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

To construct the confidence interval and perform the hypothesis test, we use formulas involving the sample means, standard deviations, and sample sizes of the two trains at different times. The confidence interval gives us a range estimate for the difference in mean number of daily travelers. The hypothesis test helps us determine if there is sufficient evidence to conclude that more passengers ride the 8:30 train. The p-value further supports our decision.

Step-by-step explanation:

To construct the 90% confidence interval for the difference in the mean number of daily travelers on the 8:15 a.m. train and the 8:30 a.m. train, we can use the formula:

Confidence Interval = (X1 - X2) ± t * sqrt(S1^2/n1 + S2^2/n2)

Where X1 and X2 are the sample means, S1 and S2 are the sample standard deviations, and n1 and n2 are the sample sizes. In this case, X1 = 323, X2 = 356, S1 = 41, S2 = 45, n1 = 30, and n2 = 45. By substituting these values into the formula and using a t-table or software to find the t-value for a 90% confidence level, we can calculate the confidence interval.

To test whether the data provide sufficient evidence to conclude that more passengers ride the 8:30 train, we can perform a two-sample t-test. The null hypothesis is that there is no difference in the mean number of daily travelers between the two trains, while the alternative hypothesis is that more passengers ride the 8:30 train. We calculate the test statistic using the formula:

Test Statistic = (X1 - X2) / sqrt(S1^2/n1 + S2^2/n2)

Where X1, X2, S1, S2, n1, and n2 have the same values as in the confidence interval calculation. By comparing the test statistic to the critical value for a 5% significance level, we can determine if there is sufficient evidence to reject the null hypothesis.

The rejection region depends on the alternative hypothesis and the specific software or t-table used for calculations. To make a decision, we compare the test statistic to the critical value and either reject or fail to reject the null hypothesis.

The p-value can be calculated using software or a t-table. It indicates the probability of obtaining a test statistic as extreme as the observed value, assuming the null hypothesis is true. A p-value less than the significance level (5% in this case) would lead to rejecting the null hypothesis.

Answer 2

The 90% confidence interval
`-49.8 <\mu_1 -\mu_2 < -16.16`

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
`(\alpha )/(2)` obtained from the normal distribution table is

`Z_{(\alpha )/(2) } = 1.645`

Generally the pooled variance is mathematically represented as

`s^2 = ((n_1 - 1)s_1^2 + (n_2 -1)s_2^2 )/(n_1 + n_2 -2)`

`s^2 = ((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{(s^2)/(n_1) + (s^2)/(n_2) }`

=>
`SE = \sqrt{(1888.34)/(30) + (1888.34)/(45) }`

=>
`SE = 10.24`

Generally the margin of error is mathematically evaluated as

`E = Z_{(\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 <\mu_1 -\mu_2 < 323 -356 +16.84`

`-49.8 <\mu_1 -\mu_2 < -16.16`

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 = (\= x_1 - \=x_2 )/(SE)`

=>
`t = (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