Answer to Question #180636 in Statistics and Probability for Joshua Cornfield

Question

Benoit is opening a new pizza shop and trying to estimate how much pepperoni he'll need to order. A recent randomized survey of 80 families in his area found that 29% of them ordered pepperoni on their pizza. Develop a 95% confidence statement.

The margin of error for the 95% confidence interval is 

Answer for part 1

%

Benoit can be 95% confident that between 

Answer for part 2 and coordinate 1

% and 

Answer for part 2 and coordinate 2

% of families in his area order pepperoni on their pizzas. 

Answer 1

n=80, p=0.29,q=0.71

95% confidence statement is=$p\pm Z_{0.025}\sqrt{\dfrac{pq}{n}}=0.29\pm 1.96\sqrt{\dfrac{0.29\times 0.71}{80}}$

$=0.29\pm 0.0994=(0.19056,0.3894)$

Margin of error $=z\times \sqrt{\dfrac{pq}{n}}=1.96\times \sqrt{\dfrac{0.29\times 0.71}{80}}=0.0994$