P(Men) = 0.80, P(Women) = 0.20, P(Promoted) = 0.27, P(Not Promoted) = 0.73, these are known asmarginal probabilities.
c) P(Promoted/Men)=288/960.
d) P(Not Promoted/ Female) =204/240.
e) P(Male/Promotion) = 288/324.
f) P(Female/not Promoted) = 204/876.
HW: Text, Ex: 22-27, pp.169-170 and Ex: 32-35, pp.176-177
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Computing Probabilities for Any Normal Probability Distribution
The reason for discussing the standard normal distribution so extensively is that probabilities for allnormal distributions are computed by using the standard normal distribution. That is, when we have anormal distribution with any mean and standard deviation, we answer probability questions about thedistribution by first converting to the standard normal distribution. Then we can use Table and the
appropriate Z values to find the desired probabilities.
ProblemConsider according to a survey, subscribers to The Wall Street Journal Interactive Edition spend averageof 27 hours per week using the computer at work. Assume the normal distribution applies and that thestandard deviation is 8 hours.
a) What is the probability a randomly selected subscriber spends less than 11 hours using thecomputer at work?
b) What percentage of the subscribers spends more than 40 hours per week using the computer atwork?
c)
A person is classified as a heavy user if he or she is in the upper 20% in terms of hours of usage.
How many hours must a subscriber use the computer in order to be classified as a heavy user?
SolutionDenote X = No. of hours per week using the computer at work, X~N(27, SD = 8)
a) Need to find p(X<11) = p(Z<-2) = 0.028
b) p(X>40) = p(Z>1.62) = 0.0526 i.e. 5.26%.
c) Need to find X when p = 0.20. Thus, X = μ + Zσ = 27 + 8Z. When p = 0.20, then find Z and substitutethe Z-value to get the value of X.
HW: Ex: 10-25, pp.248-250
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Lecture 17Chapter 8_Interval estimation (Estimation of Parameters)
Aim
Be familiar how to construct a confidence interval for the population parameter.
The sample statistic is calculated from the sample data and the population parameter is inferred (orestimated) from this sample statistic. In alternative words, statistics are calculated; parameters areestimated.
Two types of estimates we find: point estimate and interval estimate.
Point Estimate – It is the single best value. For example, mean and SD of total marks for a course of IUB
students are point estimates because these are single value.
Interval Estimate - Confidence Interval
The point estimate is varying for sample to sample and going to be different from the population
parameter because due to the sampling error. There is no way to know who close it is to the actual parameter. For this reason, statisticians like to give an interval estimate (confidence interval), which is arange of values used to estimate the parameter.
A confidence interval is an interval estimate with a specific level of confidence. A level of confidence is
the probability that the interval estimate will contain the parameter. The level of confidence is 1 - α. 1-α area lies within the confidence interval.
Confidence interval for based on large samples
Problem
Suppose, total marks for a course of 35 randomly selected IUB students is normally distributed with mean
78 and SD 9. Find 90%, 95% and 99% confidence intervals for population mean μ. Make a summary
based on findings.
Solution:
We have given X~N(78,9), where X - total marks for a course of 10 randomly selected IUB students and
n=35.
90% confidence interval for :
α
/ √
α
/ √
Here =78, =9, n=35, α=1-0.90 = 0.10, α/2 = 0.05 and α /=.=1.65
Thus, 781.65 9√ 35 78 1.65 9√ 35
78 2.5101 78 2.5101
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Summary: Based on our findings, we are 90% confident that population mean is ranging 75.5 to 80.5.
95% confidence interval for :
α / √ α / √
Here =78, =9, n=35, α=1-0.95 = 0.05, α/2 = 0.025 and α /=.=1.96
Thus, 781.96 9√ 35 78 1.96 9√ 35
78 2.9817 78 2.9817 75.0183 80.9817
Summary:
Based on our findings, we are 95% confident that population mean is ranging 75.01 to 80.98.
