Step 1

Sample mean \(\displaystyle=\overline{{{X}}}={2.12}\)

Sample standard deviation \(\displaystyle{s}={0.51}\)

Sample size \(\displaystyle{n}={30}\)

Level of significance \(\displaystyle={0.01}\)

Degree of freedom \(\displaystyle={n}-{1}\)

\(\displaystyle={30}-{1}={29}\)

Critical value \(\displaystyle={t}_{{\frac{\alpha}{{2}};{d}{f}}}={t}_{{{0.005},\ {29}}}={2.76}\)

Step 2

a) To find \(\displaystyle{99}\%\) confidence level

\(\displaystyle{C}.{I}.=\overline{{{X}}}\pm{t}_{{\frac{\alpha}{{2}};{d}{f}}}\times{\frac{{{s}}}{{\sqrt{{{n}}}}}}\)

\(\displaystyle={2.12}\pm{2.76}\times{\frac{{{0.51}}}{{\sqrt{{{30}}}}}}\)

\(\displaystyle={2.12}\pm{0.2570}\)

\(\displaystyle={\left({1.863},\ {2.377}\right)}\)

Lower limit \(\displaystyle={1.863}\)

Upper limit \(\displaystyle={2.377}\)

Step 3

b) Option A is the correct answer

We can be 99% confident that the mean gross earnings for this sample of 30 Rolling Stones concerts lies in the interval \(\displaystyle{\left({1.863},\ {2.377}\right)}\)

Sample mean \(\displaystyle=\overline{{{X}}}={2.12}\)

Sample standard deviation \(\displaystyle{s}={0.51}\)

Sample size \(\displaystyle{n}={30}\)

Level of significance \(\displaystyle={0.01}\)

Degree of freedom \(\displaystyle={n}-{1}\)

\(\displaystyle={30}-{1}={29}\)

Critical value \(\displaystyle={t}_{{\frac{\alpha}{{2}};{d}{f}}}={t}_{{{0.005},\ {29}}}={2.76}\)

Step 2

a) To find \(\displaystyle{99}\%\) confidence level

\(\displaystyle{C}.{I}.=\overline{{{X}}}\pm{t}_{{\frac{\alpha}{{2}};{d}{f}}}\times{\frac{{{s}}}{{\sqrt{{{n}}}}}}\)

\(\displaystyle={2.12}\pm{2.76}\times{\frac{{{0.51}}}{{\sqrt{{{30}}}}}}\)

\(\displaystyle={2.12}\pm{0.2570}\)

\(\displaystyle={\left({1.863},\ {2.377}\right)}\)

Lower limit \(\displaystyle={1.863}\)

Upper limit \(\displaystyle={2.377}\)

Step 3

b) Option A is the correct answer

We can be 99% confident that the mean gross earnings for this sample of 30 Rolling Stones concerts lies in the interval \(\displaystyle{\left({1.863},\ {2.377}\right)}\)