Chapter 10

Section 2Z Test for Mean

1

z Test for a Mean

The z test is a statistical test for the mean of a population. It can be used when n > 30, or when the population is normally distributed and is known.

The formula for the z test is

where

= sample mean

μ = hypothesized population mean

= population standard deviation

n = sample size2

X

zn

X

Days on Dealers LotsA researcher wishes to see if the mean number

of days that a basic, low-price, small automobile sits on a dealer’s lot is 29. A sample of 30 automobile dealers has a man of 30.1 days for basic, low-price, small automobiles. At a = 0.05m teat the claim that the mean time is greater than 29 days. The standard deviation of the population is 3.8 days.

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Cost of RehabilitationThe Medical Rehabilitation Education Foundation reports that the average cost of rehabilitation for stroke victims is $24,672. To see if the average cost of rehabilitation is different at a particular hospital, a researcher selects a random sample of 35 stroke victims at the hospital and finds that the average cost of their rehabilitation is $25,226. The standard deviation of the population is $3251. At α = 0.01, can it be concluded that the average cost of stroke rehabilitation at a particular hospital is different from $24,672?

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Cost of Rehabilitation“reports that the average cost of rehabilitation for stroke victims is $24,672”

“a random sample of 35 stroke victims”

“the average cost of their rehabilitation is $25,226”

“the standard deviation of the population is $3251”

“α = 0.01”

Step 3: Find the test value.

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Hypothesis TestingThe PP-value-value (or probability value) is the probability of getting a sample statistic (such as the mean) or a more extreme sample statistic in the direction of the alternative hypothesis when the null hypothesis is true.

6

Test Value

P-Value

Cost of Men’s ShoesA researcher claims that the average cost of men’s athletic shoes is less than $80. He selects a random sample of 36 pairs of shoes from a catalog and finds the following costs (in dollars). (The costs have been rounded to the nearest dollar.) Is there enough evidence to support the researcher’s claim at α = 0.10? Assume = 19.2.

60 70 75 55 80 55 50 40 80 70 50 95

120 90 75 85 80 60 110 65 80 85 85 45

75 60 90 90 60 95 110 85 45 90 70 70

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Hypothesis Testing In this section, the traditional method for

solving hypothesis-testing problems compares z-values:

critical value test value

The P-value method for solving hypothesis-testing problems compares areas:

alpha P-value

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Procedure Table

Solving Hypothesis-Testing Problems (P-Value Method)

Step 1 State the hypotheses and identify the claim.

Step 2 Compute the test value.

Step 3 Find the P-value.

Step 4 Make the decision.

Step 5 Summarize the results.

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Cost of College TuitionA researcher wishes to test the claim that the average cost of tuition and fees at a four-year public college is greater than $5700. She selects a random sample of 36 four-year public colleges and finds the mean to be $5950. The population standard deviation is $659. Is there evidence to support the claim at a 0.05? Use the P-value method.

10

Wind SpeedA researcher claims that the average wind speed in a certain city is 8 miles per hour. A sample of 32 days has an average wind speed of 8.2 miles per hour. The standard deviation of the population is 0.6 mile per hour. At α = 0.05, is there enough evidence to reject the claim? Use the P-value method.

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Guidelines for P-Values With No α If P-value 0.01, reject the null hypothesis. The

difference is highly significant.

If P-value > 0.01 but P-value 0.05, reject the null hypothesis. The difference is significant.

If P-value > 0.05 but P-value 0.10, consider the consequences of type I error before rejecting the null hypothesis.

If P-value > 0.10, do not reject the null hypothesis. The difference is not significant.

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Significance The researcher should distinguish between

statistical significance statistical significance and practical practical significancesignificance.

When the null hypothesis is rejected at a specific significance level, it can be concluded that the difference is probably not due to chance and thus is statistically significant. However, the results may not have any practical significance.

It is up to the researcher to use common sense when interpreting the results of a statistical test.

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A report in USA TODAY stated that the average age of commercial jets in the United States is 14 years. An executive of a large airline company selects a sample of 36 planes and finds the averageage of the planes is 11.8 years. The standard deviation of the sample is 2.7 years. At = 0.01, can it be concluded that the average age of the planes in his Company is less than the national average?

Example 5

The average one-year-old (both sexes) is 29 inches tall. A random sample of 30 one-year-olds in a large day care franchise resulted in the following heights. At = 0.05, can it be concluded that the average height differs from 29 inches?

25 32 35 25 30 26.5 26 25.5 29.5 32

30 28.5 30 32 28 31.5 29 29.5 30 34

29 32 27 28 33 28 27 32 29 29.5

Example 6

To see if young men ages 8 through 17 years spend more or less than the national average of $24.44 per shopping trip to a local mall, the managersurveyed 33 young men and found the average amount spent per visit was $22.97.The standard deviation of the sample was $3.70. At = 0.02, can it be concluded that the average amount spent at a local mall is not equal to the national average of $24.44.

Example 7

A study found that the average stopping distance of a school bus traveling 50 miles per hour was 264 feet (Snapshot, USA TODAY, March12, 1992). A group of automotive engineers decided to conduct a study of itsschool buses and found that for 20 buses, the average stopping distance of buses traveling 50 miles per hour was 262.3 feet.The standard deviation of the populationwas 3 feet. Test the claim that the average stopping distance of the company’s buses is actually less than 264 feet. Find the P-value. On the basis of the P-value, should the null hypothesis be rejected at = 0.01? Assume that the variable isnormally distributed.

Example 8