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

Answer:

The 99% confidence interval for the mean gross earnings of all Rolling Stones concerts (in millions) is (2.5690, 3.0110).

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.005 = 0.995$, so $z = 2.575$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.575*\frac{0.47}{\sqrt{30}} = 0.2210$

The lower end of the interval is the mean subtracted by M. So it is 2.79 - 0.2210 = 2.5690 million dollars.

The upper end of the interval is the mean added to M. So it is So it is 2.79 + 0.2210 = 3.0110 million dollars.

The 99% confidence interval for the mean gross earnings of all Rolling Stones concerts (in millions) is (2.5690, 3.0110).