Question

a) For 30 randomly selected Rolling Stones concerts, the mean gross earnings is 2.36 million dollars. Assuming a population standard deviation gross earnings of 0.48 million dollars, obtain a 99% confidence interval for the mean gross earnings of all Rolling Stones concerts (in millions).
Confidence interval: ( , ).

b) Which of the following is the correct interpretation for your answer in part (a)?
A. We can be 99% confident that the mean gross earnings of all Rolling Stones concerts lies in the interval
B. There is a 99% chance that the mean gross earnings of all Rolling Stones concerts lies in the interval
C. We can be 99% confident that the mean gross earnings for this sample of 30 Rolling Stones concerts lies in the interval
D. None of the above

Answer 1

Solution :

Given that,

a) Point estimate = sample mean = = 2.36

Population standard deviation =    = 0.48

Sample size = n = 30

At 99% confidence level

= 1 - 99%  

= 1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z0.005 = 2.576

Margin of error = E = Z/2 * ( /n)

= 2.576 * ( 0.48 /  30 )

= 0.23

At 99% confidence interval estimate of the population mean is,

  ± E

2.36 ± 0.23

( 2.13, 2.59 )  

b) A. We can be 99% confident that the mean gross earnings of all Rolling Stones concerts lies in the interval