Customers who participate in a store’s free loyalty card program save money on their purchases but allow the store to keep track of their shopping habits and potentially sell these data to third parties. A Pew Internet & American Life Project Survey (January 2016) revealed that half (225) of a random sample of 250 U.S. adults would agree to participate in a store loyalty card program, despite the potential for information sharing.

a. Estimate the true proportion of all U.S. adults who would agree to participate in a store loyalty card program, despite the potential for information sharing.

b. Form a 90% confidence interval around the estimate, part a.

c. Provide a practical interpretation of the confidence interval, part b. Your answer should begin with, “We are 90% confident . . .”

d. Explain the theoretical meaning of the phrase, “We are 90% confident.”

A Pew Internet & American Life Project Survey (January 2016) revealed that half (225) of a random sample of 250 U.S. adults would agree to participate in a store loyalty card program, despite the potential for information sharing. We need to calculate the following

a. an estimate of the true proportion of all U.S. adults who would agree to participate in a store loyalty card program, despite the potential for information sharing.

b.90% confidence interval around the estimate, part a

c.a practical interpretation of the confidence interval, part b

d.the theoretical meaning of the phrase, “We are 90% confident.”

Sample size n=250

A number of people would agree to participate in a store loyalty card program, despite the potential for information sharing. We need to calculate the following

X=125

$\begin{array}{rcl}\hat{p}& =& \frac{x}{n}\\ & =& \frac{125}{250}\\ & =& 0.5\\ & & \end{array}$

Hence,the true proportion of all U.S. adults who would agree to participate in a store loyalty card program, despite the potential for information sharing, is 0.5

Here the confidence coefficient is 90%, hence$\alpha =10\%$ .Now from the standard normal distribution table,$z_{\frac{\alpha }{2}}=1.645$

The margin of error

$\begin{array}{rcl}E& =& z_{\frac{\alpha }{2}}\times \sqrt{\frac{\hat{p}\left(1-\hat{p}\right)}{n}}\\ & =& 1.645\times \sqrt{\frac{0.5\times \left(1-0.5\right)}{250}}\\ & =& 0.0520\\ & & \end{array}$

Hence 90% lower limit:$\hat{p}-E=0.5-0.0520=0.8314$

90% upper limit:$\hat{p}+E=0.5+0.0520=0.9286$

90% confidence interval around the estimate, part a is$\left(0.8314,0.9286\right)$

We are confident that the proportion of all U.S. adults who would agree to participate in a store loyalty card program, despite the potential for information sharing, is between interval$\left(0.8314,0.9286\right)$

The meaning of theoretical meaning of the phrase, “We are 90% confident.” is that 90% of all possible values contain in the interval$\left(0.8314,0.9286\right)$