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.

Respuesta :

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 [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]

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

So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]

Now, find the margin of error M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

[tex]M = 1.96*\frac{12000}{36} = 3920[/tex]

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.

[tex]M = 2.17*\frac{12000}{36} = 4340[/tex]

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.