Company A is trying to sell its website to Company B. As part of the sale, Compa

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 10 users. Assume normality.

Construct a 95% confidence interval for the true mean time spent on the web site.

a) What is the lower limit of the 95% 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 95% interval? Give your answer to three decimal places.

Explanation / Answer

A)

Note that              
              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    12.38          
t(alpha/2) = critical t for the confidence interval =    2.262157163          
s = sample standard deviation =    7.747372314          
n = sample size =    10          
df = n - 1 =    9          
Thus,              
              
Lower bound =    6.837863712   [ANSWER]

**********************************

b)
As

Upper Bound = X + t(alpha/2) * s / sqrt(n)      

Then
      
Upper bound =    17.92213629   [ANSWER]

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