Question

Keep on Rolling. The Rolling Stones, a rock group formed in the $1960\mathrm{s}$, have toured extensively in support of new albums, Pollstar has collected data on the earnings from the Stones's North American tours. For thirty randomly selected Rolling Stones concerts, the mean gross earnings is $\backslash (2.27$ million. Assuming a population standard deviation gross earnings of $\backslash )0.5$ million, obtain a $99\%$ confidence interval for the mean gross earnings of all Rolling Stones concerts. Interpret your answer in words.

Short answer

$99\%$confidence level of$\mu$is$(2.0349,2.5051)$

Step-by-step solution

Given Information

It is given that

$\sigma =0.5\text{million}$

$n=30$

$\overline{x}=2.27$

Determination of Confidence Level

It is given that sample mean earning is $\overline{x}=2.27$

Now,

Confidence level for $90\%$confidence interval of $\mu$$=99\%=100\times 0.99\%$

$\Rightarrow 1-\alpha =0.99$

$\Rightarrow \alpha =0.01$

Hence,$z_{\frac{a}{2}}=z_{\frac{0.01}{2}}=z_{0.005}=2.575$

Determination of 99% confidence interval of μ

It is given by formula

$\overline{x}-z_{0.005}\times \frac{\sigma }{\sqrt{n}},\overline{x}+z_{0.002}\times \frac{\sigma }{\sqrt{n}}$

$=2.27-2.575\times \frac{0.5}{\sqrt{30}},2.27+2.575\times \frac{0.5}{\sqrt{30}}$

$=2.27-0.2351,2.27+0.2351$

$=2.0349,2.5051$

Therefore,$99\%$confidence interval of$\mu$is$\left(2.0349,2.5051\right)$