Question

I just want to make sure these answers are right. In 2018, as a way of commemorating the 10-year anniversary of the release of the first smartphone, an undergraduate student polled a random sample of 26 her peers in hopes of estimating the average time (in hours) such students spend on their smartphone each day, on average. Using the data she collected, she produced a 95% confidence interval for the true mean time college students spend on their phone each day: (5.035, 6.165).

Question 16 (1 point)

What was the margin of error of this confidence interval?

Your Answer: 0.565

Question 17 (1 point)

What was the sample mean time (in hours) for the n = 26 students?

Your Answer: 5.6
Question 18 (1 point)

What was the sample standard deviation (in hours) for the n = 26 students? Submit your answer rounded to four decimals.

Your Answer: 1.4699

Answer 1

We are given

n = 26 and 95% confidence interval for population mean is ( 5.035 , 6.135)

xbar - E = 5.035

xbar + E = 6.135

16) The margin of error E = 0.55

17) sample mean Xbar = 5.585

18) sample standard deviation s = (E * √n )/ ta/2

For a= 0.05 and d.f = n-1 = 25

t​​​​​​a/2,n-1 = t​​​​​​0.025,25 = 2.06

s =( 0.55 * √26) /2.06

S = 1.3614