\fFind the critical value(s). V = (Round to three decimal places as needed. Use a comma to separate answers as needed.) What is he conclusion of the hypothesis test? V the null hypothesis, meaning that the data V sufcient evidence to conclude that the mean number of acute postoperative days in the hospital is smaller with the dynamic system than with the static system. A scientist investigated the effect of crossfertilization on the heights of plants. In one study, the scientist planted 15 pairs of a species of plant. Each pair consisted of one cross-fertilized plant and one selffertilized plant grown in the same pot. The accompanying table gives the height differences, in eighths of an inch, for the 15 pairs. Each difference is obtained by subtracting the height of the selffertilized plant from that of the cross-fertilized plant. Complete parts (a) through (e) below. Click here to view the data. Click here to View a table of critical values of t. b a. Identify the variable under consideration. What is the variable? Height Mean height Mean paired-height difference Paired-height difference OOO b. Identify the two populations What are the populations? 0 A. The mean height of the self-fertilized plants and the mean height of the cross-fertilized plants 6) B. The self-fertilized plants and the cross-fertilized plants O C. The plants and the pots they are growing in O D. The heights of the self-fertilized plants and the heights of the cross-fertilized plants c. Identify the paired-difference variable. What is the paired-difference variable? O A. Height of the self-fertilized plant - height of the cross-fertilized plant O B. Mean height of the self-fertilized plants - mean height of the cross-fertilized plants O C. Height of the cross-fertilized plant - height of the self-fertilized plant O D. Mean height of the cross-fertilized plants - mean height of the self-fertilized plantsd. Are the numbers in the table paired differences? Why or why not? Since there are equal numbers of cross-fertilized plants and self-fertilized plants, the numbers are not paired differences. \fFind the test statistic. t = (Round to two decimal places as needed.) Find the critical value(s). The critical value{s) isiare . (Round to two decimal places as needed. Use a comma to separate answers as needed.) What is the conclusion for the hypothesis test? V H0; the data V sufcient evidence at the 5% signicance level that the a table o X Height Differences 49 - 63 11 6 28 23 48 20 29 50 28 76 63 - 50 Print DoneSuppose a study reported that the average person watched 4.72 hours of television per day. A random sample of 15 people gave the number of hours of television watched per day shown. At the 10% signicance level, do the data provide sufcient evidence to conclude that the amount of television watched per day last year by the average person differed from the value reported in the study? (Note: x= 4.200 hours and s = 1.511 hours.) a Click here to View a partial table of values of tn. 5.6 3.9 5.7 5.5 4.1 5.3 2.4 5.7 3.5 4.6 1.7 E 1.8 4.4 6.2 2.6 Set up the hypotheses for the onemean t-test. H041 = 4.72 Ha:p it 4.72 The test statistic is t = . (Round to two decima places as needed.) Identify the critical value(s). Select the correct choice below and ll in the answer box within your choice. (Round to three decimal places as needed.) O A- The critical value is tn = The test statistic is t = |_|. (Round to two decimal places as needed.) Identify the critical value(s). Select the correct choice below and ll in the answer box within your choice. (Round to three decimal places as needed.) O A- The critical value is tIx = O B- The critical value is A\" = H- O C. The critical values are ux/2 = |:| the null hypothesis. Twe data : sufcient evidence to conclude that the amount of television watched by the average person is V the value reported in the study. For a right-tailed hypothesis test at the signicance level a, the null hypothesis HO: p = no will be rejected in favor of the alternative hypothesis Ha: p. > [to if and only if on is less than or equal to the (1 - uHevel lower condence bound for p. Illustrate the preceding relationship by obtaining the lower condence bound for the summary statistics provided. Click here to View page 1 of the t-table. Click here to view page 2 of the t-table. a. The owner of a particular golf club has a company test the effectiveness of the golf club. A robot is used to test a particular golf club. The golfer wanted a club that, on average, would hit the ball more than 180.2 yards. A random sample of 6 identical balls were hit at the same speed. At the 5% signicance level, the data provide sufcient evidence to conclude that the club does what the golfer wants. (Note: x = 182.8 yards and s = 2.8 yards.) The lower condence bound is . (Round to three decimal p aces as needed.) Since the value of no lies V the (1 u)-level lower condence bound for u, these results V consistent with the hypothesis test V the null hypothesis. Repeat part (a) at the at the 1% signicance level. At the 1% signicance level, the data do not provide sufcient evidence to conclude that the club does what the golfer wants. The lower condence bound is . (Round to three decimal p aces as needed.) Since the value of no lies V the (1 u)-level lower condence bound for u, these results V consistent with the hypothesis test V the null hypothesis. b. Alligators perform a spinning maneuver, referred to as a "death roll," to subdue their prey. Videos were taken of juvenile alligators performing this maneuver in a study. One of the variables measured was the degree of the angle between the body and head of the alligator while performing the roll. Asample of 20 rolls were taken. At the 1% signicance level, the data do not provide sufcient evidence to conclude that, on average, the angle between the body and head of an alligator during a death roll is greater than 47. (Note: x =48.6 and s = 10.8) The lower condence bound is . (Round to three decimal places as needed.) Since the value of \"0 lies V the ('l - u)-level lower oondenoe bound for u, these results V oonsistent with the hypothesis test V the null hypothesis