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

Answer:

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

Step-by-step explanation:

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

$\alpha = \frac{1-0.95}{2} = 0.025$

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

$M = z*\frac{\sigma}{\sqrt{n}}$

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

$M = 1.96\frac{36}{\sqrt{60}} = 9.11$

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