dependent random samples are selected from two populations and are used to test the hypothesis Ho: (4, -42)= 0 against the alternative H: (H1 -2) #0. An analysis of 232 observations from population 1 and 311 from population 2 yielded p-value of 0.113. Complete parts a and b below. Interpret the results of the computer analysis. Use as0.10. A. Since this p-value exceeds the given value of a, there is insufficient evidence indicate that the population means are different. O B. Since the given value exceeds this p-value, there is insufficient evidence to indicate the population means are different. OC. Since the given value exceeds this p-value, there is sufficient evidence to indicate that the population means are different. OD. Since this p-value exceeds the given value of a, there is sufficient evidence to indicate that the population means are different. .If the alternative hypothesis had been H: (H, -H2) <0, how would the p-value change? Interpret the p-value for this one-tailed test. -value = Type an integer or a decimal,)

A physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 54 men and 70 women to participate in the study. Each subject was required to step up and down a 6-inch platform. The pulse of each subject was then recorded. The following results were obtained. Two sample T for Men vs Women N Mean StDev SE Mean Men Women 98% CI for mu Men - mu Women (- 12.20, - 1.00) T-Test mu Men = mu Women (vs H2 O C. Ho: H1 = H2; Ha: H1 #H2 (b) Identify the P-value and state the researcher's conclusion if the level of significance was a = 0.01. What is the P-value? P-value =

The given quantities below were calculated from a sample data set. Στο 338 I Lu=308 Ση2 = 4678 2 y = 4236 Determine the least squares regression line. Round values to four decimal places, if necessary. Σry = 3595 n = 30 Determine the correlation coefficient. Round the solution to four decimal places, if necessary.

4.2 A sample of 19 female bears was measured for chest girth (y) and neck girth (x), both in inches. The least squares regression equation relating the two variables is = 5.32 + 1.53 x. The standard deviation of the regression is s = 3.692. y^ 4.2.1 What is a 95% confidence interval for the average (or mean) chest girth for bears with a neck girth of 20 inches? The standard error of the fit is s.e.(fit) = 0.880 a. (27.9, 43.9) b. (28.1, 43.7) c. (34.1, 37.8) d. None of the above 4.2.2 Interpret the confidence interval obtained in 4.2.1