7.2 Hypothesis Testing for the Mean (Large Samples)

1. 7.2 Hypothesis Testing for the Mean (Large Samples) Find P-value and use them to test a mean μ Use P-values for a z-test Find critical values and rejection regions Use rejection regions for a z-test

2. To use a P-value to make a conclusion in a hypothesis test, compare the P-value with α. • If P ≤ α, then reject the null hypothesis. • If P > α, then fail to reject the null hypothesis. Decision Rule based on P-value

3. The P-value for a hypothesis test is P = 0.0347. What is your decision if the level of significance is α = 0.01 α = 0.05 P = 0.0347 > 0.01 = α Fail to reject the null hypothesis. P = 0.0347 < 0.05 = α Reject the null hypothesis. Try it yourself 1 Interpreting a P-value

4. After determining the hypothesis test’s standardized test statistic and the test statistic’s corresponding area, do one of the following to find the P-value. • For a left-tailed test, P = (Area in left tail) • For a right-tailed test, P = (Area in right tail) • For a two-tailed test, P = 2(Area in tail of test statistic) Finding the P-value for a hypothesis test

5. Find the P-value for a left-tailed hypothesis test with a test statistic of z = -1.71. Decide whether to reject the null hypothesis if the level of significance is α = 0.05. P = 0.0436 Reject the null hypothesis. Try it yourself 2 Finding a P-value for a Left-Tailed Test

6. Find the P-value for a two-tailed hypothesis test with a test statistic of z = 1.64. Decide whether to reject the null hypothesis if the level of significance is α = 0.10. P = 2(Area) = 2(0.0505) = 0.1010 Fail to reject the null hypothesis. Try it yourself 3 Finding a P-value for a Two-Tailed Test

7. The z-test for a mean is a statistical test for a population mean. The z-test can be used when the population is normal and σ is known, or for any population when the sample size n is at least 30. The test statistic is the sample mean and the standardized test statistic is z-Test for a mean μ

8. Homeowners claim that the mean speed of automobiles traveling on their street is greater than the speed limit of 35 miles per hour. A random sample of 100 automobiles has a mean speed of 36 miles per hour and a standard deviation of 4 miles per hour. Is there enough evidence to support the claim at α = 0.05? Use a P-value. Reject the null hypothesis. z = 2.5 P = 0.0062 There is enough evidence at the 5% level of significance to support the claim that the average speed is greater than 35 miles per hour. Try it yourself 4 Hypothesis Testing Using P-values

9. One of your distributors reports an average of 150 sales per day. You suspect that this average is not accurate, so you randomly select 35 days and determine the number of sales each day. The sample mean is 143 daily sales with a standard deviation of 15 sales. At α = 0.01, is there enough evidence to doubt the distributor’s reported average? Use a P-value. z = -2.76 Reject the null hypothesis. P = 2(0.0029) = 0.0059 There is enough evidence at the 1% level of significance to reject the claim that the distributorship averages 150 sales per day. Try it yourself 5 Hypothesis Testing Using P-values

10. For the TI-83/84 Plus hypothesis test shown, make a decision at the α = 0.01 level of significance. P = 0.0440 > 0.01 = α Fail to reject the null hypothesis Try it yourself 6 Using a Technology Tool to Find a P-value

11. A rejection region (or critical region) of the sample distribution is the range of values for which the null hypothesis is rejected. Rejection region (critical region)

12. A critical value separates the rejection region from the nonrejection region. Critical value

13. Find the critical value and rejection region for a left-tailed test with α = 0.10. Critical value: -1.28 Rejection region: z < -1.28 Try it yourself 7 Finding a Critical Value for a Left-Tailed Test

14. Find the critical values and rejection regions for a two-tailed test with α = 0.08. Critical values: -1.75 and 1.75 Rejection regions: z < -1.75, z > 1.75 Try it yourself 8 Finding a Critical Value for a Two-Tailed Test

15. To use a rejection region to conduct a hypothesis test, calculate the standardized test statistic z. If the standardized test statistic • is in the rejection region, then reject the null hypothesis. • is not in the rejection region, then fail to reject the null hypothesis. Decision Rule Based on Rejection Region

16. The CEO of the company claims that the mean work day of the company’s mechanical engineers is less than 8.5 hours. A random sample of 35 of the company’s mechanical engineers has a mean work day of 8.2 hours with a standard deviation of 0.5 hour. At the α = 0.01, test the CEO’s claim. Critical value: -2.33; Rejection region: z < -2.33 z = -3.55 Reject the null hypothesis There is enough evidence at the 1% level of significance to support the claim that the mean work day is less than 8.5 hours. Try it yourself 9 Testing μ with a Large Sample

17. The U.S. Department of Agriculture claims that the mean cost of raising a child from birth to age 2 by husband-wife families in the United States is $13,120. A random sample of 500 children (age 2) has a mean cost of $12,925 with a standard deviation of $1745. At α = 0.01, is there enough evidence to reject the claim? Critical values: ±2.575; Rejection regions: z < -2.575, z > 2.575 z = -2.50 Fail to reject the null hypothesis. There is not enough evidence at the 1% level of significance to reject the claim that the mean cost of raising a child from birth to age 2 by husband-wife families in the United States is $13,120. Try it yourself 10 Testing μ with a Large Sample