Question

The Municipal Transit Authority wants to know if, on weekdays, more passengers ride the northbound blue line train towards the city center that departs at 8:30 a.m. than the one that departs at 8:15 a.m. The following sample statistics are assembled by the Transit Authority.n x(bar) s8:15 a.m. train 30 323 passengers 418:30 a.m. train 45 356 passengers 45First, construct the 90% confidence interval for the difference in the mean number of daily travelers on the 8:15 train and the mean number of daily travelers on the 8:30 train:Next, test at the 5% level of significance whether the data provide sufficient evidence to conclude that more passengers ride the 8:30 train:State null and alternative hypotheses:Determine distribution of test statistic and compute its value:Construct the rejection region:Make your decision:State your conclusion:Compute the p-value (observed level of significance) for this test:

Answer 1

Answer:

The 90% confidence interval  $-49.8$

The null hypothesis is  $H_o : \mu_1 = \mu_2$

The alternative hypothesis $H_a : \mu_1 < \mu_2$

The distribution test statistics is $t = -3.222$

The rejection region is  p-value < $\alpha$

The decision rule is reject the null hypothesis

The conclusion is

      There is sufficient evidence to conclude that there are more passengers riding the 8:30  train

The p-value  is  $p-value =0.000951$

Step-by-step explanation:

From the question we are told that

    The first sample size $n_1 = 30$

    The first sample mean is  $\= x _1 = 323$

    The first standard deviation is $s_1 = 41$

    The second sample size is $n_2 = 45$

    The second sample mean is  $\=x_2 = 356$

    The second standard deviation is $s_2 = 45$

given that the confidence level is 90% then the level of significance is mathematically represented as

          $\alpha = (100 -90)\%$

         $\alpha = 0.10$

Generally the critical value of $\frac{\alpha }{2}$ obtained from the normal distribution table is  

   $Z_{\frac{\alpha }{2} } = 1.645$

Generally the pooled variance is mathematically represented as

        $s^2 = \frac{(n_1 - 1)s_1^2 + (n_2 -1)s_2^2 }{n_1 + n_2 -2}$

      $s^2 = \frac{(30 -1)(41^2) + (45-1)45^2}{30+45 -2}$

     $s^2 = 1888.34$

Generally the standard error is mathematically represented as

     $SE = \sqrt{\frac{s^2}{n_1} + \frac{s^2}{n_2} }$

=>  $SE = \sqrt{\frac{1888.34}{30} + \frac{1888.34}{45} }$

=>   $SE = 10.24$

Generally the margin of error is mathematically evaluated as  

      $E = Z_{\frac{\alpha }{2} } * SE$

       $E = 1.645* 10.24$

       $E = 16.85$

Generally the 90% confidence interval is mathematically represented as

     $\=x_1 -\=x_2 -E < \mu_1 -\mu_2 < \=x_1 -\=x_2 +E$

     $323 -356 -16.84$

     $-49.8$

The null hypothesis is  $H_o : \mu_1 = \mu_2$

The alternative hypothesis $H_a : \mu_1 < \mu_2$

Generally the test statistics is mathematically represented as

     $t = \frac{\= x_1 - \=x_2 }{SE}$

=>   $t = \frac{323-356}{10.24}$

=>   $t = -3.222$

Generally the degree of freedom is mathematically represented as

     $df = n_1+n_2 -2$

      $df = 30 + 45 -2$

     $df = 73$

The p-value is obtained from the student t distribution table at degree of freedom of 73 at 0.05 level of significance

    The value is  $p-value =0.000951$

Here the level of significance is  $\alpha = 5\% = 0.05$

Given that the p-value < $\alpha$ then we  reject the null hypothesis

Then the conclusion is  

  There is sufficient evidence to conclude that there are more passengers riding the 8:30  train