91Ó°ÊÓ

The data for the times (sec) for the discrepancy between the real time and the time indicated on the wristwatch is provided.

The level of confidence is 95%.

Let x represent thetimes (sec) for the discrepancy between the real time and the time indicated on the wristwatch.

The sample mean is computed as follows:

\(\begin{array}{c}\bar x = \frac{{\sum x }}{n}\\ = \frac{{ - 85 + 325 + 20 + 305 - 93 + ... + 36}}{{12}}\\ = 143\end{array}\)

Therefore, the sample mean is 143.

The sample standard deviation is computed as follows:

\(\begin{array}{c}{\bf{s = }}\sqrt {\frac{{\sum {{{\left( {{\bf{x - \bar x}}} \right)}^{\bf{2}}}} }}{{{\bf{n - 1}}}}} \\ = \sqrt {\frac{{{{\left( { - 85 - 143} \right)}^2} + {{\left( {325 - 143} \right)}^2} + ... + {{\left( {36 - 143} \right)}^2}}}{{12 - 1}}} \\ = 259.775\end{array}\)

Therefore, the standard deviation is 259.775.

The level of confidence is 95%, whichimplies the level of significance is 0.05.

The degrees of freedom are computed as follows:

\(\begin{array}{c}df = n - 1\\ = 12 - 1\\ = 11\end{array}\)

From the ttable, the critical value at 11 degrees of freedom and at 0.05 level of significance is 2.201.

The margin of error (E) is computed as follows:

\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}}\frac{s}{{\sqrt n }}\\ = 2.201 \times \frac{{259.775}}{{\sqrt {12} }}\\ = 165.054\end{array}\)

Therefore, the margin of error is 165.054.

The 95% confidence interval is given as follows:

\(\begin{array}{c}\left( {{\bf{\bar x - E,\bar x + E}}} \right) = \left( {143 - 165.054,143 + 165.054} \right)\\ = \left( { - 22.054,308.054} \right)\\ \approx \left( { - 22.1,308.1} \right)\end{array}\)

Therefore, the 95% confidence interval estimate of the mean discrepancy for the population of wristwatches is\(\left( { - 22.1\;sec,308.1\;sec} \right)\).