Question

(1 point) a) For 30 randomly selected Rolling Stones concerts, the mean gross earnings is 2.46 million dollars. Assuming a population standard deviation gross earnings of 0.52 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. There is a 99% chance that the mean gross earnings of all Rolling Stones concerts lies in the interval
B. We can be 99% confident 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,

= 2.46

= 0.52

n = 30

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * (0.52 / 30)

= 0.24

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

- E < < + E

2.46 - 0.24 < < 2.46 + 0.24

2.22 < < 2.70

Confidence interval :(2.22 , 2.70)

B)

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