Question: 6.87 NFL player survey. Researchers at the University of Pennsylvania’s Wharton Sports Business Initiative collaborated with the National Football League Players Association (NFLPA) to produce the first NFL Player Survey. Of the 1,696 active NFL players, 1,355 (almost 80%) responded to the survey. One of the survey questions asked, “Who is the coach—professional, college, or high school—that has been the most influential in your career?” Of the 1,355 respondents, 759 selected an NFL (professional) coach.

a. Construct a 95% confidence interval for the true proportion of active NFL players who select a professional coach as the most influential in their careers.

b. Why is it necessary to use the continuity correction factor in the construction of the interval, part a?

c. Give a practical interpretation of the interval, part a.

The number of active National Football League (NFL) Players is

The sample size is

Let x denotes the (NFL) candidates who were selected by the coach that is

The proportion of active NFL players who select a professional coach as the most influential in their careers is calculated using the following formula,

$\hat{p}=\frac{x}{n}$

Therefore

$$\begin{gathered} \hat{p}=\frac{759}{1355} \\ =0.5601 \end{gathered}$$

The 95% confidence interval for the true proportion is obtained by using the following formula

$\hat{p}\pm 2\times \sqrt{\frac{\hat{p}\left(1-\hat{p}\right)}{n}}\times \sqrt{\frac{N-n}{N}}$

$$\begin{gathered} \hat{p}\pm 2\times \sqrt{\frac{\hat{p}\left(1-\hat{p}\right)}{n}}\times \sqrt{\frac{N-n}{N}}=0.5601\pm 2\times \sqrt{\frac{0.5601\left(1-0.5601\right)}{1355}}\times \sqrt{\frac{1696-1355}{1696}} \\ =0.5601\pm 0.012 \\ =\left(0.5481,0.5721\right) \end{gathered}$$

Hence, the required confidence interval is (0.5481, 0.5721).

When the sample size is large relative to the number of measurements in the population, the standard error of the estimators of p should be multiplied with a finite population correction factor. Let’s find the value of ; if it is greater than the finite population correction factor, it should be included in the standard error calculation.

Here it is observed that the$\frac{n}{N}=0.7989>0.05$finite population correction factor should be included in the calculation of the standard error for true proportion.

The 95% confidence interval is (0.548, 0.572). Thus, it is 95% confident that the true proportion of active NFL players who selected a professional coach as most influential is between 0.548 and 0.572.