In order to test whether average commuting time in city A is different from city

Question

In order to test whether average commuting time in city A is different from city B, 10 households were chosen from city A and 10 households were chosen from city B and their commuting times were recorded. The average commuting time for 10 commuters (x bar_1) in city A was 10.12 minutes with sample standard deviation (s_1) being 4.90 minutes. The average commuting time for 10 commuters (x bar_2) in city B was 18.78 minutes with sample standard deviation (s_2) being 4.64 minutes. Assuming normality of commuting time, do the data provide sufficient evidence to conclude the population mean commuting time in city A is different from population mean commuting time in city B at 5% level of significance? (Population standard deviations of commuting time in both cities are not equal)

Explanation / Answer

Data:     

n1 = 10    

n2 = 10    

x1-bar = 10.12    

x2-bar = 18.78    

s1 = 4.9    

s2 = 4.64    

Hypotheses:     

Ho: 1 = 2     

Ha: 1 2     

Decision Rule:     

=   0.05   

Degrees of freedom =    17.9467584

Lower Critical t- score = -2.109815559    

Upper Critical t- score = 2.109815559    

Reject Ho if |t| > 2.109815559    

Test Statistic:     

SE = {(s1^2 /n1) + (s2^2 /n2)} = (((4.9^2)/10) + ((4.64^2)/10)) = 2.134000937   

t = (x1-bar -x2-bar)/SE = (10.12 - 18.78)/2.13400093720692 = -4.058105059   

p- value = 0.000817572    

Decision (in terms of the hypotheses):     

Since 4.058105059 > 2.10981556 we reject Ho and accept Ha

Conclusion (in terms of the problem):

There is sufficient evidence that 1 2.   

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