Question

Online courses: A sample of 264 students who were taking online courses were asked to describe their overall impression of online learning on a scale of −17, with 7 representing the most favorable impression. The average score was 5.57, and the standard deviation was 0.93.

Construct a 99.8% interval for the mean score. Round the answers to two decimal places. A 99.8% confidence interval for the mean score is <μ<

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 5.57

sample standard deviation = s = 0.93

sample size = n = 264

Degrees of freedom = df = n - 1 = 264 -1 = 263

At 99.8% confidence level the t is ,

= 1 - 99.8% = 1 - 0.998 = 0.002

/ 2 = 0.002 / 2 = 0.001

t /2,df = t0.001,263 = 3.122

Margin of error = E = t/2,df * (s /n)

= 3.122 * (0.93 / 264)

   E = 0.18

The 99.8% confidence interval estimate of the population mean is,

- E < < + E

5.57 - 0.18 < < 5.57 + 0.18

5.39 < < 5.75

(5.39,5.75) ,  A 99.8% confidence interval for the mean score is 5.39 < μ < 5.75