6.52 What’s wrong? Here are several situations where there is an incorrect application of the ideas presented in this section. Write a short paragraph explaining what is wrong in each situation and why it is wrong.

6.53 What’s wrong? Here are several situations where there is an incorrect application of the ideas presented in this section. Write a short paragraph explaining what is wrong in each situation and why it is wrong.

6.54 Determining hypotheses. State the appropriate null hypothesis H₀ and alternative hypothesis H_a in each of the following cases.

6.55 More on determining hypotheses. State the null hypothesis H₀ and the alternative hypothesis H_a in each case. Be sure to identify the parameters that you use to state the hypotheses.

6.56 Even more on determining hypotheses. In each of the following situations, state an appropriate null hypothesis H₀ and alternative hypothesis H_a. Be sure to identify the parameters that you use to state the hypotheses. (We have not yet learned how to test these hypotheses.)

6.57 Translating research questions into hypotheses. Translate each of the following research questions into appropriate H₀ and H_a.

6.58 Computing the P-value. A test of the null hypothesis H₀: μ = μ₀ gives test statistic z = 1.89.

6.59 More on computing the P-value. A test of the null hypothesis H₀: μ = μ₀ gives test statistic z = −1.33.

6.60 Timing of food intake and weight loss. A study found that a large group of late lunch eaters lost less weight over a 20-week observation period than a large group of early lunch eaters (P = 0.002).¹⁶ Explain what this P = 0.002 means in a way that could be understood by someone who has not studied statistics.

6.61 Average starting salary. Refer to Exercise 6.22 (page 358). Use the information presented in the exercise to test that the average income of graduates from your institution is different from the national average (α = 0.01). Write a short paragraph summarizing your conclusions.

6.62 Change in consumption of sweet snacks? Refer to Exercise 6.23 (page 358). A similar study performed four years earlier reported the average consumption of sweet snacks among healthy weight children aged 12 to 19 years to be 369.4 kilocalaries per day (kcal/d). Does this current study suggest a change in the average consumption? Perform a significance test using the 5% significance level. Write a short paragraph summarizing the results.

6.63 Peer pressure and choice of major. A study followed a cohort of students entering a business/economics program.¹⁷ All students followed a common track during the first three semesters and then chose to specialize in either business or economics. Through a series of surveys, the researchers were able to classify roughly 50% of the students as either peer driven (ignored abilities and chose major to follow peers) or ability driven (ignored peers and chose major based on ability). When looking at entry wages after graduation, the researchers conclude that a peer-driven student can expect an average wage that is 13% less than that of an ability-driven student. The report states that the significance level is P = 0.09. Can you be confident of the researchers’ conclusion statement regarding the wage decrease? Explain your answer.

6.64 Symbol of wealth in ancient China? Every society has its own symbols of wealth and prestige. In ancient China, it appears that owning pigs was such a symbol. Evidence comes from examining burial sites. If the skulls of sacrificed pigs tend to appear along with expensive ornaments, that suggests that the pigs, like the ornaments, signal the wealth and prestige of the person buried. A study of burials from around 3500 B.C. concluded that “there are striking differences in grave goods between burials with pig skulls and burials without them... A test indicates that the two samples of total artifacts are significantly different at the 0.01 level.”¹⁸ Explain clearly why “significantly different at the 0.01 level” gives good reason to think that there really is a systematic difference between burials that contain pig skulls and those that lack them.

6.65 Alcohol awareness among college students. A study of alcohol awareness among college students reported a higher awareness for students enrolled in a health and safety class than for those enrolled in a statistics class.¹⁹ The difference is described as being statistically significant. Explain what this means in simple terms and offer an explanation for why the health and safety students had a higher mean score.

6.66 Change in eighth-grade average mathematics score. A report based on the 2015 National Assessment of Educational Progress (NAEP)²⁰ states that the average score on their mathematics test for eighth-grade students attending public schools is significantly higher than in 2011. The report also states that the average score for eighth-grade students attending private schools is not significantly different from the average score in 2011. A footnote states that comparisons are determined by two-sided statistical tests with 0.05 as the level of significance. Explain what this footnote means in language understandable to someone who knows no statistics. Do not use the word “significance” in your answer.

6.67 More on change in eighth-grade average mathematics score. Refer to the previous exercise. On the basis of the NAEP study, a friend who works for the school newspaper wants to report that between 2011 and 2013 the average mathematics score improved for students attending public schools but stayed the same for students attending private schools. Do you agree with this statement? Explain your answer.

6.68 Background television in homes of U.S. children. In one study, U.S. parents were surveyed to determine the amount of background television their children were exposed to. A total of n = 1454 families with one child between the ages of 8 months and 8 years participated.²¹ For those families in which the caregiver had a high school degree or less, the child was exposed to an average of 313.0 minutes of background television per day. For those families in which the caregiver had some college or a college degree, the child was exposed to an average of 218.8 minutes per day. These average times were reported to be significantly different with P < 0.05. The actual P-value is 0.003. Explain why the actual P-value is more informative than the statement of significance at the 0.05 level.

6.69 Sleep quality and elevated blood pressure. A study looked at n = 238 adolescents, all free of severe illness.²² Subjects wore a wrist actigraph, which allowed the researchers to estimate sleep patterns. Those subjects classified as having low sleep efficiency had an average systolic blood pressure that was 5.8 millimeters of mercury (mm Hg) higher than that of other adolescents. The standard deviation of this difference is 1.4 mm Hg. Based on these results, test whether this difference is significant at the 0.01 level.

