Question
In: Statistics and Probability
The marketing director for a cereal company would like to know what proportion of households that...
The marketing director for a cereal company would like to know what proportion of households that receive free samples of cereal with their newspapers later purchase the cereal. A random sample showed 55 out of 100 purchased the cereal after receiving the free sample. Find a 95% confidence interval for the true proportion.
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Given That:-
The marketing director for a cereal company would like to know what proportion of households that receive free samples of cereal with their newspapers later purchase the cereal. A random sample showed 55 out of 100 purchased the cereal after receiving the free sample. Find a 95% confidence interval for the true proportion.
Given that,
n = 100
x = 55
Point estimate = sample proportion = 
= 1-0.55 =0.45
At 95% confidence level the z is ,
%
= 1 - 0.95
= 0.05
( Using z table )
Margin of error = E = 
E = 0.098
A 95% confidence interval for proportion p is ,
0.55-0.098 < p < 0.55+0.098
0.452< p < 0.648
(0.452 , 0.468)
Plz like it.......,
orchestra answered 1 year agoRelated Solutions
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