Section 9.2 ~ Hypothesis Tests for Population Means Introduction to Probability and Statistics Ms. Young.

Section 9.2 ~ Hypothesis Tests for Population

MeansIntroduction to Probability and Statistics

Ms. Young

Objective

Sec. 9.2

After this section you will understand and interpret one- and two-tailed hypothesis tests for claims made about population means, and learn to recognize and avoid common errors (type I and type II errors) in hypothesis tests.

Background Info Recall that there are two possible outcomes of a

hypothesis test; to either reject or not reject the null hypothesis To determine whether to reject or not, a level of significance

(0.05 or 0.01) needs to be found To find the level of significance, a P-value needs to be calculated

To calculate a P-value, you must first understand the concepts of a normal distribution (introduced in ch.5): Recall that if a distribution is normal, you can use z-scores

along with a z-score table to find probabilities of certain values occurring

Also recall that a distribution begins to take the shape of a normal distribution when the sample size is at least 30 and becomes more and more normal as the sample size increases (Central Limit Theorem)

In essence, a P-value (probability value) is the probability that is found using z-scores and the z-score table

Be sure that you are using the sample standard deviation, , when calculating the z-score since you are comparing a sample (group mean or group proportion) to the entire population

Sec. 9.2

s

One-Tailed Hypothesis Tests As mentioned in section 9.1, hypothesis tests can either be one-

tailed (left or right) or two-tailed The process for conducting a left-tailed test is the same as the process for

conducting a right-tailed test, but a two-tailed test varies slightly

Example 1 ~ Left-Tailed Hypothesis Test: Columbia College advertises that the mean starting salary of its

graduates is $39,000. The Committee for Truth in Advertising suspects that this claim is exaggerated and that the mean starting salary for graduates is actually lower. They decide to conduct a hypothesis test to seek evidence to support this suspicion.

Step 1: State the null and alternative hypotheses

Step 2: Draw a sample and come up with a sample statistic and the standard deviation of that sample:

Suppose that the committee gathered a sample of 100 graduates and the sample mean is

Also suppose that the standard deviation for that sample is $6,150

Sec. 9.2

0 : $39,000H : $39,000aH

$37,000x

Example 1 Cont’d… Step 3: Calculate the P-value (using the normal distribution

and z-scores) and determine the level of significance In order to calculate the P-value, we need to find the z-score using the

Central Limit Theorem since we are dealing with the mean of a group. Since we do not know the population standard deviation, we will use the standard deviation found for the sample as an estimate.

Using the z-score table we find that a z-score of -3.25 correlates with a probability of .0006, or .06%. This is the P-value.

Since this value is less than .05 it is significant at the .05 level, but even better, this value is less than .01 which means that it is significant at the .01 level

Step 4: Decide if you should reject or not reject the null hypothesis

Since the P-value is significant at both levels (.05 and .01), we should reject the null hypothesis of $39,000

What this means is that we have strong evidence to believe that Columbia College exaggerated about the mean starting salary of their graduates being $39,000 and that it is most likely lower.

Sec. 9.2

37,000 39,000

6150100

z

x

zn

3.25z

One-Tailed Hypothesis TestsExample 2 ~ Right-Tailed Hypothesis Test

In the United States, the average car is driven about 12,000 miles each year. The owner of a large rental car company suspects that for his fleet, the mean distance is greater than 12,000 miles each year. He selects a random sample of n = 225 cars from his fleet and finds that the mean annual mileage for this sample is miles. Suppose that the standard deviation for that sample is 2,415 miles. Interpret this claim by conducting a hypothesis test.

Step 1: State the null and alternative hypotheses

Step 2: Draw a sample and come up with a sample statistic and the standard deviation of that sample

This information was already given: The sample mean is The standard deviation for that sample is 2,415 miles

Step 3: Calculate the P-value and determine the level of significance: The z-score is:

Sec. 9.2

0 : 12,000 milesH : 12,000aH

12,375x

12,375x

12,375 12,0002.33

2415225

z

One-Tailed Hypothesis TestsExample 2 Cont’d…

Step 3 cont’d… The z-score was found to be 2.33 which corresponds to a probability

of .9901 on the z-score table, but that represents the area below 12,375 and we are interested in knowing the probability of a car being driven more than that value so we subtract .9901 from 1 (1 - .9901) and get a probability of .0099

The P-value is .0099 which is less than .01, meaning that it is significant at the .01 level

Step 4: Decide if you should reject or not reject the null hypothesis Since the P-value is significant at both levels (.05 and .01), we should

reject the null hypothesis of 12,000 miles What this means is that we have strong evidence to believe that the

mean distance traveled for the rental car fleet is greater than 12,000 miles

Sec. 9.2

Critical Values for Statistical Significance Since we can decide to reject the null hypothesis if the P-value

is .05 or lower (or .01 or lower), we can use critical values as a quick guideline to decide if we should reject the null hypothesis or not

Critical values for .05 significance level: For a left-tailed test, the z-score that corresponds to a probability of .05

is -1.645, so any z-score that is less than or equal to -1.645 will be statistically significant at the .05 level

For a right-tailed test, the z-score that corresponds to a probability of .05 (which we would look for .95 on the chart) is 1.645, so any z-score greater than or equal to 1.645 will be statistically significant at the .05 level

Critical values for the .01 significance level: For a left-tailed test, the z-score that corresponds to a probability of .01

is -2.33, so any z-score that is less than or equal to -2.33 will be statistically significant at the .01 level

For a right-tailed test, the z-score that corresponds to a probability of .01 (which we would look for .99 on the chart) is 2.33, so any z-score greater than or equal to 2.33 will be statistically significant at the .01 level

