Question

Wristwatch AccuracyStudents of the author collected data measuring the accuracy ofwristwatches. The times (sec) below show the discrepancy between the real time and the timeindicated on the wristwatch. Negative values correspond to watches that are running ahead ofthe actual time. The data appear to be from a normally distributed population. Construct a 95%

confidence interval estimate of the mean discrepancy for the population of wristwatches.

-85 325 20 305 -93 15 282 27 555 570 -241 36

Short answer

The 95% confidence interval estimate of the mean discrepancy for the population of wristwatches is (22.1 sec, 308.1 sec).

Step-by-step solution

Given information

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%.

Compute the mean and standard deviation

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:

$$\begin{gathered} \overline{\mathbf{x}}\mathbf{=}\frac{\mathbf{\sum }\mathbf{x}}{\mathbf{n}} \\ =\frac{-85+325+20+305-93+...+36}{12} \\ =143 \end{gathered}$$

Therefore, the sample mean is 143.

The sample standard deviation is computed as follows:

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

Therefore, the standard deviation is 259.775.

Compute the margin of error

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:

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

Therefore, the margin of error is 165.054.

Construct the confidence interval

The 95% confidence interval is given as follows:

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

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