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 (in years) are listed below. Use a 0.05

significance level to test the claim that in Dublin, car ages and taxi ages have the same variation.

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

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

It is claimed that the variation in the car ages is equal to the variation in the taxi ages.

Let\({\sigma _1}\)and\({\sigma _2}\)be the population standard deviations of thecar ages and the taxi ages,respectively.

Nullhypothesis: The population standard deviation of the car ages is equal to the population standard deviation of the taxi ages.

Symbolically,

\({H_0}:{\sigma _1} = {\sigma _2}\)

Alternativehypothesis: The population standard deviation of the car ages is not equal to the population standard deviation of the taxi ages.

Symbolically,

\({H_1}:{\sigma _1} \ne {\sigma _2}\)

The sample variance has the following formula:

\({s^2} = \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {x - \bar x} \right)}^2}} \)

The sample mean age of carsequals the following:

\(\begin{array}{c}{{\bar x}_1} = \frac{{4 + 0 + ....... + 8}}{{27}}\\ = 5.56\end{array}\)

The sample variance of car ages is computed below:

\(\begin{array}{c}s_{car}^2 = \frac{{\sum\limits_{i = 1}^n {{{({x_i} - {{\bar x}_1})}^2}} }}{{{n_1} - 1}}\\ = \frac{{{{\left( {4 - 5.56} \right)}^2} + {{\left( {0 - 5.56} \right)}^2} + ....... + {{\left( {8 - 5.56} \right)}^2}}}{{27 - 1}}\\ = 15.03\end{array}\)

Thus, the sample variance of car ages is equal to 15.03 years squared.

The sample mean age of taxisis the following:

\(\begin{array}{c}{{\bar x}_1} = \frac{{8 + 8 + ....... + 0}}{{27}}\\ = 5.85\end{array}\)

The sample variance of taxi ages is computed below:

\(\begin{array}{c}s_{taxi}^2 = \frac{{\sum\limits_{i = 1}^n {{{({x_i} - {{\bar x}_2})}^2}} }}{{{n_2} - 1}}\\ = \frac{{{{\left( {8 - 5.85} \right)}^2} + {{\left( {8 - 5.85} \right)}^2} + ....... + {{\left( {0 - 5.85} \right)}^2}}}{{20 - 1}}\\ = 8.03\end{array}\)

Thus, the sample variance of taxi ages is equal to 8.03 years squared.

Since two independent samples involve a claim about the population standard deviation, apply an F-test.

Consider the larger sample variance to be\(s_1^2\)and the corresponding sample size to be\({n_1}\).

Here,\(s_1^2\)is the sample variance corresponding to car ages and has a value equal to 15.03 years squared.

\(s_2^2\)is the sample variance corresponding to taxi ages and has a value equal to 8.03 years squared.

Substitute the respective values to calculate the F statistic:

\(\begin{array}{c}F = \frac{{s_1^2}}{{s_2^2}}\\ = \frac{{15.03}}{{8.03}}\\ = 1.871\end{array}\)

Thus, F is equal to 1.871.

The value of the numerator degrees of freedomequals the following:

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

The value of the denominator degrees of freedomequals the following:

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

For the F test, the critical value corresponding to the right-tail is considered.

The critical value can be obtained using the F-distribution table with numerator degrees of freedom equal to 26 and denominator degrees of freedom equal to 19 for a right-tailed test.

The level of significance is equal tothe following:

\(\begin{array}{c}\frac{\alpha }{2} = \frac{{0.05}}{2}\\ = 0.025\end{array}\)

Thus, the critical value is equal to 2.43.

The two-tailed p-value for F equal to 2.927 is equal to 0.1629.

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

Thus, there is not enough evidence to rejectthe claimthat car ages and taxi ages have the same variation.