99% confidence interval for :
α / √ α / √
Here =78, =9, n=35, α=1-0.99 = 0.01, α/2 = 0.005 and α /=.=2.58
Thus, 782.58 9√ 35 78 2.58 9√ 35
78 3.9249 78 3.9249 74.0751 81.9249
Summary:
Based on our findings, we are 99% confident that population mean is ranging 74.07 to 81.92.
Practice problems
1. In an effort to estimate the mean amount spent per customer for dinner at a major Atlanta restaurant,data were collected for a sample of 49 customers over a three-week period. Assume a populationdeviation of $2.50.a. At a 95% confidence level, what is the margin of error? b. If the sample mean is $22.6, what is the 95% confidence interval for the population mean?
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X- Amount spent per customer for dinner at a major Atlanta restaurant. Here n=49, SD = $2.50
a) Find Margin of error (ME) =
α /
√ , here α=1-0.95 = 0.05, α/2 = 0.025 and
α /=
.=1.96
b)
95% confidence interval for the population mean: α / √ α / √
(Solve it)
2. Have a machine filling bags of popcorn; weight of bags known to be normally distributed with meanweight 14.1 oz and SD 0.3 oz. Take sample of 40 bags, what’s a 95% confidence interval for population
mean μ?
Guideline:
a) X - weight of bags. Here n=40,
=14.1,
=0.3 α=1-0.95 = 0.05, α/2 = 0.025 and
α /=.=1.96
95% confidence interval for population mean μ:
α / √ α / √
(Solve it)
3. The National Quality Research Center at the University of Michigan provides a quarterly measure ofconsumer opinions about products and services (The Wall Street Journal, February 18, 2013). A survey of40 restaurants in the Fast Food/ Pizza group showed a sample mean customer satisfaction index of 71.
Past data indicate that the population standard deviation of the index has been relatively stable with σ=5.
a. Using 95% confidence, determine the margin of error. b. Determine the margin of error if 99% confidence is desired.
Guideline:
Follow 1 and 2 questions guideline
4. The undergraduate GPA for students admitted to the top graduate business schools is 3.37. Assume thisestimate is based on a sample of 120 students admitted to the top schools. Using past years' data, the population standard deviation can be assumed known with .28. What is the 95% confidence interval
estimate of the mean undergraduate GPA for students admitted to the top graduate business schools?Guideline:
Follow 1 and 2 questions guideline
HW: Text,
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When sample size is less than 30 i.e. n<30, the mean has a Student's t distribution. The Student's tdistribution was created by William S. Gosset, an Irish worker. He wouldn't allow him to publish hiswork under his name, so he used the pseudonym "Student".
The Student's t distribution is very similar to the standard normal distribution.
• It is symmetric about its mean• As the sample size increases, the t distribution approaches the normal distribution.• It is bell shaped.• The t-scores can be negative or positive, but the probabilities are always positive.
(1-α)100% confidence interval for :
α
,
√ α
,
√
Problem
Suppose we have given sample heights of 20 IUB students, where = 67.3", SD = 3.6" and the
distribution is symmetric. Develop 95% confidence interval for μ and make a summary based on yourfindings.
Solution:
We have given X~N(67.3,3.6), where X - heights of 20 randomly selected IUB students and n=20.
95% confidence interval for :
α , √ α , √
Here =67.3, =3.6, n=25, α=1-0.95 = 0.05, α/2 = 0.025 and α ,=.=2.093
Thus, 67.3 2.093 3.620 67.3 2.093 3.6√ 20
Summary: Based on our findings, we are 95% confident that population mean is ranging 65.61 to 68.98.
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6, 4, 6, 8, 7, 7, 6, 3, 3, 8, 10, 4, 8, 7, 8, 7, 5, 9, 5, 8, 4, 3, 8, 5,5Develop a 95% confidence interval estimate of the population mean rating for Miami.
