Wristwatch Accuracy Use the sample data from Exercise 7 “Wristwatch Accuracy” and construct a 95% confidence interval estimate of $\sigma$.

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 represents thetimes (sec) for the discrepancy between the real time and the time indicated on the wristwatch.

The sample mean is computed as,

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

Therefore, the sample mean is 143.

The sample standard deviation is computed as,

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

Therefore, the standard deviation is 259.775.

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

The degrees of freedom is computed as,

$$\begin{gathered} \mathit{d}\mathit{f}\mathbf{=}\mathit{n}\mathbf{-}\mathbf{1} \\ =12-1 \\ =11 \end{gathered}$$

From the ${\chi }^2$ table, the critical values at 11 degrees of freedom and at 0.05 level of significance are 3.8157 and 21.92.

The 95% confidence interval is given as,

$$\begin{gathered} \sqrt{\frac{\left(n-1\right){\mathbf{s}}^{\mathbf{2}}}{{\mathbf{\chi }}_{\mathbf{R}}^{\mathbf{2}}}}\mathbf{<}\mathit{\sigma }\mathbf{<}\sqrt{\frac{\left(n-1\right){\mathbf{s}}^{\mathbf{2}}}{{\mathbf{\chi }}_{\mathbf{L}}^{\mathbf{2}}}} \\ \sqrt{\frac{\left(12-1\right){259.775}^2}{21.92}}<\sigma <\sqrt{\frac{\left(12-1\right){259.775}^2}{3.8157}} \\ 184.0<\sigma <441.1 \end{gathered}$$

Therefore, the confidence interval estimate for $\sigma$ is $\left(184.0\sec ,441.1\sec \right)$.