Question

In a random sample of 60 computers, the mean repair cost was $150. Assume the population standard deviation is $36. Construct a 95% confidence interval for the population mean.

Answer 1

The 95% confidence interval for the population mean is between $140.89 and $159.11.

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.95)/(2) = 0.025`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.025 = 0.975`, so
`z = 1.96`

Now, find the margin of error M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 1.96(36)/(√(150)) = 9.11`

The lower end of the interval is the sample mean subtracted by M. So it is 150 - 9.11 = $140.89

The upper end of the interval is the sample mean added to M. So it is 150 + 9.11 = $159.11

The 95% confidence interval for the population mean is between $140.89 and $159.11.