Asurvey 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?

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


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


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.

About 38.

(-8,5) bu tun not sure ok