Question

According to the U.S Census Bureau, the average travel time to work in the U.S. is 25.4 minutes. a simple random sample of 10 people reported the following travel times, 10,11,9,24,37,22,75,23,51,48.

at a 5% significance level, do the data provide sufficient evidence to conclude that people in Bryan, tx spend less time traveling to work than the national average? calc the p-value.

a. p= 0.7838

b. p= 0.4325

c. p= 0.8217

d. p= 0.8067

e. p= 0.1933

Answer 1

Ho :   µ =   25.4                  
Ha :   µ <   25.4       (Left tail test)          
                          
Level of Significance ,    α =    0.050                  
sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   21.5510                  
Sample Size ,   n =    10                  
Sample Mean,    x̅ = ΣX/n =    31.0000                  
                          
degree of freedom=   DF=n-1=   9                  
                          
Standard Error , SE = s/√n =   21.5510   / √    10   =   6.8150      
t-test statistic= (x̅ - µ )/SE = (   31.000   -   25.4   ) /    6.8150   =   0.82
                                
p-Value   =   0.7838   [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value>α, Do not reject null hypothesis

                       Conclusion: There is not enough evidence to  conclude that people in Bryan, tx spend less time traveling to work than the national average