2. Text book, Ex.15-17, p.324
3. Have a machine filling bags of popcorn; weight of bags known to be normally distributed with meanweight 10.5 oz and SD 0.8 oz. Take sample of 10 bags, what’s a 90% confidence interval for population
mean μ?
Confidence interval for variance and standard deviation
We have learned that estimates of population means can be made from sample means, and confidenceintervals can be constructed to better describe those estimates. Similarly, we can estimate a populationstandard deviation from a sample standard deviation, and when the original population is normallydistributed, we can construct confidence intervals of the standard deviation as well
Variances and standard deviations are a very different type of measure than an average, so we can expectsome major differences in the way estimates are made.
We know that the population variance formula, when used on a sample, does not give an unbiasedestimate of the population variance. In fact, it tends to underestimate the actual population variance. Forthat reason, there are two formulas for variance, one for a population and one for a sample. The samplevariance formula is an unbiased estimator of the population variance.
Also, both variance and standard deviation are nonnegative numbers. Since neither can take on a negativevalue, thus the normal distribution cannot be the distribution of a variance or a standard deviation. It can
be shown that if the original population of data is normally distributed, then the expression
has a
chi-square distribution with n−1 degrees of freedom.
The chi-square distribution of the quantity allows us to construct confidence intervals for the
variance and the standard deviation (when the original population of data is normally distributed).
(1-α)100% confidence interval for 2:
/
where the χ
/ values are based on a chi-square distribution with n-1 degress of freedom and 1-α is the
confidence coefficient (Details see, Text, p.440)
(1-α)100% confidence interval for :
/
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A statistician chooses 27 randomly selected dates and when examining the occupancy records of a particular motel for those dates, finds a standard deviation of 5.86 rooms rented. If the number of roomsrented is normally distributed, find the 95% confidence interval for the population standard deviation ofthe number of rooms rented.
Solution:
Here X - Number of rooms rented, S = 5.86 and n=27
95% confidence interval for the population standard deviation (σ):
/
Here
.. ..
. .
Summary: Based on our findings, we are 95% confident that population standard deviation is ranging
4.615 to 8.031.
Problem-2
A statistician chooses 27 randomly selected dates and when examining the occupancy records of a
particular motel for those dates, finds a standard deviation of 5.86 rooms rented. If the number of roomsrented is normally distributed, find the 95% confidence interval for the population variance of the numberof rooms rented.
Solution:
Here X - Number of rooms rented, S = 5.86 and n=27
95% confidence interval for the population variance (σ2):
/
Here .. . .
21.297 64.492
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Summary: Based on our findings, we are 95% confident that population variance is ranging 21.297 to
64.492
Practice problems
1. The variance in drug weights is critical in the pharmaceutical industry. For a specific drug,with weights measured in grams, a sample of 18 units provided a sample variance of s2=0.36.
a. Construct a 90% confidence interval estimate of the population variance for the weight of thisdrug. b. Construct a 90% confidence interval estimate of the population standard deviation.
2. The daily car rental rates for a sample of eight cities follow.
City Daily Car Rental Rate ($)
Atlanta 69
Chicago 72Dallas 75
New Orleans 67
Phoenix 62
Pittsburgh 65
San Francisco 61
Seattle 59
a. Compute the sample variance and the sample standard deviation for these data.
b. What is the 95% confidence interval estimate of the variance of car rental rates for the population?
c. What is the 90% confidence interval estimate of the standard deviation for the population?
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In general, we do not know the true value of population parameters (mean, proportion, variance,SD and others). They must be estimated based on random samples. However, we do have
hypotheses about what the true values are.
The major purpose of hypothesis testing is to choose between two competing hypotheses aboutthe value of a population parameter.
Actually, in hypothesis testing we begin by making a tentative assumption about a population
parameter. This tentative assumption is called the null hypothesis and is denoted by H0.
It is needed then to define another hypothesis, called the alternative hypothesis, which is the
opposite in H0. It is denoted by Ha or H1.
Both the null and alternative hypothesis should be stated before any statistical test of significanceis conducted.
