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. Use α= 0.05 significance level to test the claim that in Dublin, car ages and taxi ages have the same variation.

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 6 1 8
Taxi Ages 8 8 0 3 8 4 3 3 6 11 7 7 6 9 5 10 8 4 3 4

Answer:

The  Decision Rule

Fail to reject the null hypothesis

The conclusion

 There is no sufficient evidence to support the claim that the mean age of the cars is greater than that of taxi

Step-by-step explanation:

From the question we are told that

   The data is  

      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 6 1 8

     Taxi Ages 8 8 0 3 8 4 3 3 6 11 7 7 6 9 5 10 8 4 3 4

      The  level of significance [tex]\alpha = 0.05[/tex]

 Generally the null hypothesis  is  [tex]H_o  :  \mu_1 - \mu_2  = 0[/tex]

                  the alternative hypothesis is   [tex]H_a  :  \mu_1 - \mu_2 >  0[/tex]

Generally the sample mean for the age of  cars is mathematically represented as

        [tex]\= x_1 = \frac{\sum x_i }{n}[/tex]

=>     [tex]\= x_1 = \frac{4+ 0+ 8 +11 + \cdots + 8 }{27}[/tex]

=>     [tex]\= x_1 = 5.56[/tex]

Generally the standard deviation of age of  cars

     [tex]\sigma _1  = \sqrt{\frac{\sum (x_i - \= x)^2}{n_1} }[/tex]

=>  [tex]\sigma _1  = \sqrt{\frac{(4 - 5.56)^2 + (0 - 5.56)^2+ (8 - 5.56)^2 + \cdots + 8}{ 27} }[/tex]

=>  [tex]\sigma _1  =  3.88 [/tex]

Generally the sample mean for the age of taxi is mathematically represented as

        [tex]\= x_2 = \frac{\sum x_i }{n}[/tex]

=>     [tex]\= x_2 = \frac{8 +8 +0  + \cdots + 4 }{20}[/tex]

=>     [tex]\= x_2 = 5.85 [/tex]

Generally the standard deviation of age of  taxi

[tex]\sigma _2  = \sqrt{\frac{\sum (x_i - \= x)^2}{n_1} }[/tex]

=>  [tex]\sigma _2  = \sqrt{\frac{(8 - 5.85)^2 + (8 - 5.85)^2+ (0 - 5.85)^2 + \cdots + 8}{ 20} }[/tex]

=>  [tex]\sigma _2  = 2.83 [/tex]

Generally the test statistics is mathematically represented as

   [tex]t = \frac{(\= x_ 1 - \= x_2 ) - 0}{\sqrt{\frac{\sigma^2_1}{n_1}  + \frac{\sigma^2_2}{n_2} }  }[/tex]

=> [tex]t = \frac{(5.56 - 5.85 ) - 0}{\sqrt{\frac{(3.88)^2}{27}  + \frac{(2.83)}{20} }}[/tex]  

=> [tex]t = -0.30 [/tex]  

Generally the degree of freedom is mathematically  represented as

   [tex]df =  n_1 + n_2 -2[/tex]

    [tex]df =  27 +  20 -2[/tex]

    [tex]df =  45[/tex]

From the t distribution table  the [tex]P(t >  t )[/tex] at the obtained degree of freedom = 45 is  

   [tex]P(t >  -0.30 ) = 0.61722067[/tex]

So  the  p-value  is

    [tex]p-value  =  P(t >  T) =  0.61722067[/tex]

From the obtained values we see that the  p-value  >  [tex]\alpha [/tex] hence we fail to reject the null hypothesis

Hence the there is no sufficient evidence to support the claim that the mean age of the cars is greater than that of taxi

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