Question

a) For 30 randomly selected Rolling Stones concerts, the mean gross earnings is 2.79 million dollars. Assuming a population standard deviation gross earnings of 0.47 million dollars, obtain a 99% confidence interval for the mean gross earnings of all Rolling Stones concerts (in millions).

Confidence interval: ( __________________ , __________________ ).

Answer 1

Solution :

Given that,

= 2.79

= 0.47

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 * (0.47 / 30)

= 0.22

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

- E < < + E

2.79 - 0.22 < < 2.79 + 0.22

2.57 < < 3.01

Confidence interval :(2.57 , 3.01)