In general, it is most convenient to always have the null hypothesis contain an equal sign, e.g.
(1) H0: μ = 100
H1: μ ≠ 100
(2) H0: μ ≥ 100H1: μ < 100
(3) H0: μ ≤ 100H1: μ > 100
Thus, note that
under H0, signs are =, ≤ and ≥
under H1, signs are ≠, < and >
In general, a hypothesis tests about the values of the population mean μ take one of the following
three forms:
H0: μ = μ0 H0: μ ≥ μ0 H0: μ ≤ μ0
H1: μ ≠ μ0 H1: μ < μ0 H1: μ > μ0
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For example, consider the following problems in choosing the proper form for a hypothesis test:
Problem 1
The manager of an automobile dealership is considering a new bonus plan designed to increase
sales volume. Currently, the mean sales volume is 14 automobiles per month. The managerwants to conduct a research study to see whether the new bonus plan increases sales volume. To
collect data on the plan, a sample of sales personnel will be allowed to sell under the new bonus
plan for a 1-month period. Define the null and the alternative hypotheses.
Solution: Here H0: μ ≤ 14 and H1: μ > 14.
Problem 2
The manager of an automobile dealership is considering a new bonus plan designed to increase
sales volume. Currently, the mean sales volume is 14 automobiles per month. The manager
wants to conduct a research study to see whether the new bonus plan decreases sales volume. Tocollect data on the plan, a sample of sales personnel will be allowed to sell under the new bonus
plan for a 1-month period. Define the null and the alternative hypotheses.
Solution: Here H0: μ ≥ 14 and H1: μ < 14.
Problem 3
The manager of an automobile dealership is considering a new bonus plan designed to increase
sales volume. Currently, the mean sales volume is 14 automobiles per month. The manager
wants to conduct a research study to see whether the new bonus plan changes sales volume. To
collect data on the plan, a sample of sales personnel will be allowed to sell under the new bonus
plan for a 1-month period. Define the null and the alternative hypotheses.
Solution: Here H0: μ = 14 and H1: μ ≠ 14.
Steps for conducting a of hypothesis test
1. Develop H0 and H1.
2. Specify the level of significance, α, which defines unlikely values of sample statistic if the
null hypothesis is true. It is selected by the researcher at start. The common values of α are 0.01,
0.05 and 0.10 and is most common 0.05.
3. Select the test statistic (a quantity calculated using the sample values that is used to perform
the hypothesis test) that will be used to test the hypothesis.
Guidelines to select test statistic:
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1) Scatter Diagram – To guess relationship between two variables2) Correlation coefficient (r xy) will indicate us percent of relation exists between two
variables.
Let’s consider the following problem to understand it very clearly!
Problem
Consider two variables
x (No. of TV commercials): 2,5,1,3,4,1,5,3,4,2
y(Total sales): 50,57,41,54,54,38,63,48,59,46
Find the relationship between two variables and make a summary based on your findings.
Solution:
Denote x - No. of TV commercials and y- Total sales because it is believable that sales dependson No. of commercials
Make a shape of Scatter diagram to see what sorts of relation exist between and x and y.
Summary: We see that there is a positive relation exists between no. of TV commercials and totalsales.
To understand very clearly what percent relation exist between x and y, we will apply the followingformula (known as correlation coefficient) is defined as
0
10
20
30
40
50
60
70
0 1 2 3 4 5 6
Total Sales
No. of TV Commercials
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Thus, estimated model y on x becomes: yi= 36.15+4.95xi
Summary
=36.15 means that if there are no commercials (i.e. x=0), then expected sales may be 36.15$.=4.95 means that when no. of TV commercials increases there is a chance that total sales may beincreased.
(ii) We know that the predicted model is: y p = +x p, i = 1,2,….
According to question, we have to predict total sales, when x=5.
Thus y p =36.15+(4.95x5)=60.9$.
So, we can expect when there are 5 commercials in a week, company can expect total sales 60.9$.