A 2010 article reported that 93% of Americans feel Starbucks coffee is overprice
ID: 3152734 • Letter: A
Question
A 2010 article reported that 93% of Americans feel Starbucks coffee is overpriced. Consider the population of all Americans.
A random sample of n = 100 couples will be selected from this population and , the proportion of Americans who believe Starbucks is overpriced will be computed. What are the mean and standard error of the sampling distribution of the sample proportion ?
Is it reasonable to assume that the sampling distribution of the sample proportion is approximately Normal for random samples of size n = 100? Explain.
Suppose that the sample size is n = 200 rather than n = 100 as in Part b. Does the change in sample size change the mean and standard error of the sampling distribution of the sample proportion ? If so, what are the new values for the mean and standard error? If not, explain why not.
Is it reasonable to assume that the sampling distribution of the sample proportion is approximately Normal for random samples of size n = 200? Explain.
When n = 200, what is the approximate probability that the sample proportion of Americans who believe Starbucks is overpriced will be greater than 0.90? Show your work (i.e., z-score and value obtained from table).
Verify your answer to Part e using StatCrunch. Copy the StatCrunch results (graph plus answer) into your solutions document.
When n = 200, what is the approximate probability that the sample proportion will fall within 1.5 standard errors of the population proportion. Show your work (i.e., z-scores and values obtained from table).
Verify your answer to Part g using StatCrunch. Copy the StatCrunch results (graph plus answer) into your solutions document.
A 2010 article reported that 93% of Americans feel Starbucks coffee is overpriced. Consider the population of all Americans.
A random sample of n = 100 couples will be selected from this population and , the proportion of Americans who believe Starbucks is overpriced will be computed. What are the mean and standard error of the sampling distribution of the sample proportion ?
Is it reasonable to assume that the sampling distribution of the sample proportion is approximately Normal for random samples of size n = 100? Explain.
Suppose that the sample size is n = 200 rather than n = 100 as in Part b. Does the change in sample size change the mean and standard error of the sampling distribution of the sample proportion ? If so, what are the new values for the mean and standard error? If not, explain why not.
Is it reasonable to assume that the sampling distribution of the sample proportion is approximately Normal for random samples of size n = 200? Explain.
When n = 200, what is the approximate probability that the sample proportion of Americans who believe Starbucks is overpriced will be greater than 0.90? Show your work (i.e., z-score and value obtained from table).
Verify your answer to Part e using StatCrunch. Copy the StatCrunch results (graph plus answer) into your solutions document.
When n = 200, what is the approximate probability that the sample proportion will fall within 1.5 standard errors of the population proportion. Show your work (i.e., z-scores and values obtained from table).
Verify your answer to Part g using StatCrunch. Copy the StatCrunch results (graph plus answer) into your solutions document.
Explanation / Answer
proportion of Americans feel Starbucks coffee is overpriced =p=93%=0.93
n=100
answer part 1) mean of the sample proportion =n*p=100*0.93=93
answer part 2)and standard error of the sampling distribution =sqrt(p(1-p)/n)=sqrt(0.93(1-0.93)/100)=0.0255
answer part 3)yes, it is reasonable to assume that the sampling distribution of the sample proportion is approximately Normal for random samples of size n = 100 because when usualy sample size is more than 30 are taken as a approximately Normal for random samples
answer part 4) when n=200 , them mean=200*0.93=186 and standard error=sqrt(0.93*0.07/200)=0.018
answer part 5) here P=0.9
we go for z calculation and z=|(P-p)|/SE(p)=|(0.9-0.93)|/0.018=1.6667
P(z>1.6667)=1-P(z<1.6667)=0.0478
approximate probability that the sample proportion of Americans who believe Starbucks is overpriced will be greater than 0.90=0.0478
answer part 6)
P(that the sample proportion will fall within 1.5 standard errors of the population proportion)=
P(|z|<1.5)=P(-1.5<z<1.5)=2*P(z<-1.5)=2*0.0668=0.1336
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