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In preparation for an earnings report, a large retailer wants to estimate p= the proportion of annual sales

that occur during the month of December. A SRS of sales from last year revealed that 37 of

the randomly selected sales occurred during the month December out of 161 sales.

Sagot :

Using the z-distribution, it is found that the 95% confidence interval for the proportion of sales that occured in December is (0.1648, 0.2948).

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which:

  • [tex]\pi[/tex] is the sample proportion.
  • z is the critical value.
  • n is the sample size.

In this problem, we have a 95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so the critical value is z = 1.96.

The sample size and the estimate are given by:

[tex]n = 161, \pi = \frac{37}{161} = 0.2298[/tex]

Hence:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2298 - 1.96\sqrt{\frac{0.2298(0.7702)}{161}} = 0.1648[/tex]

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2298 + 1.96\sqrt{\frac{0.2298(0.7702)}{161}} = 0.2948[/tex]

The 95% confidence interval for the proportion of sales that occured in December is (0.1648, 0.2948).

More can be learned about the z-distribution at https://brainly.com/question/25890103

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