Sec. 9.2

Critical Values for Statistical Significance

Sec. 9.2

Two-Tailed Hypothesis Tests The process for conducting a two-tailed hypothesis test is very

similar to the one-tailed tests, except the critical values are slightly different

Since a two tailed test tests both above and below the claimed value, a .05 significance level would have to be split between the two extremes thus looking for a z-score that corresponds to a probability of .025

The z-scores that correspond to a probability of .025 are -1.96 and 1.96, so for a two-tailed test, it is significant at the .05 level if the z-score is less than or equal to -1.96 or greater than or equal to 1.96

Sec. 9.2

Two-Tailed Hypothesis Tests For a two-tailed test, a .01 significance level would mean

that the z-score needs to correspond to a probability of .005 (.01 split in half) The z-scores that correspond to a probability of .005 are -2.575

and 2.575, so if the z-score is less than or equal to -2.575 or greater than or equal to 2.575, then it is statistically significant at the .01 level

Summary of critical values for two-tailed tests: .05 significance level:

.01 significance level:

Sec. 9.2

1.96 or 1.96z z

2.575 or 2.575z z

Two-Tailed Hypothesis TestsExample 3 ~ Two-Tailed Hypothesis Test:

Consider the study in which University of Maryland researchers measured body temperatures in a sample of n = 106 healthy adults, finding a sample mean body temperature of with a sample standard deviation of 0.62°F. We will assume that the population standard deviation is the same as the standard deviation found from the sample. Determine whether this sample provides evidence for rejecting the common belief that the mean human body temperature is

Step 1: State the null and alternative hypotheses

Step 2: Draw a sample and come up with a sample statistic and the standard deviation of that sample

This information was already given: The sample mean is The standard deviation for that sample is 0.62°F

Sec. 9.2

98.20 Fx °

98.60 F °

0 : 98.6 FH ° : 98.6 FaH °

98.20 Fx °

Two-Tailed Hypothesis TestsExample 3 Cont’d…

Step 3: Calculate the P-value and determine the level of significance

To calculate the P-value for a two tailed test, you must find the z-score like you would with a one-tailed test, but the probability that corresponds to it must then be multiplied by 2

The z-score is:

The P-value is less than .02 (.01 * 2), and since the z-score of -6.64 is significantly lower than -1.96 and -2.575, this would be statistically significant at both levels

Step 4: Decide if you should reject or not reject the null hypothesis The null hypothesis should be rejected which provides strong evidence

that the mean human body temperature is not 98.6°F. It may be either higher or lower.

Sec. 9.2

98.2 98.66.64

0.62106

z

Common Errors in Hypothesis Testing A hypothesis test guides us in the right direction of either

supporting or not supporting a claim, but even if a hypothesis test is carried out correctly, two common types of errors may affect the conclusions Type I error ~ when the null hypothesis is wrongly rejected

Ex. ~ Consider a drug company that seeks to be sure that its “500-milligram” aspirin tablets really contain 500 milligrams. If the tablets contain less than 500 milligrams, the consumers are not getting the advertised dose. If the tablets contain more than 500 milligrams, the consumers are getting more than the advertised dose (which could be potentially dangerous).

Suppose that a hypothesis test gave evidence to reject the null hypothesis (thus stating that the drug does not contain 500 milligrams, when in reality it does contain 500 milligrams

This would be an example of a type I error; it is appearing as though the drug has either more or less aspirin than it should, when in reality it has the correct amount

Type II error ~ when the null hypothesis should be rejected, but it wasn’t

Consider the drug scenario above. Suppose that a hypothesis test gave evidence to not reject the null

hypothesis, stating that the drug does contain 500 milligrams, when in reality it does not contain 500 milligrams

This would be an example of a type II error; it is appearing as though the drug has the correct amount of aspirin when in reality it doesn’t

Sec. 9.2

Common Errors in Hypothesis Testing Cont’d…

Example 4 ~ Type I and Type II Errors The success of precious metal mines depends on the purity (or

grade) of ore removed and the market price for the metal. Suppose the purity of gold ore must be at least 0.5 ounce of gold per ton of ore in order to keep the mine open. Samples of gold ore are used to estimate the purity of the ore for the entire mine. Discuss the impact of type I and type II errors on two of the possible alternative hypotheses:

For the first alternative hypothesis Wrongly rejecting the null hypothesis (type I error) would result in closing

the mine, when in reality they were up to standard This would be a terrible loss for the mine and all workers

Wrongly “accepting” the null hypothesis (type II error) would result in the mine staying open when they should really be closed

This would impact the consumers in that they are not getting the quality in which they believe they are getting

Sec. 9.2

: purity < 0.5 ounce per tonaH

: purity > 0.5 ounce per tonaH

: purity < 0.5 ounce per tonaH

Common Errors in Hypothesis Testing Cont’d…

Example 4 ~ Type I and Type II Errors The success of precious metal mines depends on the purity (or

grade) of ore removed and the market price for the metal. Suppose the purity of gold ore must be at least 0.5 ounce of gold per ton of ore in order to keep the mine open. Samples of gold ore are used to estimate the purity of the ore for the entire mine. Discuss the impact of type I and type II errors on two of the possible alternative hypotheses:

For the second alternative hypothesis Wrongly rejecting the null hypothesis (type I error) would result in the

mine staying open when it shouldn’t Wrongly accepting the null hypothesis (type II error) would result in the

mine closing when it should be able to stay open

Sec. 9.2

: purity < 0.5 ounce per tonaH

: purity > 0.5 ounce per tonaH

: purity > 0.5 ounce per tonaH