QuestionVegetarians: In a recent poll of 1046 adults in the United States, 47 said they were vegetarians.
Part: $0 / 3$
Part 1 of 3 (a) Construct a $99 \%$ confidence interval for the proportion of adults in the United States who are vegetarians. (Round the answers to three decimal places.)
A $99 \%$ confidence interval for the proportion of adults in the United States who are vegetarians is $\square$ $<p<$ $\square$

Studdy Solution
Construct the confidence interval:
$\hat{p} \pm z^* \times SE = 0.0449 \pm 2.576 \times 0.0064$
Calculate the margin of error:
$2.576 \times 0.0064 \approx 0.0165$
Calculate the confidence interval:
$0.0449 - 0.0165 < p < 0.0449 + 0.0165$
$0.0284 < p < 0.0614$
The $99\%$ confidence interval for the proportion of adults in the United States who are vegetarians is:
$\boxed{0.028} < p < \boxed{0.061}$