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.
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 ,
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 the 1440 men who responded, only 24 said that their most important measure of success was their work. a) Estimate the percentage of all males who measure success primarily from their work. Use a 99.9% confidence interval. Don't forget to check the conditions first.
A magazine reported the results of a random telephone poll
commissioned by a television network. Of the 1359 men who
responded, only 18 said that their most important measure of
success was their work.
Complete parts a through c.
a) Estimate the percentage of all males who measure success
primarily from their work. Use a 95% confidence interval. Don't
forget to check the conditions first.
(__________%,_______________%)
(Type integers or decimals. Round to one decimal place as
needed.)
b) Some believe...
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