Question

What proportion of adults like pineapple as a pizza topping? A 95% confidence interval for the proportion of adults who say they like pineapple on their pizza was constructed and it was found to be from 0.14 to 0.20. Based on this interval, we know the margin of error must be 0.02. 0.03 It's impossible to answer this question without more information. O 0.06. O 0.05.

Answer 1

95% confidence interval for the proportion of adults who say they like pine apple on their pizza will be given by -

$[p - z_{0.05/2}* \sqrt{\frac{p(1-p)}{n}}, \ p + z_{0.05/2}* \sqrt{\frac{p(1-p)}{n}}]$

= P - M.O.E, P+M.O.E

= [0.14, 0.20]

where, p is the sample proportion and

M.O.E is the margin of errror

So, we have

p - M.O.E = 0.14 ________ eqn (1)

p + M.O.E = 0.20 ________ eqn (2)

Substracting eqn (1) from eqn (2), we get,

2 * M.O.E = 0.06

M.O.E = 0.03

Hence, based on the give interval, we know that margin of errror must be 0.03

Hence, option (b) is the correct option.