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

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.

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

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.

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.

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.

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.

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