Question

A random sample of 20 purchases showed the amounts in the table (in $). The mean is $49.57 and the standard deviation is $20.28. Construct a 98​% confidence interval for the mean purchases of all​ customers, assuming that the assumptions and conditions for the confidence interval have been met.

Answer 1

Answer:

The 98​% confidence interval for the mean purchases of all​ customers is ($37.40, $61.74).

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:

$\alpha = \frac{1-0.98}{2} = 0.01$

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 $1-0.01 = 0.99$, so $z = 2.325$

Now, find 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.

$M = 2.325*\frac{20.28}{\sqrt{20}} = 12.17$

The lower end of the interval is the mean subtracted by M. So it is 49.57 - 12.17 = $37.40.

The upper end of the interval is the mean added to M. So it is 49.57 + 12.17 = $61.74.

The 98​% confidence interval for the mean purchases of all​ customers is ($37.40, $61.74).