Question

A survey claims that the percent of customers who favor Cheerios cereal over Fruit Loops cereal is likely between 78.01% and 45.99%. How many people were surveyed?

Answer 1

no

Explanation:

Answer 2

Given:
The proportion of customers who favor Cheerios over Fruit Loops is
(45.99%, 78.01%) = (0.4599, 0.7801).

Estimate the sample proportion as
(0.4599 + 0.7801)/2 = 0.62

Assume that the result was obtained at 95% confidence interval.
Half of the interval is
(0.7801 - 0.4599)/2 = 0.1601

Because we do not have population information, use the student's t-distribution.
Half of the interval is calculated as

`t^(*) \sqrt{ \frac{\hat{p}(1-\hat{p}}{n} } =0.1601`

Use the maximum value of t* at 95% confidence level of 1.96 to obtain

`\sqrt{ ((0.62)(0.38))/(n) }=( (0.1601)/(1.96))^(2) \\ n=(0.62)(0.38)( (1.96)/(0.1601))^(2) = 35.3`

When n = 35, the df (degrees of freedom) s 34, and t* = 2.03 approximately.
Recalculate n to obtain
n = 9.1916*(t*)² = 9.1916*(2.03²) = 37.9

We can conclude that n = 38 approximately.

Answer: About 38.