A manager of a local store wants to estimate the mean amount spent per shopping visit by customers. Summary statistics from a sample taken reveal the following: The store manager wonders whether the population mean could have been $50 or $60.
A) Since $60 is within the 95% confidence interval, the population mean is likely to be $60.
B) Since $60 is not within the 95% confidence interval, the population mean is not likely to be $60.
C) Since $50 is within the 95% confidence interval, the population mean is likely to be $50.
D) Since $50 is not within the 95% confidence interval, the population mean is not likely to be $50.
E) B & C are true

Question

A manager of a local store wants to estimate the mean amount spent per shopping visit by customers. Summary statistics from a sample taken reveal the following:  The store manager wonders whether the population mean could have been $50 or $60.
A) Since $60 is within the 95% confidence interval, the population mean is likely to be $60. 
B) Since $60 is not within the 95% confidence interval, the population mean is not likely to be $60. 
C) Since $50 is within the 95% confidence interval, the population mean is likely to be $50. 
D) Since $50 is not within the 95% confidence interval, the population mean is not likely to be $50. 
E) B & C are true

Quizplus · Accepted Answer

The Answer of A manager of a local store wants to estimate the mean amount spent per shopping visit by customers. Summary statistics from a sample taken reveal the following:  The store manager wonders whether the population mean could have been $50 or $60.
A) Since $60 is within the 95% confidence interval, the population mean is likely to be $60. 
B) Since $60 is not within the 95% confidence interval, the population mean is not likely to be $60. 
C) Since $50 is within the 95% confidence interval, the population mean is likely to be $50. 
D) Since $50 is not within the 95% confidence interval, the population mean is not likely to be $50. 
E) B & C are true

A Manager of a Local Store Wants to Estimate the Mean