Question

1. One hundred eight Americans were surveyed to determine the number of hours they spent watching television each month. The average time was 75 hours a month (that’s approximately 2.5 hours per day). In the population of all Americans, it is estimated that the variation of the time spent watching television per month is 15.2 hours. Use this information to construct a 98% confidence interval to estimate the average time spent watching television per month for all Americans.

A) 71.592 to 78.408

B) 72.133 to 77.867

C) 73.567 to 76.433

D) 74.971 to 75.029   

2. One hundred eight Americans were surveyed to determine the number of hours they spent watching television each month. The average time was 75 hours a month (that’s approximately 2.5 hours per day). In the population of all Americans, it is estimated that the variation of the time spent watching television per month is 15.2 hours. What is the margin of error in the estimated 98% confidence interval?

A) 0.328

B) 1.463

C) 2.867

D) 3.408

3. One hundred eight Americans were surveyed to determine the number of hours they spent watching television each month. The average time was 75 hours a month (that’s approximately 2.5 hours per day). In the population of all Americans, it is estimated that the variation of the time spent watching television per month is 15.2 hours. What is the standard error in the estimated 98% confidence interval?

A) 0.328

B) 1.463

C) 2.867

D) 3.408

Answer 1

Answer Date: 10/10/2019 A 98% confidence interval to estimate the average time spent watching television per month for all Americans is, First, compute z-critical value then find confidence interval. From the standard normal table, the z-critical value at 98% confidence is 2.33. 98%C.I. = u tz, xo/In = 75+ 2.33x15.20/1/108 = 75+3.408 = 71.592 to 78.408 A 98% confidence interval to estimate the average time spent watching television per month for all Americans is 71.592 and 78.408. The answer is A

At 98% confidence, the margin of error is, From the standard normal table, the z-critical value at 98% confidence is 2.33. The information represent n=108, 0 = 15.20 and ze = 2.33 for finding margin of error. E =ze xemene = 2.33x 15.20 1108 = 3.408 At 99% confidence, the margin of error is 3.408. The answer is D. 3)

The standard error is, SE = 일 15.20 V108 = 1.463 The standard error is 1.463. The answer is B.