ࡱ> SURkq` ubjbjqPqP .:: #hXhXhXhXYd!^@YZZZZo[o[o[y]{]{]{]{]{]{]$_hVb]o[o[o[o[o[]ZZh^\\\o[FZZy]\o[y]\\\ZY p`k$_hX[\y]~^0^\bs\db\b\o[o[\o[o[o[o[o[]]\ o[o[o[^o[o[o[o[d!d!d!7hXd!d!d!hX$$ STAT 515 -- Chapter 8: Hypothesis Tests CIs are possibly the most useful forms of inference because they give a range of reasonable values for a parameter. But sometimes we want to know whether one particular value for a parameter is reasonable. In this case, a popular form of inference is the hypothesis test. We use data to test a claim (about a parameter) called the null hypothesis. Example 1: We claim the proportion of USC students who travel home for Christmas is 0.95. Example 2: We claim the mean nightly hotel price for hotels in SC is no more than $65. Null hypothesis (denoted H0) often represents status quo, previous belief or no effect. Alternative hypothesis (denoted Ha) is usually what we seek evidence for. We will reject H0 and conclude Ha if the data provide convincing evidence that Ha is true. Evidence in the data is measured by a test statistic. A test statistic measures how far away the corresponding sample statistic is from the parameter value(s) specified by H0. If the sample statistic is extremely far from the value(s) in H0, we say the test statistic falls in the rejection region and we reject H0 in favor of Ha. Example 2: We assumed the mean nightly hotel price in SC is no more than $65, but we seek evidence that the mean price is actually greater than $65. We randomly sample 64 hotels and calculate the sample mean price  EMBED Equation.3 . Let  EMBED Equation.3  be our test statistic here. Note: If this Z value is much bigger than zero, then we have evidence against H0: m d" 65 and in favor of Ha: m > 65. Suppose we ll reject H0 if Z > 1.645. If m really is 65, then Z has a standard normal distribution. (Why?) Picture: If we reject H0 whenever Z > 1.645, what is the probability we reject H0 when H0 really is true? P(Z > 1.645 | m = 65) = This is the probability of making a Type I error (rejecting H0 when it is actually true). P(Type I error) =  level of significance of the test (denoted a). We don t want to make a Type I error very often, so we choose a to be small: The a we choose will determine our rejection region (determines how strong the sample evidence must be to reject H0). In the previous example, if we choose a = .05, then Z > 1.645 is our rejection region. Hypothesis Tests of the Population Mean In practice, we don t know s, so we don t use the Z-statistic for our tests about m. Use the t-statistic:  EMBED Equation.3 , where m0 is the value in the null hypothesis. This has a t-distribution (with n  1 d.f.) if H0 is true (if m really equals m0). Example 2: Hotel prices: H0: m = 65 Ha: m > 65 Sample 64 hotels, get EQ \o (X,EQ \s\up18(_)) = $67 and s = $10. Let s set a = .05. Rejection region: Reject H0 if t is bigger than 1.67. Conclusion: We never accept H0; we simply fail to reject H0. This example is a one-tailed test, since the rejection region was in one tail of the t-distribution. Only very large values of t provided evidence against H0 and for Ha. Suppose we had sought evidence that the mean price was less than $72. The hypotheses would have been: H0: m = 72 Ha: m < 72 Now very small values of  EMBED Equation.3  would be evidence against H0 and for Ha. Rejection region would be in left tail: Rules for one-tailed tests about population mean H0: m = m0 H0: m = m0 Ha: m < m0 or Ha: m > m0 Test statistic:  EMBED Equation.3  Rejection t < -ta t > ta Region: (where ta is based on n  1 d.f.) Rules for two-tailed tests about population mean H0: m = m0 Ha: m `" m0 Test statistic:  EMBED Equation.3  Rejection t < -ta/2 or t > ta/2 (both tails) Region: (where ta/2 is based on n  1 d.f.) Example: We want to test (using a = .05) whether or not the true mean height of male USC students is 70 inches. Sample 26 male USC students. Sample data: EQ \o (X,EQ \s\up18(_)) = 68.5 inches, s = 3.3 inches. Assumptions of t-test (and CI) about m " We assume the data come from a population that is approximately normal. " If this is not true, our conclusions from the hypothesis test may not be accurate (and our true level of confidence for the CI may not be what we specify). How to check