Answer to Question #105681 in Statistics and Probability for suresh

Question

A sports training camp has 60% boys. 20 random players win a trophy each day. 68% of the winners can be expected to be between ---- boys

A> 59.8905% and 60.1095%
B> 49.05% and 70.95%
C> 59.8658% and 60.1342%
D> 46.58% and 73.42%
E> None of the above

Answer 1

We need to construct the 68% confidence interval for the population proportion. We have been provided with the following information about the sample proportion: $\hat{p}=0.60, n=20.$

The critical value for $\alpha=0.32$ is $z_c=z_{1-\alpha/2}=0.994.$ The corresponding confidence interval is computed as shown below:

$$CI(Proportion)=(\hat{p}-z_c\sqrt{{\hat{p}(1-\hat{p}) \over n}},\hat{p}+z_c\sqrt{{\hat{p}(1-\hat{p}) \over n}})$$

$$CI(Proportion)=(0.6-0.994\sqrt{{0.6(1-0.6) \over 20}},0.6-0.994\sqrt{{0.6(1-0.6) \over 20}})\approx$$

$$\approx(0.491, 0.709)$$

B) 49.05% and 70.95%