A sports league introduced a video referee to review certain critical plays. However, the system has not been without its controversies. One particular unfortunate incident led to an outcry among fans. Nevertheless, a poll after this event found that 57% of fans were still in favor of the continued use of the video referee. Suppose the size of this sample was 300 people. Complete parts a and b.

(a) Estimate the true proportion of fans in favor of keeping the video referee using a 99% confidence interval.

(b) What is the margin of error for this sample?

Answer :

Given :

Sample proportion $(p\widehat{)}=0.57$

Sample size $(n)=300$

Confidence interval level $=99$

Solution :

A) The 99% confidence interval :

The level of significance is calculated as :

$$\alpha =1–0.99=0.01$$

The critical value is calculated as :

$$z_c=Z_{\frac{\alpha }{2}}=Z_{\frac{0.01}{2}}=2.576$$

The confidence interval is calculated as :

$$\text{CI}=\hat{p}\pm z_c\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ $$=0.57\pm 2.576\times \sqrt{\frac{0.57(1-0.57)}{300}}$$ $$=(0.496,0.644)$$

B) The margin of error :

$$e=z_c\cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ $$=2.576\cdot \sqrt{\frac{0.57(1-0.57)}{300}}$$ $$=0.0737$$