Answer:
Correct option:
The 95% confidence interval is wider than the 90% confidence interval.
Explanation:
The (1 - α)% confidence interval for the population proportion is:
![CI=\hat p\pm z_(\alpha/2)* \sqrt{(\hat p(1-\hat p))/(n)}](https://img.qammunity.org/2021/formulas/health/high-school/cw1er8fafijbbx787piyvhs1xl8gavtncw.png)
The width of this interval is:
![W=UL-LL\\=2* z_(\alpha/2)* \sqrt{(\hat p(1-\hat p))/(n)}](https://img.qammunity.org/2021/formulas/mathematics/college/86f0kyaxbwe0p0aqp514nq2phlf0buv40w.png)
The width of the interval is directly proportional to the critical value.
The critical value of a distribution is based on the confidence level.
Higher the confidence level, higher will be critical value.
The z-critical value for 95% and 90% confidence levels are:
![90\%: z_(\alpha /2)=z_(0.05)=1.645\\95\%: z_(\alpha /2)=z_(0.025)=1.96](https://img.qammunity.org/2021/formulas/mathematics/college/c8mu5oz4pv8gw4vuc7a1vwgpohrdmy13j1.png)
*Use a z-table.
The critical value of z for 95% confidence level is higher than that of 90% confidence level.
So the width of the 95% confidence interval will be more than the 90% confidence interval.
Thus, the correct option is:
"The 95% confidence interval is wider than the 90% confidence interval."