Question

In Exercises 5–20, assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of\({n_1} - 1\)and\({n_2} - 1\).) Car and Taxi Ages When the author visited Dublin, Ireland (home of Guinness Brewery employee William Gosset, who first developed the t distribution), he recorded the ages of randomly selected passenger cars and randomly selected taxis. The ages can be found from the license plates. (There is no end to the fun of traveling with the author.) The ages (in years) are listed below. We might expect that taxis would be newer, so test the claim that the mean age of cars is greater than the mean age of taxis.

Car Ages	4	0	8	11	14	3	4	4	3	5	8	3	3	7	4	6	6	1	8	2	15	11	4	1	1	8
Taxi Ages	8	8	0	3	8	4	3	3	6	11	7	7	6	9	5	10	8	4	3	4

Short answer

There is insufficient evidence to support the claim that the mean age of passenger cars is greater than the mean age of taxis.

Step-by-step solution

Step 1: Given information

The given table contains the ages of randomly selected passenger cars and the ages of taxis.

Formulation of the hypotheses

Null hypothesis:The mean age of passenger cars is equal to the mean age of taxis.

\({H_0}\):\({\mu _1} = {\mu _2}\).

Alternative Hypothesis:The mean age of passenger cars is greater than the mean age of taxis.

\({H_1}\):\({\mu _1} > {\mu _2}\).

Calculation of sample means

The mean age of cars is equal to:

\(\begin{array}{c}{{\bar x}_1} = \frac{{\sum\limits_{i = 1}^{{n_1}} {{x_i}} }}{{{n_1}}}\\ = \frac{{4 + 0 + ..... + 8}}{{27}}\\ = 5.56\end{array}\)

Therefore, the mean age of cars is equal to 5.56 years.

The mean age of taxis is computed below:

\(\begin{array}{c}{{\bar x}_2} = \frac{{\sum\limits_{i = 1}^{{n_2}} {{x_i}} }}{{{n_2}}}\\ = \frac{{8 + 8 + .... + 4}}{{20}}\\ = 5.85\end{array}\)

Therefore, the mean age of taxis is equal to 5.85 years.

Calculation of sample standard deviations

The standard deviation of the ages of cars is computed as follows:

\(\begin{array}{c}{s_1} = \sqrt {\frac{{\sum\limits_{i = 1}^{{n_1}} {{{({x_i} - {{\bar x}_1})}^2}} }}{{{n_1} - 1}}} \\ = \sqrt {\frac{{{{\left( {4 - 5.56} \right)}^2} + {{\left( {0 - 5.56} \right)}^2} + .... + {{\left( {8 - 5.56} \right)}^2}}}{{27 - 1}}} \\ = 3.88\end{array}\)

Therefore, the standard deviation of the ages of cars is equal to 3.88 years.

The standard deviation for the ages of taxis is computed as follows:

\(\begin{array}{c}{s_2} = \sqrt {\frac{{\sum\limits_{i = 1}^{{n_2}} {{{({x_i} - {{\bar x}_2})}^2}} }}{{{n_2} - 1}}} \\ = \sqrt {\frac{{{{\left( {8 - 5.85} \right)}^2} + {{\left( {8 - 5.85} \right)}^2} + .... + {{\left( {4 - 5.85} \right)}^2}}}{{20 - 1}}} \\ = 2.83\end{array}\)

Therefore, the standard deviation of the ages of taxis is equal to 2.83.

Calculation of the test statistic

Under null hypothesis,\({\mu _1} - {\mu _2} = 0\).

The test statistic is equal to:

\(\begin{array}{c}t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}\\ = \frac{{\left( {5.56 - 5.85} \right) - \left( 0 \right)}}{{\sqrt {\frac{{{{\left( {3.88} \right)}^2}}}{{27}} + \frac{{{{\left( {2.83} \right)}^2}}}{{20}}} }}\\ = - 0.301\end{array}\)

The value of the test statistic is -0.301.

Computation of critical value

Degrees of freedom: The smaller of the two values\(\left( {{n_1} - 1} \right)\)and\(\left( {{n_2} - 1} \right)\)is considered as the degreesof freedom.

\(\begin{array}{c}\left( {{n_1} - 1} \right) = \left( {27 - 1} \right)\\ = 26\end{array}\)

\(\begin{array}{c}\left( {{n_2} - 1} \right) = \left( {20 - 1} \right)\\ = 19\end{array}\)

The value of the degrees of freedom is the minimum of (26,19) which is equal to19.

Now see the t-distribution table for the right-tailed test with 0.05 level of significance and for 19 degrees of freedom.

The critical value is equal to,\({t_{0.05,19}} = 1.729\). The corresponding p-value is equal to 0.6167.

The value of the test statistic is less than the critical value, and the p-value is greater than 0.05. Therefore, the null hypothesis is failed to reject.

Conclusion

There is not sufficient evidence to support the claim that the mean age of passenger cars is greater than the mean age of taxis.