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.

Fast Food Drive-Through Service Times Listed below are drive-through service times (seconds) recorded at McDonald’s during dinner times (from Data Set 25 “Fast Food” in Appendix B). Assuming that dinner service times at Wendy’s have standard deviation\(\sigma = 55.93\,\,sec\), use a 0.01 significance level to test the claim that service times at McDonald’s have the same variation as service times at Wendy’s. Should McDonald’s take any action?

121 119 146 266 333 308 333 308

Drive-through service times are observed from McDonald’s.

The standard deviation of the dinner service times at Wendy’s is 55.93 sec.

The level of significance is 0.01.

The claim states that the variation of service times is the same for McDonald’s and Wendy’s.

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 actual standard deviation of the dinner times of McDonald’s.\(\)

From the claim, the null and alternative hypotheses are as follows.

\(\begin{array}{l}{H_0}:\sigma = 55.93\\{H_1}:\sigma \ne 55.93\end{array}\)

LetX be the simple random sample ofthe service times recorded at McDonald’s during dinner times.

121 119 146 266 333 308 333 308

The sample mean is computed as follows.

\(\begin{array}{c}\bar x = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{{\rm{n}}}\\ = \frac{{121 + 119 + ... + 308}}{8}\\ = 241.75\end{array}\)

Thus, the sample mean is 241.75 sec.

The sample standard deviation is calculated as follows.

\(\begin{array}{c}s = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{{\rm{n}} - 1}}} \\ = \sqrt {\frac{{{{\left( {121 - 241.75} \right)}^2} + {{\left( {119 - 241.75} \right)}^2} + ... + {{\left( {308 - 241.75} \right)}^2}}}{{8 - 1}}} \\ = 96.2404\end{array}\)

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

\(\begin{array}{c}{\chi ^2} = \frac{{\left( {{\rm{n}} - 1} \right) \times {s^2}}}{{{\sigma ^2}}}\\ = \frac{{\left( {10 - 1} \right) \times {{96.24}^2}}}{{{{55.93}^2}}}\\ = 20.726\end{array}\).

Thus, the value of the test statistic is 20.726.

The degree of freedom is as follows.

\(\begin{array}{c}df = n - 1\\ = 8 - 1\\ = 7\end{array}\)

For a two-sided hypothesis, divide the\(\alpha \)equally between two tails as\(\frac{\alpha }{2}\)and\(1 - \frac{\alpha }{2}\).

Mathematically,

\(\begin{array}{l}P\left( {{\chi ^2} < \chi _{\frac{\alpha }{2}}^2} \right) = 1 - \frac{\alpha }{2}\\P\left( {{\chi ^2} > \chi _{\frac{{0.01}}{2}}^2} \right) = 1 - \frac{{0.01}}{2}\\P\left( {{\chi ^2} > \chi _{\frac{{0.01}}{2}}^2} \right) = 0.995\end{array}\)

Also,

\(\begin{array}{l}P\left( {{\chi ^2} > \chi _{\frac{\alpha }{2}}^2} \right) = \frac{\alpha }{2}\\P\left( {{\chi ^2} > \chi _{\frac{{0.01}}{2}}^2} \right) = \frac{{0.01}}{2}\\P\left( {{\chi ^2} > \chi _{\frac{{0.01}}{2}}^2} \right) = 0.005\end{array}\)

Referring to the chi-square table, the critical value with 7 degrees of freedom corresponding to the right-tailed area of 0.005 is observed as 20.278.

Referring to the chi-square table, the critical value with 7 degrees of freedom corresponding to the right-tailed area of 0.995 is observed as 0.989.

The decision rule for the test is as follows.

If the test statistic lies between the critical values, reject the null hypothesis at the given level of significance. Otherwise, fail to reject the null hypothesis.

As it is observed that \({\chi ^2} = 20.726\, > 20.278\), the null hypothesis is rejected.

Thus, there is not enough evidence to support the claim thatservice times at McDonald’s have the same variation as service times at Wendy’s.

The variation between the service time at McDonald’s is 96.24 sec, which is significantly larger than that at Wendy’s. Thus, McDonald’s should consider taking some action.