Respuesta :
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
[tex]t^{*} \sqrt{ \frac{\hat{p}(1-\hat{p}}{n} } =0.1601[/tex]
Use the maximum value of t* at 95% confidence level of 1.96 to obtain
[tex] \sqrt{ \frac{(0.62)(0.38)}{n} }=( \frac{0.1601}{1.96})^{2} \\ n=(0.62)(0.38)( \frac{1.96}{0.1601})^{2} = 35.3[/tex]
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.
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
[tex]t^{*} \sqrt{ \frac{\hat{p}(1-\hat{p}}{n} } =0.1601[/tex]
Use the maximum value of t* at 95% confidence level of 1.96 to obtain
[tex] \sqrt{ \frac{(0.62)(0.38)}{n} }=( \frac{0.1601}{1.96})^{2} \\ n=(0.62)(0.38)( \frac{1.96}{0.1601})^{2} = 35.3[/tex]
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.
