Question

4. Hydrocarbons in the cab of an automobile were measured during trips on the New Jersey Turnpike and trips through the Lincoln Tunnel connecting New York and New Jersey. The concentrations (± standard deviations) of m- and p-xylene were: Turnpike: 31.4 ± 30.0 g/m3 (32 measurements) Tunnel: 52.9 ± 29.8 g/m3 (32 measurements) Do these results differ at the 95% confidence level?

Answer 1

Answer:

No these these result do not differ at 95% confidence level  

Step-by-step explanation:

From the question we are told that

  The first concentrations is  $c _1= 30.0 \ g/m^3$

      The second concentrations is  $c _2 = 52.9 \ g/m^3$

  The first sample size is  $n_1 = 32$

    The second sample size is  $n_2 = 32$

   The  first standard deviation is $\sigma_1 = 30.0$

     The  first standard deviation is $\sigma_1 = 29.0$

The mean for  Turnpike is  $\= x _1 = \frac{c_1}{n} = \frac{31.4}{32} = 0.98125$

The mean for   Tunnel is  $\= x _2 = \frac{c_2}{n} = \frac{52.9}{32} = 1.6531$

The  null hypothesis is  $H_o : \mu _1 - \mu_2 = 0$

The  alternative hypothesis is  $H_a : \mu _1 - \mu_2 \ne 0$

Generally the test statistics is mathematically represented as

              $t = \frac{\= x_1 - \= x_2}{ \sqrt{\frac{\sigma_1^2}{n_1} +\frac{\sigma_2^2}{n_2} }}$

         $t = \frac{0.98125 - 1.6531}{ \sqrt{\frac{30^2}{32} +\frac{29^2}{32} }}$

        $t = - 0.0899$

Generally the degree of freedom is mathematically represented as

     $df = 32+ 32 - 2$

     $df = 62$

The  significance $\alpha$  is  evaluated as

      $\alpha = (C - 100 )\%$

=>   $\alpha = (95 - 100 )\%$

=>   $\alpha =0.05$

The  critical value  is evaluated as

      $t_c = 2 * t_{0.05 , 62}$

From the student t- distribution table  

        $t_{0.05, 62} = 1.67$

So

     $t_c = 2 * 1.67$

=>  $t_c = 3.34$

given that

       $t_c > t$ we fail to reject the null hypothesis so  this mean that the result do  not differ