Stat 240 Quiz 6 Study Guide

Section 13.3  Know that the test for comparing two means is called the two-sample t-test.  Know how to write the null and alternative hypotheses for comparing two means. As always, notice that hypotheses are about population values, so notation for population means is used.  Know that the test statistic for comparing two means is a t-statistic (but you won’t have to do any computations by hand).  Know how to read Minitab output for the two-sample t-test. (as on pages 453-454)  Suggestion: Look carefully at the two examples given beginning at the bottom of page on to page 455.  Don’t worry about the “pooled” procedure. Section 13.4  Know how to write the null and alternative hypotheses for comparing two proportions. As always, notice that hypotheses are about population values.  Know that the test statistic for comparing two proportions is a z-statistic (but you won’t have to do any computations by hand).  Notice how the p-value is determined in Example 13.8-Continued (See Steps 3,4,5 at bottom of p.462)  Know how to read the Minitab output for comparing two proportions. (as on p.463) Section 13.5  For a two-sided alternative hypothesis about a parameter, If the null value is within the confidence interval, we cannot reject the null hypothesis. If the null value is not within the confidence interval, we can reject the null hypothesis. The confidence level and the level of significance must be complements – for instance, .95 confidence level and .05 level of significance.  When the alternative hypothesis is one-sided, we adjust the confidence level of the two-sided interval. If level of significance = .05, and alternative hypothesis is one-sided, the confidence level for a two- sided interval should be .90. We decide for the alternative hypothesis, when this interval falls entirely in the alternative hypothesis region. Section 13.6 . Know how to decide whether a confidence interval, a test, or both should be used (pgs 467-68). . Know how to pick the right parameter (see chart at top of page 469) Section 13.7 It might help to review the definitions of Type 1 and Type 2 error in Section 11.7  What is the power of a statistical test? How is power related to the risk of making a Type 2 error?  How does sample size affect the power of a statistical test?  How does the level of significance for a test affect power? Section 13.8 . You can more or less skim this section. Highlights that you should know about, though, are: . Effect size = standardized difference between two means, or a standardized difference between a mean and s null value. . Power is related to effect size. Bigger true effect size --.> higher power. Basically, it’s easier to find a big difference than a small difference. . See Table 13.4 on page 479. Power increases with sample size, and also increases with true effect size. Section 13.9 Emphasize Items 2,3, and 4  Item 2 means: statistical significance doesn’t guarantee real world importance  Item 3: Don’t trust findings of “no statistical significance” when the sample size is small.  Item 4. Big sample sizes might lead us to say that very small differences are “significant” . An Example for Section 13.5

The data are amounts spent on textbooks for a semester reported by a sample of n = 75 students. Suppose we wish to test H0:  = 300.

Two-sided alternative hypothesis

The following output is for a test of H0:  = 300 versus Ha:   300. The p-value is 0.084, so we cannot reject the null hypothesis using the  = 0.05 significance level. The p-value is not less than  Notice also that the 95% confidence interval for m includes the value 300, evidence that the null value 300 cannot be rejected.

Test of mu = 300 vs mu not = 300

Variable N Mean StDev SE Mean books 75 277.5 111.2 12.8

Variable 95.0% CI T P books 251.9, 303.1 -1.75 0.084

One-sided alternative hypothesis Suppose now that we use a one-sided alternative hypothesis Ha:  < 300. The output is below. The p- value is 0.042 (1/2 the two-sided p-value) so we can reject the null hypothesis. Notice that the confidence interval given is a one-sided 95% confidence interval. We can be 95% confident the population mean is at most 298.9. Thus, we have evidence the mean is less than 300.

Test of mu = 300 vs mu < 300

Variable N Mean StDev SE Mean books 75 277.5 111.2 12.8

Variable 95.0% Upper Bound T P books 298.9 -1.75 0.042

How Minitab finds the one-sided interval The “95% Upper Bound” is the upper end of a 90% two-sided confidence interval. Below is a 90% confidence interval in this situation. If we move the lower end of the interval down to 0 (smallest possible amount spent), we get a 95% confidence interval that the mean is between 0 and 298.9. In other words, we’re 95% confident that the mean is less than 298.9 (so it’s less than 300).

Variable N Mean StDev SE Mean 90.0% CI books 75 277.5 111.2 12.8 ( 256.1, 298.9)

A Basic Principle for One-Sided Testing (Based on Textbook) If a 90% confidence interval is entirely in the alternative hypothesis region, we can reject the null hypothesis at the .05 significance level. The one-sided alternative hypothesis is that  < 300. The 90% confidence interval 256.1 to 298.9 is in entirely within the region  < 300, so we reject the null hypothesis at the .05 significance level. . Selected Solutions For Section 13.3 13.14 a. Paired. b. Two-sample. c. Paired.

13.16 a. H0: 1   2 = 0, or equivalently 1 =  2

Ha: 1   2  0, or equivalently, 1   2 b. 2.16 c. Don’t worry about this part. d. Do not reject the null hypothesis. There is insufficient evidence to conclude that hardcover and softcover books in the populations represented by these books have different mean number of pages.

