Question

You are thinking about moving to Los Angeles to get a job, but want to know whether salaries are higher in LA than Sacramento. Luckily you were able to randomly select five people who moved from Sacramento to LA and find out their salaries in each place. Testing at the p<.05 level, are salaries higher in LA than Sacramento? The salaries you collected are as follows (in thousands, with Sacramento the first salary on the left of the parentheses, LA is the right salary): Person A (48, 52), Person B (50, 56), Person C (46, 46), Person D (58, 60), Person E (35, 40).

1. Restate the question:

   • Population 1:

   • Population 2 (comparison):

   • Research hypothesis:

   • Null hypothesis:

   • Determine the characteristics of the comparison distribution to compare to sample distribution:

     • μ 1 (Population 1):

     • N (Sample 1):

     • df (Sample 1):

     • μ2 (Population 2):

     • S22(Population 2):

     • S2 (Population 2):

     • μ2M (Population 2, distribution of means):

     • σ22M (Population 2, distribution of means):

     • σ2M (Population 2, distribution of means):

c. Determine the critical t-value (cutoff score) on the comparison distribution at which the null hypothesis should be rejected

d. Determine the t score of your sample

e.Decide whether to reject the null hypothesis

f. Explain the findings to your coworkers who are also thinking about moving

Answer 1

mean of Sacramento, x1-bar:     47.4
standard deviation of Sacramento, s1:   8.294576541
size of Sacramento, n1:     5
mean of LA, x2-bar:     50.8
standard deviation of LA, s2:   7.949842766
difference in sample means, x1-bar - x2-bar:   -3.4

Ho :   µd=   0                  
Ha :   µd <   0                  
                          
Level of Significance ,    α =    0.05       claim:µd=0          
                          
sample size ,    n =    5                  
                          
mean of Sacramento,    x̅1=   47.400                  
                          
mean of LA,    x̅2=   50.800                  
                          
mean of difference ,    D̅ =ΣDi / n =   -3.400                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    2.4083                  
                          
std error , SE = Sd / √n =    2.4083   / √   5   =   1.0770      
                          
t-statistic = (D̅ - µd)/SE = (   -3.4   -   0   ) /    1.0770   =   -3.157
                          
Degree of freedom, DF=   n - 1 =    4                  
t-critical value , t* =        -2.132   [excel function: =t.inv(α,df) ]              
                          
p-value =        0.017144   [excel function: =t.dist(t-stat,df) ]              
Conclusion:     p-value <α , Reject null hypothesis              

Yes the salary has increased when one moved to LA   

Please revert back in case of any doubt.

Please upvote. Thanks in advance.