Question

A local electronics firm wants to determine their average daily sales (in dollars.) A sample of the sales for 36 days revealed an average sales of $139,000. Assume that the standard deviation of the population is known to be $12,000.
a) provide a 95% confidence interval estimate for average daily sales.
b) provide a 97% confidence interval estimate for average daily sales.

Answer 1

Answer:

a) The 95% confidence interval estimate for average daily sales is between $135,080 and $142,920.

b) The 97% confidence interval estimate for average daily sales is between $134,660 and $143,340.

Step-by-step explanation:

a) provide a 95% confidence interval estimate for average daily sales.

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.

$M = 1.96*\frac{12000}{36} = 3920$

The lower end of the interval is the sample mean subtracted by M. So it is 139000 - 3920 = $135,080

The upper end of the interval is the sample mean added to M. So it is 139000 + 3920 = $142,920

The 95% confidence interval estimate for average daily sales is between $135,080 and $142,920.

b) provide a 97% confidence interval estimate for average daily sales.

By the same logic as above, now Z = 2.17.

$M = 2.17*\frac{12000}{36} = 4340$

The lower end of the interval is the sample mean subtracted by M. So it is 139000 - 4340 = $134,660

The upper end of the interval is the sample mean added to M. So it is 139000 + 4340 = $143,340

The 97% confidence interval estimate for average daily sales is between $134,660 and $143,340.