Question

A random sample of 21 desktop PCs is selected. The mean life span is 6.8 years with a standard deviation of 2.4 years. Construct a 95% confidence interval for the mean life span of all desktop PCs. Assume that the life spans of all desktop PCs are approximately normally distributed (a) (5.85, 7.75) (b) (1.68, 3.12) (c) (5.60, 8.00) (d) (5.71, 7.89) (e) (5.77, 7.83)

Answer 1

(d) (5.71, 7.89)

Explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 21 - 1 = 20

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 20 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.95)/(2) = 0.975`. So we have T = 2.086

The margin of error is:

`M = T(s)/(√(n)) = 2.086(2.4)/(√(21)) = 1.09`

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 6.8 - 1.09 = 5.71 years

The upper end of the interval is the sample mean added to M. So it is 6.8 + 1.09 = 7.89 years

So the confidence interval is (5.71, 7.89), and the correct answer is given by option b.