13.83 Step 1: H0: 1   2 = 0, or equivalently, 1 =  2

Ha: 1   2 > 0, or equivalently, 1 >  2

1 = mean GPA for population of students who typically sit in the front of a classroom, and

 2 = mean GPA for population of students who typically sit in the back of a classroom Step 2: There are two independent samples, which we assume to represent random samples from larger populations of students. The sample sizes are sufficiently large to proceed. Minitab output for a t-test is. Output for Exercise 13.83 Two-sample T for Front vs Back N Mean StDev SE Mean Front 52 3.260 0.565 0.078 Back 45 2.978 0.498 0.074

Difference = mu Front - mu Back Estimate for difference: 0.282 95% CI for difference: (0.067, 0.496) T-Test of difference = 0 (vs not =): T-Value = 2.61 P-Value = 0.011 DF = 94

Step 2 continued and Steps 3, 4, and 5: The test statistic is t = 2.61(with df = 94) and the p-value is 0.011. Reject the null hypothesis and conclude that in the population of students represented by this sample, mean GPA is higher for students who typically sit in the front. The observed difference in means is

x1  x2  3.26  2.978  0.28 .

For Section 13.4

13.27 a. H0: p1  p2 = 0, or equivalently, p1 = p2 (no difference in proportions)

Ha: p1  p2 < 0, or equivalently, p1 < p2 (proportion lower for men with no anger) b. From the Minitab output, z = 2.79. c. Because the p-value = 0.003 < 0.05, reject the null hypothesis. Conclude that the probability of developing heart disease for men with no anger is lower then it is for men with the most anger.

13.63 a. H0: p1p2 = 0, or equivalently, p1 = p2

Ha: p1p2  0, or equivalently, p1  p2 p1 = proportion of adult Canadians in February 2000 who favor marriages between people of the same sex , and p2 = proportion of adult Canadians in April 1999 who favor marriages between people of the same sex b. Step 2: Sample sizes are sufficiently large so that observed counts in both categories (favored or not) are greater than 10 for both years. Assume that the samples represent random samples from the populations in the two years, and that the samples were independently selected. Test statistic is z =3.19 Steps 3. 4, and 5: p-value = 0.001. Reject the null hypothesis. Conclude that the proportion of adult Canadians favoring marriages between people of the same sex was different in February 2000 than in April 1999. c. The confidence interval does not cover 0, so it is evidence in favor of the alternative hypothesis. For Section 13.5

13.33 a. No. The null value is covered by the 95% confidence interval. b. Yes. The null value is not covered by the 95% confidence interval. c. No. The null value is covered by the 90% confidence interval. d. No. The null value is covered by the 90% confidence interval. e. Yes. The entire 90% confidence interval falls above the null value, so it is likely that  > 25. f. No. We would be able to reject the null hypothesis if the entire 90% confidence interval fell below the null value, but in this case the entire interval falls above the null value.

For Section 13.6

13.42 a. Use a confidence interval to estimate the difference in proportions of college men and women who

would answer yes, p1  p2, and also test whether p1  p2 = 0 or not.

b. Test H0: 1  2 = 0 versus Ha: 1  2 < 0, where 1 and 2 are the mean IQs for children whose mothers smoke at least 10 cigarettes a day during pregnancy and for those whose mothers don't smoke during pregnancy. c. Test H0: p = 31/365 versus Ha: p > 31/365 where p is the probability of being born in October.

d. It is likely that the researchers want to find a confidence interval estimate for d and also test whether it nd th is 0 (where d is the difference in cholesterol on the 2 and 4 days after having a heart attack.)

For Sections 13.7, 13.8, 13.9

13.43 The power of any hypothesis tests done will be greater with the larger sample size.

13.44 The power will be less if the researcher uses the 0.01 level of significance rather than 0.05. The lower level of significance makes it harder to reject the null hypothesis. As a result, there is an increased risk of failing to reject the null in situations where the alternative hypothesis is in fact true.

13.45 a. Type 1. It would happen if the null hypothesis is rejected. b. Type 2. It would happen if the null hypothesis is not rejected. c. Type 1. It would happen if the null hypothesis is true. d. Type 2. It would happen if the alternative hypothesis is true.

13.48 a. Higher. It is easier to detect a larger effect. b. Higher. It is easier to detect an effect with a larger sample size. c. Lower. It is more difficult to reject the null hypothesis with a smaller value for level of significance.

13.49 Items 2, 4 and 5 should be of concern. As stated in item 2, with a large sample size a statistically significant result may not have practical significance. Item 4 is also concerned with this issue, and item 5 helps to resolve it because the magnitude of the effect can be used to determine practical significance.

13.50 Item 3 is the most relevant because it addresses the main problem with interpreting results when the sample size is small.

13.64 The consequence of a type 1 error would be that children might be given xylitol syrup to prevent ear infections, when in fact it is not effective for that purpose. The consequence of a type 2 error would be that xylitol syrup might not be given to children to prevent ear infections, when in fact it would be effective for that purpose. Assuming xylitol has no harmful side effects, a type 2 error would be more serious.