Question

A 90% confidence interval for the true percentage of college students who like Brussels sprouts is (1.8%, 4.6%). What is the point estimator of the true percentage of college students who like Brussels sprouts?

Answer 1

Answer:

$\hat p = \frac{Lower+Upper}{2}$

And replacing the info from the problem we have:

$\hat p = \frac{0.018+0.046}{2}= 0.032$

So then the best estimator for the true proportion p is given by $\hat p = 0.032$ or equivalent to 3.2 %

Step-by-step explanation:

We want to find a confidence interval for a proportion p who represent the parameter of interest.

The confidence interval would be given by this formula:

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

For this case the 90% confidence interval is given by (1.8%=0.018, 4.6%=0.046) after apply the last formula

Since the confidence interval is symmetrical we can estimate the point estimator of the true percentage with this formula:

$\hat p = \frac{Lower+Upper}{2}$

And replacing the info from the problem we have:

$\hat p = \frac{0.018+0.046}{2}= 0.032$

So then the best estimator for the true proportion p is given by $\hat p = 0.032$ or equivalent to 3.2 %