Question

In: Statistics and Probability

A comparison is made between two bus lines to determine if arrival times of their regular...

A comparison is made between two bus lines to determine if arrival times of their regular buses from Denver to Durango are off schedule by the same amount of time. For 51 randomly selected runs, bus line A was observed to be off schedule an average time of 53 minutes, with standard deviation 19 minutes. For 60 randomly selected runs, bus line B was observed to be off schedule an average of 61 minutes, with standard deviation 13 minutes. Do the data indicate a significant difference in average off-schedule times? Use a 5% level of significance.

What are we testing in this problem?

paired differencedifference of means    single meandifference of proportionssingle proportion

What is the level of significance?


State the null and alternate hypotheses.

H0: μ1μ2; H1: μ1 > μ2H0: μ1μ2; H1: μ1 = μ2    H0: μ1μ2; H1: μ1 < μ2H0: μ1 = μ2; H1: μ1μ2


What sampling distribution will you use? What assumptions are you making?

The Student's t. We assume that both population distributions are approximately normal with unknown population standard deviations.The standard normal. We assume that both population distributions are approximately normal with unknown population standard deviations.    The Student's t. We assume that both population distributions are approximately normal with known population standard deviations.The standard normal. We assume that both population distributions are approximately normal with known population standard deviations.


What is the value of the sample test statistic? (Test the difference μ1μ2. Round your answer to three decimal places.)


Estimate the P-value.

P-value > 0.5000.250 < P-value < 0.500    0.100 < P-value < 0.2500.050 < P-value < 0.1000.010 < P-value < 0.050P-value < 0.010


Sketch the sampling distribution and show the area corresponding to the P-value.


Will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α?

At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.    At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.


Interpret your conclusion in the context of the application.

There is sufficient evidence at the 0.05 to conclude that there is a difference in average off schedule times.There is insufficient evidence at the 0.05 to conclude that there is a difference in average off schedule times.   

Solutions

Expert Solution

difference of means test

...............

H0: μ1 = μ2

H1: μ1 ≠ μ2

....................

The Student's t. We assume that both population distributions are approximately normal with unknown population standard deviations

.............

Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   53.00                  
standard deviation of sample 1,   s1 =    19.00                  
size of sample 1,    n1=   51                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   61.00                  
standard deviation of sample 2,   s2 =    13.00                  
size of sample 2,    n2=   60                  
                          
difference in sample means =    x̅1-x̅2 =    53.0000   -   61.0   =   -8.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    16.0335                  
std error , SE =    Sp*√(1/n1+1/n2) =    3.0537                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -8.0000   -   0   ) /    3.05   =   -2.620

...................


                          
Degree of freedom, DF=   n1+n2-2 =    109                  
  
p-value =        0.010053   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value <α , Reject null hypothesis      

...................

There is sufficient evidence at the 0.05 to conclude that there is a difference in average off schedule times

...........................


THANKS

revert back for doubt

please upvote


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