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

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.

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 = (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*(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*(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.