Keep on Rolling. In Exercise $8.74$, you found a $99\%$ confidence interval of $\backslash (2.03$ million to $\backslash )2.51$ million for the mean gross earnings of all Rolling Stones concerts.

a. Determine the margin of error, $E$

b. Explain the meaning of $E$ in this context in terms of the accuracy of the estimate.

c. Find the sample size required to have a margin of error of $\backslash (0.1$ million and a $95\%$ confidence level. (Recall that $\sigma =\backslash )0.5$ million.)

d. Obtain a $95\%$ confidence interval for the mean gross earnings if a sample of the size determined in part (c) has a mean of $\$2.35$ million.

The mean gross earnings of all Rolling Stones performances have a $99\%$ confidence interval of $\$2.03$ million to $\$2.51$ million.

The formula used:$n={\left(\frac{Z_{\frac{a}{2}}\sigma }{E}\right)}^2$and$n={\left\{\frac{Z_{0.025}\times \sigma }{E}\right\}}^2$

The $99\%$ confidence interval for all Rolling Stones concerts' mean gross earnings, $\mu$ is $\$2.03$ million to $\$2.51$ million.

$\therefore$Length of the confidence interval

$=\$2.51$million $-\$2.03$million

$=\$0.48$million

$\therefore$The margin of error $=\frac{1}{2}\times$length of the C.I for $\mu$$=\frac{1}{2}\times \$0.48$million

$=\$0.24$million

The margin of error $E=\$0.24$ million for a $99\%$confidence interval of the population mean gross earnings $\mu$indicates that we are $99\%$sure that the error in estimating population mean $\mu$by sample mean $\overline{x}$ is at most $\$0.24$ million.

Or, to put it another way, if we take numerous random samples of size $n$ from a population with a mean of $\mu$we may anticipate around $99\%$of the sample means to depart from $\mu$by at most $\$0.24$million.

The sample size $n$required for a $100(1-\alpha )\%$confidence interval for $\mu$with a certain margin of error $\left(E\right)$is calculated as follows:

$n={\left(\frac{Z_{\frac{a}{2}}\sigma }{E}\right)}^2$

Here confidence level $=95\%$

$$\begin{gathered} =100\times 0.95\% \\ \therefore 1-\alpha =0.95 \\ \Rightarrow \alpha =0.05 \\ \Rightarrow \frac{\alpha }{2}=0.025 \\ \therefore Z_{\frac{a}{2}}=Z_{0.025}=1.96 \end{gathered}$$

The margin of error, $E=\$0.1$ million

Population S.D $\sigma =\$0.5$million

$\therefore n={\left\{\frac{Z_{0.025}\times \sigma }{E}\right\}}^2$$={\left\{\frac{1.96\times 0.5}{0.1}\right\}}^2$$=96.04$$\approx 97$

$\therefore$The required sample size is $97$

For the population mean gross earnings $\mu$we need a $95$percent confidence interval.

Given that,

Sample mean $\overline{x}=2.35$

Sample size $n=97$

$95\%$confidence interval is given by,

$\left(\overline{x}-Z_{0.025}\frac{\sigma }{\sqrt{n}},\overline{x}+Z_{0.025}\frac{\sigma }{\sqrt{n}}\right)=(\overline{x}-E,\overline{x}+E)$

Where margin of error $E=Z_{0.025}\frac{\sigma }{\sqrt{n}}$

$$\begin{gathered} \therefore E=Z_{0.025}\frac{\sigma }{\sqrt{n}} \\ =\frac{1.96\times 0.5}{\sqrt{97}} \\ =0.0995 \end{gathered}$$

$\therefore 95\%$confidence interval of $\mu$

$$\begin{gathered} =(\overline{x}-E,\overline{x}+E) \\ =(2.35-0.0995,2.35+0.0995) \\ =(2.2505,2.4495) \end{gathered}$$