In Exercises 9–16, assume that each sample is a simple

random sample obtained from a population with a normal distribution.

Comparing Waiting Lines

a. The values listed below are waiting times (in minutes) of customers at the Jefferson Valley Bank, where customers enter a single waiting line that feeds three teller windows. Construct a95% confidence interval for the population standard deviation .

6.5 6.6 6.7 6.8 7.1 7.3 7.4 7.7 7.7 7.7

b. The values listed below are waiting times (in minutes) of customers at the Bank of Providence, where customers may enter any one of three different lines that have formed at three teller windows. Construct a 95% confidence interval for the population standard deviation .

4.2 5.4 5.8 6.2 6.7 7.7 7.7 8.5 9.3 10.0

c. Interpret the results found in parts (a) and (b). Do the confidence intervals suggest a difference in the variation among waiting times? Which arrangement seems better: the single-line system or the multiple-line system?

The number of customers in a single waiting line is $n=10$.

The number of customers in multiple waiting linesis $n=10$.

The level of confidence is 95%.

The degrees of freedom are computed as follows:

$$\begin{gathered} df=n-1 \\ =10-1 \\ =9 \end{gathered}$$

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

Using the Chi-square table, the critical values at 0.05 level of significance and 9 degrees of freedom are ${\mathit{\chi }}_{\mathbf{L}}^{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{7004}$and ${\mathit{\chi }}_{\mathbf{R}}^{\mathbf{2}}\mathbf{=}\mathbf{19}\mathbf{.}\mathbf{0228}$.

a.

Let x represents the waiting times (in minutes) of customers in a single waiting line.

The mean value is computed asfollow:

$$\begin{gathered} \overline{x}=\frac{\sum x}{n} \\ =\frac{6.6+6.6+6.7+...+7.7+7.7}{10} \\ =7.15 \end{gathered}$$

The standard deviation is computed asfollow:

role="math" localid="1648116547536" $$\begin{gathered} \mathit{s}\mathbf{=}\sqrt{\frac{\mathbf{\sum }{\mathbf{(}\mathbf{x}\mathbf{-}\overline{\mathbf{x}}\mathbf{)}}^{\mathbf{2}}}{\mathbf{n}\mathbf{-}\mathbf{1}}} \\ =\sqrt{\frac{{\left(6.5-7.15\right)}^2+{\left(6.6-7.15\right)}^2+{\left(6.7-7.15\right)}^2+...+{\left(7.7-7.15\right)}^2}{10-1}} \\ =0.477 \end{gathered}$$

The 95% confidence interval estimate of the standard deviation of the population from which the sample was obtained is computed as follow:

$$\begin{gathered} \sqrt{\frac{\left(n-1\right){\mathbf{s}}^{\mathbf{2}}}{{\mathbf{\chi }}_{\mathbf{R}}^{\mathbf{2}}}}\mathbf{<}\mathit{\sigma }\mathbf{<}\sqrt{\frac{\left(n-1\right){\mathbf{s}}^{\mathbf{2}}}{{\mathbf{\chi }}_{\mathbf{L}}^{\mathbf{2}}}} \\ \sqrt{\frac{\left(10-1\right){0.477}^2}{19.0228}}<\sigma <\sqrt{\frac{\left(10-1\right){0.477}^2}{2.7004}} \\ 0.328<\sigma <0.871 \\ 0.3<\sigma <0.9 \end{gathered}$$

Therefore, the 95% confidence interval estimate of the standard deviation of the population from which the sample was obtained is role="math" localid="1648116695167" $\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{<}\mathit{\sigma }\mathbf{<}\mathbf{0}\mathbf{.}\mathbf{9}$.

b.

Let y represents the sample observations.

The mean value is computed as follows:

$$\begin{gathered} \overline{y}=\frac{\sum y}{n} \\ =\frac{4.2+5.4+5.8+...+9.3+10}{10} \\ =7.15 \end{gathered}$$

The standard deviation is computed asfollow:

$$\begin{gathered} s=\sqrt{\frac{\sum {\left(x-\overline{x}\right)}^2}{n-1}} \\ =\sqrt{\frac{{\left(4.2-7.15\right)}^2+{\left(5.4-7.15\right)}^2+{\left(5.8-7.15\right)}^2+...+{\left(10-7.15\right)}^2}{10-1}} \\ =1.822 \end{gathered}$$

The 95% confidence interval estimate of the standard deviation of the population from which the sample was obtained is computed as follows:

c.

Single line: We are 95% of the time confident that the true population standard deviation of waiting time of customers in single line will lie between the values 0.3 minutes and 0.9 minutes.

Multiple lines: We are 95% of the times confident that the true population standard deviation of waiting time of customers in multiple lines will lies between the values 1.3 minutes and 3.3 minutes.

It can be observed that the variation of the confidence interval constructed in part (b) is more as compared to the confidence interval constructed in part (a) as the confidence interval is wider in part (b).

Yes, the confidence intervals suggest a difference in the variation among waiting timesas both estimated confidence intervals are not overlapping each other

A single line system is considered to be better than multiple line system as the variability in waiting time is less in single line.