91影视

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:

$$\begin{gathered} \stackrel{\mathbf{炉}}{\mathbf{x}}\mathbf{=}\frac{\mathbf{鈭�}\mathbf{x}}{\mathbf{n}} \\ =\frac{-85+325+20+305-93+...+36}{12} \\ =143 \end{gathered}$$

Therefore, the sample mean is 143.

The sample standard deviation is computed as follows:

$$\begin{gathered} \mathbf{s}\mathbf{=}\sqrt{\frac{\mathbf{鈭�}{\left(x-\stackrel{炉}{x}\right)}^{\mathbf{2}}}{\mathbf{n}\mathbf{-}\mathbf{1}}} \\ =\sqrt{\frac{{\left(-85-143\right)}^2+{\left(325-143\right)}^2+...+{\left(36-143\right)}^2}{12-1}} \\ =259.775 \end{gathered}$$

Therefore, the standard deviation is 259.775.

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

The degrees of freedom are computed as follows:

$$\begin{gathered} df=n-1 \\ =12-1 \\ =11 \end{gathered}$$

From the t-table, 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:

$$\begin{gathered} E=t_{\frac{伪}{2}}\frac{s}{\sqrt{n}} \\ =2.201脳\frac{259.775}{\sqrt{12}} \\ =165.054 \end{gathered}$$

Therefore, the margin of error is 165.054.

The 95% confidence interval is given as follows:

$$\begin{gathered} \left(\stackrel{\mathbf{炉}}{\mathbf{x}}\mathbf{-}\mathbf{E}\mathbf{,}\stackrel{\mathbf{炉}}{\mathbf{x}}\mathbf{+}\mathbf{E}\right)=\left(143-165.054,143+165.054\right) \\ =\left(-22.054,308.054\right) \\ 鈮�\left(-22.1,308.1\right) \end{gathered}$$

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