Question

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 \(\sigma \).

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 \(\sigma \).

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?

Short answer

a. The 95% confidence interval estimate of the standard deviation of the population for the customers waiting in the single line is\(0.3\;{\rm{minutes}} < \sigma < 0.9\;{\rm{minutes}}\).

b. The 95% confidence interval estimate of the standard deviation of the population for the customers waiting in the multiple-line is\(1.3\;{\rm{minutes}} < \sigma < 3.3\;{\rm{minutes}}\).

c.We are 95% of the time confident that the true population standard deviation of customers’waiting time inasingle line will lie between the valuesof0.3 minutes and 0.9 minutes.

We are 95% of the time confident that the true population standard deviation ofcustomers' waiting timein multiple lines will lie between the values 1.3 minutes and 3.3 minutes.

Yes, the confidence intervals suggest a differencea variation inwaiting times.

A single line system seems better than amultiple line system due to less variability in waiting times.

Step-by-step solution

Given information

The number of customers inasingle waiting line is\(n = 10\).

The number of customers in multiple waiting linesis\(n = 10\).

The level of confidence is 95%.

Compute the critical values and confidence interval

The degrees of freedom are computed asfollows:

\(\begin{array}{c}df = n - 1\\ = 10 - 1\\ = 9\end{array}\)

The level of confidence is 95%,whichimplies 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 \(\chi _L^2 = 2.7004\’) and \(\chi _R^2 = 19.0228\’).

Compute the mean and standard deviation

a.

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

The mean value is computed asfollow:

\(\begin{array}{c}\bar x = \frac{{\sum x }}{n}\\ = \frac{{6.6 + 6.6 + 6.7 + ... + 7.7 + 7.7}}{{10}}\\ = 7.15\end{array}\)

The standard deviation is computed asfollow:

\(\begin{array}{c}s = \sqrt {\frac{{\sum {{{\left( {x - \bar x} \right)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {6.5 - 7.15} \right)}^2} + {{\left( {6.6 - 7.15} \right)}^2} + {{\left( {6.7 - 7.15} \right)}^2} + ... + {{\left( {7.7 - 7.15} \right)}^2}}}{{10 - 1}}} \\ = 0.477\end{array}\)

Construct the confidence interval

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

\(\begin{array}{c}\sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _R^2}}} < \sigma < \sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _L^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{array}\’)

Therefore, the 95% confidence interval estimate of the standard deviation of the population from which the sample was obtained is\(0.3 < \sigma < 0.9\).

Compute the mean and standard deviation

b.

Let y represents the sample observations.

The mean value is computed asfollows:

\(\begin{array}{c}\bar y = \frac{{\sum y }}{n}\\ = \frac{{4.2 + 5.4 + 5.8 + ... + 9.3 + 10}}{{10}}\\ = 7.15\end{array}\’)

The standard deviation is computed asfollow:

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

Construct the confidence interval

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

\(\begin{array}{c}\sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _R^2}}} < \sigma < \sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _L^2}}} \\\sqrt {\frac{{\left( {10 - 1} \right){{1.822}^2}}}{{19.0228}}} < \sigma < \sqrt {\frac{{\left( {10 - 1} \right){{1.822}^2}}}{{2.7004}}} \\1.253 < \sigma < 3.326\\1.3 < \sigma < 3.3\end{array}\’)

Step 7: Compare the above results

Compare the above results

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.