this assumption? The t-procedures are robust: If the data are close to normal, the t-test and t CIs will be quite reliable. Hypothesis Tests about a Population Proportion We often wish to test whether a population proportion p equals a specified value. Example 1: We suspect a theater is letting underage viewers into R-rated movies. Question: Is the proportion of R-rated movie viewers at this theater greater than 0.25? We test: Recall: The sample proportion  EMBED Equation.3  is approximately N EMBED Equation.3  for large n, so our test statistic for testing H0: p = p0 has a standard normal distribution when H0 is true (when p really is p0). Rules for one-tailed tests about population proportion H0: p = p0 H0: p = p0 Ha: p < p0 or Ha: p > p0 Test statistic:  EMBED Equation.3  Rejection z < -za z > za Region: Rules for two-tailed tests about population proportion H0: p = p0 Ha: p `" p0 Test statistic:  EMBED Equation.3  Rejection z < -za/2 or z > za/2 (both tails) Region: Assumptions of test (need large sample): Need: Example 1: Test H0: p = 0.25 vs. Ha: p > 0.25 using a = .01. We randomly select 60 viewers of R-rated movies, and 23 of those are underage. Example 1(a): What if we had wanted to test whether the proportion of underage viewers was different from 0.25? P-values Recall that the significance level a is the desired P(Type I error) that we specify before the test. 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The P-value gives us an indication of the strength of evidence against H0 (and for Ha) in the sample. This is a different (yet equivalent) way to decide whether to reject the null hypothesis: " A small p-value (less than a) = strong evidence against the null => Reject H0 " A large p-value (greater than a) = little evidence against the null => Fail to reject H0 How do we calculate the P-value? It depends on the alternative hypothesis. One-tailed tests Alternative P-value Ha:  <  Area to the left of the test statistic value in the appropriate distribution (t or z). Ha:  >  Area to the right of the test statistic value in the appropriate distribution (t or z). Two-tailed test Alternative P-value Ha:  `"  2 times the  tail area outside the test statistic value in the appropriate distribution (t or z). Double the tail area to get the P-value! P-values for Previous Examples Hotel Price Example: H0: m = 65 vs. Ha: m > 65 Test statistic value: Student height example: H0: m = 70 vs. Ha: m `" 70 Test statistic value: Movie theater example: H0: p = 0.25 vs. Ha: p > 0.25 Test statistic value: What if we had done a two-tailed test of H0: p = 0.25 vs. Ha: p `" 0.25 at a = .01? Relationship between a CI and a (two-sided) hypothesis test: " A test of H0: m = m* vs. Ha: m `" m* will reject H0 if and only if a corresponding CI for m does not contain the number m*. Example: A 95% CI for m is (2.7, 5.5). (1) At a = 0.05, would we reject H0: m = 3 in favor of Ha: m `" 3? (2) At a = 0.05, would we reject H0: m = 2 in favor of Ha: m `" 2? (3) At a = 0.10, would we reject H0: m = 2 in favor of Ha: m `" 2? (4) At a = 0.01, would we reject H0: m = 3 in favor of Ha: m `" 3? Power of a Hypothesis Test " Recall the significance level a is our desired P(Type I error) = P(Reject H0 | H0 true) The other type of error in hypothesis testing: Type II error = P(Type II error) = b The power of a test is " High power is desirable, but we have little control over it (different from a) Calculating Power: The power of a test about m depends on several things: a, n, s, and the true m. Example 1: Suppose we test whether the true mean nicotine contents in a population of cigarettes is greater than 1.5 mg, using a = 0.01. H0: Ha: We take a random sample of 36 cigarettes. Suppose we know s = 0.20 mg. Our test statistic is We reject H0 if: " Now, suppose m is actually 1.6 (implying that H0 is false). Let s calculate the power of our test if m = 1.6: This is just a normal probability problem! " What if the true mean were 1.65? Verify: " The farther the true mean is into the  alternative region, the more likely we are to correctly reject H0. Example 2: Testing H0: p = 0.9 vs. Ha: p < 0.9 at a = 0.01 using a sample of size 225. Suppose the true p is 0.8. 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