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Lecture 18 Dustin Lueker The p-value for testing H1: µ≠100 is p=.001. This indicates that… 1. 2. 3. 4. There There There There is is is is strong strong strong strong evidence evidence evidence evidence that that that that μ=100 μ≠100 μ>100 μ<100 STA 291 Summer 2010 Lecture 18 2 The p-value for testing H1: 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 2010 Lecture 18 3 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 STA 291 Summer 2010 Lecture 18 4 Testing µ with n large x 0 z s n ◦ 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 STA 291 Summer 2010 Lecture 18 5 Testing µ with n small x 0 t s n ◦ 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 STA 291 Summer 2010 Lecture 18 6 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 STA 291 Summer 2010 Lecture 18 7 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 5.7 and a standard deviation of 1.58. 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 2010 Lecture 18 8 Results of confidence intervals and of twosided significance tests are consistent ◦ Whenever the hypothesized mean is not in the confidence interval around the sample mean, then the p-value for testing H0: μ=μ0 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 STA 291 Summer 2010 Lecture 18 9 A survey of 35 cars that just left their metered parking spaces produced a mean of 18 minutes remaining on the meter 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 2010 Lecture 18 10