Question

Company A is trying to sell its website to Company B. As part of the sale, Company A claims that the average user of their site stays on the site for 10 minutes. To test this claim Company B collects the times (in minutes) below for a sample of 11 users. Assume normality.

Time
0.8
10
6.8
1.6
4.9
0.3
13.5
6
6
2.2
1.5

Construct a 99% confidence interval for the true mean time spent on the web site.

a) What is the lower limit of the 99% interval? Give your answer to three decimal places. Enter 0 if your lower limit is less than 0.

b) What is the upper limit of the 99% interval? Give your answer to three decimal places.

Answer 1

Solution :

Given that 0.8,10,6.8,1.6,4.9,0.3,13.5,6,6,2.2,1.5

=> Mean x = sum of terms/number of terms

= 53.6/11

= 4.8727

=> standard deviation s = 4.1678

=> df = n - 1 = 10

=> For 99% confidence interval , t = 3.169

=> For 99% confidence interval of the true mean is = x +/- t*s/sqrt(n)

a)
=> the lower limit of the 99% confidence interval of the true mean is

=> x - t*s/sqrt(n)

=> 4.8727 - 3.169*4.1678/sqrt(11)

=> 0.8904

=> 0.890 (rounded)

b)
=> the upper limit of the 99% confidence interval of the true mean is

=> x + t*s/sqrt(n)

=> 4.8727 + 3.169*4.1678/sqrt(11)

=> 8.8550

=> 8.855 (rounded)