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Taya2010 [7]
3 years ago
9

The Municipal Transit Authority wants to know if, on weekdays, more passengers ride the northbound blue line train towards the c

ity 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:
Mathematics
1 answer:
Kitty [74]3 years ago
5 0

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

               

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