Testing Claims About Variation. In Exercises 5–16, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Assume that a simple random sample is selected from a normally distributed population.

Bank Lines The Jefferson Valley Bank once had a separate customer waiting line at each teller window, but it now has a single waiting line that feeds the teller windows as vacancies occur. The standard deviation of customer waiting times with the old multiple-line configuration was 1.8 min. Listed below is a simple random sample of waiting times (minutes) with the single waiting line. Use a 0.05 significance level to test the claim that with a single waiting line, the waiting times have a standard deviation less than 1.8 min. What improvement occurred when banks changed from multiple waiting lines to a single waiting line?

6.5 6.6 6.7 6.8 7.1 7.3 7.4 7.7 7.7 7.7

The standard deviation of waiting times with the old multiple-line configuration is 1.8 min.

The observations are recorded for the single-line waiting times.

The level of significance is 0.05.

The waiting time for a single line is less than 1.8 min.

For applying the hypothesis test, first set up a null and an alternative hypothesis.

The null hypothesis is the statement about the value of a population parameter, which is equal to the claimed value. It is denoted by $H_0$.

The alternate hypothesis is a statement that the parameter has a value that is opposite to the null hypothesis. It is denoted by $H_1$.

Let $\sigma$be the standard deviation of waiting times for the single line.

As per the claim, the null and alternative hypotheses are as follows.

$$\begin{gathered} H_0:\sigma =1.8 \\ {H}_1:\sigma <1.8 \end{gathered}$$

The test is left-tailed.

LetX be the simplerandom sample of waiting times (minutes) with the single waiting lineas follows.

6.5 6.6 6.7 6.8 7.1 7.3 7.4 7.7 7.7 7.7

The samplemean is computed as follows.

$$\begin{gathered} \overline{x}=\frac{{\sum }_{i=1}^n{x}_i}{\text{n}} \\ =\frac{6.5+6.6+...+7.7}{10} \\ =7.15 \end{gathered}$$

The sample standard deviation is calculated as follows.

$$\begin{gathered} s=\sqrt{\frac{{{\sum }_{i=1}^n\left(x_i-\overline{x}\right)}^2}{\text{n}-1}} \\ =\sqrt{\frac{{\left(6.5-7.15\right)}^2+{\left(6.6-7.15\right)}^2+...+{\left(7.7-7.15\right)}^2}{10-1}} \\ =0.4767 \end{gathered}$$

To conduct a hypothesis test of a claim about a population standard deviation $\sigma$ or population variance${\sigma }^2$,the test statistics is computed as follows.

.$$\begin{gathered} {\chi }^2=\frac{\left(\text{n}-1\right)\times s^2}{{\sigma }^2} \\ =\frac{\left(10-1\right)\times {0.4767}^2}{{1.8}^2} \\ =0.6312 \end{gathered}$$

Thus, the value of the test statistic is 0.6312.

The degree of freedom is as follows.

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

The critical value ${\chi }_{0.05}^2$is obtained using the chi-square table as follows.

$$\begin{gathered} P\left({\chi }^2<{\chi }_{\alpha }^2\right)=\alpha \\ P\left({\chi }^2>{\chi }_{\alpha }^2\right)=1-\alpha \\ P\left({\chi }^2>{\chi }_{0.05}^2\right)=0.95 \end{gathered}$$

Referring to the chi-square table, the critical value, corresponding to the area of 0.05 and degree of freedom 9, is 3.325.

The decision rule for the test is as follows.

If ${\chi }^2<{\chi }_{0.05}^2$, reject the null hypothesis at a given level of significance. Otherwise, fail to reject the null hypothesis.

As it is observed that ${\chi }^2=0.6312 <{\chi }_{0.05}^2=3.325$, the null hypothesis is rejected.

Thus, there is enough evidence to support the claim that with a single waiting line, the waiting times have a standard deviation of less than 1.8 min.

The variation appears to be smaller in waiting times when a new single line is introduced than the old multiple-line configuration.It resulted in almost the same waiting time for each person.