Question

In: Statistics and Probability

The mean salary of NBA players in 1996 was $3 million. You are interested to know...

The mean salary of NBA players in 1996 was $3 million. You are interested to know if the mean salary of NBA players increased between 1996 and 2006. A simple random sample of 90 professional basketball player salaries in 2006 was recorded.

salaries from 2006

1538400
2000000
8300000
2950000
3140000
6480000
8000000
398762
2120000
8400000
1100000
2750000
398762
5390000
1150000
1890000
6000000
398762
8890000
8100000
771123
398762
5200000
6270000
1270000
2330000
1020000
1140000
6000000
879360
1138500
10970000
880000
2500000
1540000
1670000
7350000
936600
3410000
4135200
5000000
398762
5900400
1538400
1250000
1538400
900498
7230000
5610000
2950000
15950000
641748
2950000
398762
398762
5610000
18612000
1700000
7500000
2420000
1680000
6120000
1512840
9630000
641748
6480000
835810
398762
997800
873880
398762
398762
3400000
900498
3000000
398762
4000000
6180000
690960
5000000
8250000
4620000
5390000
719373
3400000
4550000
13150000
5500000
398762
1130000

Calculate the test statistic, degrees of freedom, and P-value for a test of H0:μ=$3 million against Ha:μ>$3 million. Assume the requirements are satisfied. Input your answers below.

Which hypothesis test would be most appropriate for this study?

What is the test statistic?

What are the degrees of freedom?

What is the P-value? (Round to 3 decimal places). How do you find the p value?

Based on the results of this test, is there enough evidence to say that the mean salary of NBA players increased from 1996 to 2006? Use a level of significance of α=0.05.

Solutions

Expert Solution

Ho :   µ =   3000000                  
Ha :   µ >   3000000       (Right tail test)          
                          
Level of Significance ,    α =    0.05                  
sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   3571763.6499                  
Sample Size ,   n =    90                  
Sample Mean,    x̅ = ΣX/n =    3648318.6889                  
                          
degree of freedom=   DF=n-1=   89                 
                          
Standard Error , SE = s/√n =   3571763.6499   / √    90   =   376496.9466      
t-test statistic= (x̅ - µ )/SE = (   3648318.689   -   3000000   ) /    376496.9466   =   1.72
                          
  
                          
p-Value   =   0.0443   [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value<α, Reject null hypothesis                       
Conclusion: There is enough evidence that the mean salary of NBA players increased from 1996 to 2006.

  

Thanks in advance!

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orchestra answered 1 year ago
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