Lecture 19 Dustin Lueker

 The p-value for testing H 1 : µ≠100 is p=.001. This indicates that… 1.There is strong evidence that μ=100 2.There is strong evidence that μ≠100 3.There is strong evidence that μ>100 4.There is strong evidence that μ<100 2STA 291 Summer 2008 Lecture 19

 The p-value for testing H 1 : µ≠100 is p=.001. In addition you know that the test statistic was z=3.29. This indicates that… 1.There is strong evidence that μ=100 2.There is strong evidence that μ>100 3.There is strong evidence that μ<100 STA 291 Summer 2008 Lecture 193

 Range of values such that if the test statistic falls into that range, we decide to reject the null hypothesis in favor of the alternative hypothesis ◦ Type of test determines which tail(s) the rejection region is in  Left-tailed  Right-tailed  Two-tailed 4STA 291 Summer 2008 Lecture 19

 Testing µ with n large ◦ Just like finding a confidence interval for µ with n large  Reasons for choosing test statistics are the same as choosing the correct confidence interval formula 5STA 291 Summer 2008 Lecture 19

 Testing µ with n small ◦ Just like finding a confidence interval for µ with n small  Reasons for choosing test statistics are the same as choosing the correct confidence interval formula  Note: It is difficult for us to find p-values for this test statistic because of the way our table is set up 6STA 291 Summer 2008 Lecture 19

 An assumption for the t-test is that the population distribution is normal ◦ In practice, it is impossible to be 100% sure if the population distribution is normal  It may be useful to look at histogram or stem-and-leaf plot (or normal probability plot) to check whether the normality assumption is reasonable  Good news ◦ t-test is relatively robust against violations of this assumption  Unless the population distribution is highly skewed, the hypotheses tests and confidence intervals are valid  However, the random sampling assumption must never be violated, otherwise the test results are completely invalid 7STA 291 Summer 2008 Lecture 19

 A courier service advertises that its average delivery time is less than 6 hours for local deliveries. A random sample of times for 12 deliveries found a mean of and a standard deviation of Is this sufficient evidence to support the courier’s advertisement at α=.05?  State and test the hypotheses using the rejection region method. ◦ Why wouldn’t the p-value method be good to use? STA 291 Summer 2008 Lecture 198

 Results of confidence intervals and of two- sided significance tests are consistent ◦ Whenever the hypothesized mean is not in the confidence interval around the sample mean, then the p-value for testing H 0 : μ=μ 0 is smaller than 5% (significance at the 5% level)  Why does this make sense? ◦ In general, a 100(1-α)% confidence interval corresponds to a test at significance level α 9STA 291 Summer 2008 Lecture 19

 A survey of 35 cars that just left their metered parking spaces produced a mean of 18 and a standard deviation of 22. Test the parking control officer’s claim that the average time left on meters is equal to 15 minutes.  State and test the hypotheses with a level of significance of 5% using the confidence interval method. STA 291 Summer 2008 Lecture 1910