6.70 Are the pine trees randomly distributed from north to south? In Example 6.1 (page 342), we looked at the distribution of longleaf pine trees in the Wade Tract. One way to formulate hypotheses about whether or not the trees are randomly distributed in the tract is to examine the average location in the north–south direction. The values range from 0 to 200, so if the trees are uniformly distributed in this direction, any difference from the middle value (100) should be due to chance variation. The sample mean for the 584 trees in the tract is 99.74. A theoretical calculation based on the assumption that the trees are uniformly distributed gives a standard deviation of 58. Carefully state the null and alternative hypotheses in terms of this variable. Note that this requires that you translate the research question about the random distribution of the trees into specific statements about the mean of a probability distribution. Test your hypotheses, report your results, and write a short summary of what you have found.

6.71 Are the pine trees randomly distributed from east to west? Answer the questions in the previous exercise for the east–west direction, for which the sample mean is 113.8.

6.72 Who is the author? Statistics can help decide the authorship of literary works. Sonnets by a certain Elizabethan poet are known to contain an average of μ = 8.9 new words (words not used in the poet’s other works). The standard deviation of the number of new words is σ = 2.5. Now a manuscript with six new sonnets has come to light, and scholars are debating whether it is the poet’s work. The new sonnets contain an average of words not used in the poet’s known works. We expect poems by another author to contain more new words, so to see if we have evidence that the new sonnets are not by our poet we test

H₀: μ = 8.9

H_a: μ > 8.9

Give the z test statistic and its P-value. What do you conclude about the authorship of the new poems?

6.73 Attitudes toward school. The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures the motivation, attitude toward school, and study habits of students. Scores range from 0 to 200. The mean score for U.S. college students is about 95, and the standard deviation is about 20. A teacher who suspects that older students have better attitudes toward school gives the SSHA to 25 students who are at least 30 years of age. Their mean score is .

6.74 Nutritional intake among Canadian high-performance athletes. Since previous studies have reported that elite athletes are often deficient in their nutritional intake (for example, total calories, carbohydrates, protein), a group of researchers decided to evaluate Canadian high-performance athletes.²³ A total of n = 324 athletes from eight Canadian sports centers participated in the study. One reported finding was that the average caloric intake among the n = 201 women was 2403.7 kilocalories per day (kcal/d). The recommended amount is 2811.5 kcal/d. Is there evidence that female Canadian athletes are deficient in caloric intake?

6.75 Are the measurements similar? Refer to Exercise 6.30 (page 360). In addition to the computer’s calculations of miles per gallon, the driver also recorded the miles per gallon by dividing the miles driven by the number of gallons at each fill-up. The following data are the differences between the computer’s and the driver’s calculations for that random sample of 20 records. The driver wants to determine if these calculations are different. Assume that the standard deviation of a difference is σ = 3.0.

6.76 Impact of on significance. The Statistical Significance applet illustrates statistical tests with a fixed level of significance for Normally distributed data with known standard deviation. Open the applet and keep the default settings for the null (μ = 0) and the alternative (μ > 0) hypotheses, the sample size (n = 10), the standard deviation (σ = 1), and the significance level (α = 0.05). In the “I have data, and the observed is =” box, enter the value 1. Is the difference between and μ₀ significant at the 5% level? Repeat for equal to 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9. Make a table giving and the results of the significance tests. What do you conclude?

6.77 Effect of changing a on significance. Repeat the previous exercise with significance level α = 0.01. How does the choice of α affect which values of are far enough away from μ₀ to be statistically significant?

6.78 Changing to a two-sided alternative. Repeat the previous exercise but with the two-sided alternative hypothesis. How does this change affect which values of are far enough away from μ₀ to be statistically significant at the 0.01 level?

6.79 Changing the sample size. Refer to Exercise 6.76. Suppose that you increase the sample size n from 10 to 50. Again, make a table giving and the results of the significance tests at the 0.05 significance level. What do you conclude?

6.80 Impact of on the P-value. We can also study the P-value using the Statistical Significance applet. Reset the applet to the default settings for the null (μ = 0) and the alternative (μ > 0) hypotheses, the sample size (n = 10), the standard deviation (σ = 1), and the significance level (α = 0.05). In the “I have data, and the observed is =” box, enter the value 1. What is the P-value? It is shown at the top of the blue vertical line. Repeat for equal to 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9. Make a table giving and P-values. How does the P-value change as moves farther away from μ₀?

6.81 Changing to a two-sided alternative, continued. Repeat the previous exercise but with the two-sided alternative hypothesis. How does this change affect the P-values associated with each ? Explain why the P-values change in this way.

6.82 Other changes and the P-value. Refer to the previous exercise.

6.83 Understanding levels of significance. Explain in plain language why a significance test that is significant at the 1% level must always be significant at the 5% level.

6.84 More on understanding levels of significance. You are told that a significance test is significant at the 5% level. From this information, can you determine whether or not it is significant at the 1% level? Explain your answer.

6.85 Test statistic and levels of significance. Consider a significance test for a null hypothesis versus a two-sided alternative. Give a value of z that will give a result significant at the 1% level but not at the 0.5% level.

6.86 Using Table D to find a P-value. You have performed a two-sided test of significance and obtained a value of z = 2.08. Use Table D to find the approximate P-value for this test.

6.87 More on using Table D to find a P-value. You have performed a one-sided test of significance and obtained a value of z = 1.03. Use Table D to find the approximate P-value for this test when the alternative is greater than.

6.88 Using Table A and Table D to find a P-value. Consider a significance test for a null hypothesis versus a two-sided alternative. Between what values from Table D does the P-value for an outcome z = 1.88 lie? Calculate the P-value using Table A and verify that it lies between the values you found from Table D.

6.89 More on using Table A and Table D to find a P-value. Refer to the previous exercise. Find the P-value for z = −1.88.