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: ( __________________ , __________________ ). 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 for this sample of 30 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) There is a 99% chance that the mean gross earnings of all Rolling Stones concerts lies in the interval (D) None of the above

Answer 1

Answer: Confidence interval: (2.569, 3.011)

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

Step-by-step explanation:

The confidence interval for population mean is given by :-

$\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}$

, where n= Sample size

$\overline{x}$= Sample mean

$\sigma$= Population standard deviation

z* = Critical z-value.

Let $\mu$ be the mean gross earnings of all Rolling Stones concerts (in millions).

As per given ,we have

n= 30

$\overline{x}=2.79$(in millions).

$\sigma=0.47$(in millions)

Critical value of 99% confidence level : z=2.576

Put all values in formula , we get

99% confidence interval for $\mu$  : $2.79\pm (2.576)\dfrac{0.47 }{\sqrt{30}}$

$\approx2.79\pm0.221$

$=(2.79-0.221,\ 2.79+0.221)=(2.569,\ 3.011)$

Hence, the 99% confidence interval for $\mu$  : (2.569, 3.011)

Interpretation of a% confidence interval is "A person can be a% confident that the true population parameter lies in it."

So the Interpretation for  above interval for $\mu$ :

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

Hence, the correct answer is (B) .