A magazine reported the results of a random telephone poll commissioned by a television network. Of...
A magazine reported the results of a random telephone poll commissioned by a television network. Of the1209 men who responded, only 20 said that their most important measure of success was their work. Complete parts a through c. a right parenthesis a) Estimate the percentage of all males who measure success primarily from their work. Use a 99% confidence interval. Don't forget to check the conditions first.
Homework Answers
Let , X be the number of males who measures success primarily from their work.
Given : n=1209 , x=20
a) The estimate of the percentage of all males who measures success primarily from their work. is ,
p=x/n=20/1209=0.0165
Conditions :
1) The data is simple random sample from the normal distribution.
2) The sample values are independent
3) np=1209*0.0165=20>10 and nq=1209*(1-0.0165)=1209*0.9835=1189.0515>10
b) Therefore , the 99% confidence interval for proportion is ,
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primarily from their work. Use a 95% confidence interval. Don't
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(__________%,_______________%)
(Type integers or decimals. Round to one decimal place as
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