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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

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

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

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鈥檚 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.