Rock band The Rolling Stones have played scores of concerts in the ltwenty years

postillan4 2021-09-27 Answered

Rock band The Rolling Stones have played scores of concerts in the ltwenty years. For 30 randomly selected Rolling Stones concerts, the mean gross earnings is 2.12 million dollars.
a) Assuming a population standard deviation groos earnings of 0.51 million dollars, obtain a \(\displaystyle{99}\%\) confidence interval for the mean gross earning of all Rolling Stones concerts (in millions).
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 concert lies in the interval
B) If we repeat the study many times, \(\displaystyle{99}\%\) of the calculated confidence intervals will contain the mean gross earning of all Rolling Stones concerts.
C) There is a \(\displaystyle{99}\%\) chance that the mean gross earning of all Rolling Stones concert lies in the interval
D) None of the above

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Expert Answer

liannemdh
Answered 2021-09-28 Author has 11665 answers
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 \(\displaystyle{\left({1.863},\ {2.377}\right)}\)
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