Respuesta :
Answer:
The 90% confidence interval [tex]-49.8 <\mu_1 -\mu_2 < -16.16[/tex]
The null hypothesis is [tex]H_o : \mu_1 = \mu_2[/tex]
The alternative hypothesis [tex]H_a : \mu_1 < \mu_2[/tex]
The distribution test statistics is [tex]t = -3.222[/tex]
The rejection region is p-value < [tex]\alpha[/tex]
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 [tex]p-value =0.000951[/tex]
Step-by-step explanation:
From the question we are told that
The first sample size [tex]n_1 = 30[/tex]
The first sample mean is [tex]\= x _1 = 323[/tex]
The first standard deviation is [tex]s_1 = 41[/tex]
The second sample size is [tex]n_2 = 45[/tex]
The second sample mean is [tex]\=x_2 = 356[/tex]
The second standard deviation is [tex]s_2 = 45[/tex]
given that the confidence level is 90% then the level of significance is mathematically represented as
[tex]\alpha = (100 -90)\%[/tex]
[tex]\alpha = 0.10[/tex]
Generally the critical value of [tex]\frac{\alpha }{2}[/tex] obtained from the normal distribution table is
[tex]Z_{\frac{\alpha }{2} } = 1.645[/tex]
Generally the pooled variance is mathematically represented as
[tex]s^2 = \frac{(n_1 - 1)s_1^2 + (n_2 -1)s_2^2 }{n_1 + n_2 -2}[/tex]
[tex]s^2 = \frac{(30 -1)(41^2) + (45-1)45^2}{30+45 -2}[/tex]
[tex]s^2 = 1888.34[/tex]
Generally the standard error is mathematically represented as
[tex]SE = \sqrt{\frac{s^2}{n_1} + \frac{s^2}{n_2} }[/tex]
=> [tex]SE = \sqrt{\frac{1888.34}{30} + \frac{1888.34}{45} }[/tex]
=> [tex]SE = 10.24[/tex]
Generally the margin of error is mathematically evaluated as
[tex]E = Z_{\frac{\alpha }{2} } * SE[/tex]
[tex]E = 1.645* 10.24[/tex]
[tex]E = 16.85[/tex]
Generally the 90% confidence interval is mathematically represented as
[tex]\=x_1 -\=x_2 -E < \mu_1 -\mu_2 < \=x_1 -\=x_2 +E[/tex]
[tex]323 -356 -16.84 <\mu_1 -\mu_2 < 323 -356 +16.84[/tex]
[tex]-49.8 <\mu_1 -\mu_2 < -16.16[/tex]
The null hypothesis is [tex]H_o : \mu_1 = \mu_2[/tex]
The alternative hypothesis [tex]H_a : \mu_1 < \mu_2[/tex]
Generally the test statistics is mathematically represented as
[tex]t = \frac{\= x_1 - \=x_2 }{SE}[/tex]
=> [tex]t = \frac{323-356}{10.24}[/tex]
=> [tex]t = -3.222[/tex]
Generally the degree of freedom is mathematically represented as
[tex]df = n_1+n_2 -2[/tex]
[tex]df = 30 + 45 -2[/tex]
[tex]df = 73[/tex]
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 [tex]p-value =0.000951[/tex]
Here the level of significance is [tex]\alpha = 5\% = 0.05[/tex]
Given that the p-value < [tex]\alpha[/tex] 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
