Question

When the sample size and the sample proportion remain the same, a 90 percent confidence interval for a population proportion p will be ______________ the 99 percent confidence interval for p.

Answer 1

Answer:

A 90% confidence interval for p will be narrower than the 99% confidence interval.

Step-by-step explanation:

The formula to compute the (1 - α) % confidence interval for a population proportion is:

$CI=\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

Here $\hat p$ is the sample proportion.

The margin of error of the confidence interval is:

$MOE= z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The MOE is dependent on:

1. Confidence level
2. Standard deviation
3. Sample size

The MOE is directly related to the confidence level and standard deviation.

So if any of the two increases then the MOE also increases, thus widening the confidence interval.

And the MOE is inversely related to the sample size.

So if the sample increases the MOE decreases and vice versa.

It is provided that the sample size and the sample proportion are not altered.

The critical value of z for 90% confidence level is:

$z_{\alpha/2}= z_{0.10/2}=z_{0.05}=1.645$

And the critical value of z for 99% confidence level is:

$z_{\alpha/2}= z_{0.05/2}=z_{0.05}=1.96$

So as the confidence level increases the critical value increases.

Thus, a 90% confidence interval for p will be narrower than the 99% confidence interval.