1. Daniel S. Yates The Practice of Statistics Third Edition Chapter 12: Significance Tests in Practice Copyright © 2008 by W. H. Freeman & Company

2. Chapter Objectives Conduct one-sample and paired data t significance tests. Conduct significance tests for population means and population proportions when σ is unknown. Recall that the t distribution requires degrees of freedom.

3. 12.1 – Tests about a Population Mean If you do not know the population standard deviation, you must use the sample standard deviation. In doing so, we must take into account the extra variation that occurs as the result of this. –We will use the t-distribution with n – 1 degrees of freedom. –Our test statistic changes a little.

4. Determining the P-value for t statistic Suppose you carry out a significance test of H 0 : µ = 5 and H a : µ > 5 based on a sample of size n = 20 and obtain t = 1.81. Your degrees of freedom (df) = 19 The t-statistic falls between 1.729 and 2.093 The “upper tail” probability is between 0.05 and 0.025. Since our H a : µ > 5, only positive values of t count as evidence against H 0. We can say that the P-value is between 0.025 and 0.05. Or use calculator: tcdf(1.81, 100, 19) = 0.043

5. Determining the P-value for t statistic Suppose you carry out a significance test of H 0 : µ = 5 and H a : µ ≠ 5 based on a sample of size n = 37 and obtain t = -3.17. Df = 36 so we must use 30 (closest w/out going over) Since this is a two-tailed test, we want the probability of getting a value of t less than -3.17 or greater than 3.17. The “upper tail probability p” falls between 0.0025 and 0.001 The P-value is between 2(0.0025) and 2(0.001) We can say that the P-value is between 0.002 and 0.005. Or use calculator: 2tcdf(-100, -3.17, 36) = 0.0031

6. The One-Sample t Test

7. Example. No-fee credit card offer. A bank wonders whether omitting the annual credit card fee for customers who charge at least $2400 in a year would increase the amount charged on its credit card. The bank makes this offer to an SRS of 200 of its credit card customers. It then compares how much these customers charge this year with the amount they charged last year. The mean increase is $332, and the standard deviation is $108. Is there significant evidence at the 1% level that the mean amount charged increases under the no-fee offer?

8. Step 1: Hypotheses Parameter of interest: µ d, the mean difference in the amt of money charged on credit cards in these two policies. Symbols Words H o : µ d = 0The mean amount charged has not changed with the new policy. H a : µ d > 0The mean amount charged has increased with the new no-fee policy. Choice of Test: paired t test

9. Step 2: Conditions SRS:We are told that the sample came from an SRS Normality:Since the sample size is large (n = 200) and t is very robust against non-Normality, Normal approximations will apply. We would only have to worry about outliers (we’ll assume there are none) (Also the CLT applies) Independence: We must assume that there is at least 10(200) = 2000 credit card holders at this bank.

10. Step 3: Calculations. Formula Value Test Statistic t = 43.47 with df = 199 P-value: P-value = tcdf(43.47, 1000, 199) = 0 Note: 43.47 is way off the chart for df = 100, using the chart.

11. Step 4: Interpretation Since our P-value is less than 0.01, we can reject H o. We can conclude that the mean amount charged increases under the no-fee offer. Construct and interpret a 99% confidence interval for the mean amount of increase. We are 99% confident that the true mean increase of the amount of money spent by credit card holder under the new no-fee policy is between $312.14 and $351.86.

12. A critic points out that the customers would have probably charged more this year than last even without the new offer, because the economy is more prosperous and interest rates are lower. Briefly describe the design of an experiment to study the effect of the no-fee offer that would avoid this criticism. HW: pg. 745 #12.3, 12.4 pg. 761 #12.12, 12.14

13. 12.2 – Tests about a Population Proportion

14. Example. Work stress. According to the National Institute for Occupational Safety and Health, job stress poses a major threat to the health of workers. A national survey of restaurant employees found that 75% said that work stress had a negative impact on their personal lives. A random sample of 100 employees from a large restaurant chain find that 68 answer “yes” when asked, “Does work stress have a negative impact on your personal life?” Is this good reason to think that the proportion of all employees in this chain who would say “yes” differs from the national proportion?

15. Step 1: Hypotheses Parameter of interest: p, the true proportion of this chain’s employees who would say that work stress has a negative impact on their personal lives. Symbols Words H o : p = 0.75The proportion of employees at this chain who say that work stress negatively impacts their lives equals the national average. H a : p ≠ 0.75The proportion of employees at this chain who say that work stress negatively impacts their lives differs from the national average. Choice of Test: one-proportion z-test

16. Step 2: Conditions SRS:We are told that the sample was randomly selected Normality:The expected number of “yes” and “no” responses are (100)(0.75) = 75 and (100)(0.25) = 25 respectively. Since both are at least 10, we can use the z test. Independence: We must assume that there is at least 10(100) = 1000 employees at this “large chain.”

17. Step 3: Calculations. Formula Value Test Statistic z = -1.62 P-value: P-value = 2P(z < -1.62) = 0.1052

18. Step 4: Interpretation Since our P-value is more than.10, we fail to reject H o. There is over a 10% chance of obtaining a sample result as unusual as or even more unusual as we did (0.68) when the null hypothesis is true. We have insufficient evidence that the proportion of this chain restaurant’s employees who suffer from work stress is different from the national survey result, 0.75.

19. Example continued…. If given a computer print out of this information, you will need to decide how many digits are important for your analysis.

20. Example continued…Definition of “success” doesn’t matter 32 of the 100 responded “no”. A large national survey reported that 25% of restaurant workers did not feel that stress exerted a negative impact. We test the hypotheses: –H 0 : p = 0.25 –H a : p ≠ 0.25 The test statistic is We find that the (two-tailed) P-value is 0.1052

21. When we interchanged “yes” and “no”, the only thing that changed was the sign of the test statistic z. The P-value stayed the same. Our conclusion does not depend on an arbitrary choice of success and failure. If our sample size is large enough, then we will have sufficient power to detect very small differences. However, if our sample size is small, we may be unable to detect differences that could be very important. For these reasons, we prefer to include a confidence interval as part of our analysis.

22. Example continued….Using a Confidence Interval Before we construct a confidence interval, we should verify that both n and n(1 - ) are at least 10. Since the number of successes and failures in the sample are 68 and 32, respectively, we can proceed with the calculation. Our 95% confidence interval is We are 95% confident that the true proportion of the restaurant chain’s employees that feel that work stress is damaging to their personal lives is between 59% and 77%.

23. The standard error used for the confidence interval is estimated from the data. The denominator for the test statistic z is based on the value assumed in the null hypothesis. The correspondence between a two-tailed significance test and a confidence interval for a population proportion is no longer exact. The confidence interval (0.59, 0.77) gives an approximate range of p 0 ’s that would not be rejected by a test at the α = 0.05 significance level. We would not be surprised if the true proportion was as low as 60% or as high as 75%. HW: pg. 770 #12.23, 12.24, 12.25 a, b