Answer:
Correct option:
The 95% confidence interval is wider than the 90% confidence interval.
Step-by-step explanation:
The (1 - α)% confidence interval for the population proportion is:
[tex]CI=\hat p\pm z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
The width of this interval is:
[tex]W=UL-LL\\=2\times z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
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:
[tex]90\%: z_{\alpha /2}=z_{0.05}=1.645\\95\%: z_{\alpha /2}=z_{0.025}=1.96[/tex]
*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."
