Question

Conduct a test of hypothesis to determine whether the mean salary (Salary column) of the teams was different than $75.0 million. Use the 5% level of significance. Note: We do not know the populations standard deviation.

Salary
93.6
143
108.7
61.7
95.2
67.1
103.9
71.4
189.6
79.4
106.5
24.1
68.3
89.1
66.2
87.3
99.7
68.9
54.4
30.5
87.8
108.5
71
115.2
89.4
38.5
58.1
90.2
90.3
37.3

Select one:

a. p = 19.2%. Do not reject the null. The mean salary could be $75 million.

b. p = 19.2%. Reject the null. The mean salary is different than $75 million.

c. p = 18.16%. Reject the null. The mean salary is different than $75 million.

d. p = 18.16%. Do not reject the null. The mean salary could be $75 million.

Answer 1

We do not know the populations standard deviation.

So we use t-score here , z-score is not appropriate .

Given data is as follow

Salary
93.6
143
108.7
61.7
95.2
67.1
103.9
71.4
189.6
79.4
106.5
24.1
68.3
89.1
66.2
87.3
99.7
68.9
54.4
30.5
87.8
108.5
71
115.2
89.4
38.5
58.1
90.2
90.3
37.3

Now we compute sample mean and sample standard deviation of above data

Formula :-

Sample Mean =

Standard deviation s =

After calculation we get

= = = 83.16333

s = = = 33.4724

Hence we have

= 83.16333     ;         s = 33.4724

n = 30    ( sample size )

Now , we wish test whether the mean salary (Salary column) of the teams was different than $75.0 million or not

To test :-

H0 : = 75      { mean salary may not be different from $75.0 million   }

H1 : 75      { mean salary is significantly different from $75.0 million   }

Test Statistics TS :-

TS =

TS =

TS =   1.335799

Thus calculated test statistics is TS = 1.335799

Now we calculate P-Value

Since alternative hypothesis is of " " type , so this is two-tail test , hence P-value will be given by

P-Value = Pr ( - |TS| < t ) + Pr ( |TS| > t )

P-Value = Pr ( t < - |1.335799| ) + Pr ( t > |1.335799| )

P-Value = Pr ( t < - 1.335799 ) + Pr ( t > 1.335799| )

P-Value = 2 * Pr ( t < - 1.335799 )                          [ due to symmetry ]

here t is t-distributed with n-1 = 30-1 = 29 degree of freedom , which is symmetrical distribution .

Now required probability " Pr ( t < - 1.335799 ) " , can be obtained from statistical book or more accuratly from any software like R/Excel

From R

> 2 * pt ( - 1.335799 ,df=29 )            # 2 * Pr ( t < - 1.335799 )  
[1] 0.1920043

Thus P-Value = 2 * Pr ( t < - 1.335799 )    = 0.1920043 0.192

Hence P-Value = 0.192

Rejection Criteria :-

We reject null hypothesis if P-Value is less than given level of significance .

Given 5% level of significance i.e = 0.05

Now   P-Value = 19.2 > 0.05

i.e P-Value >     ( = 0.05 )

So we do not reject null hypothesis .

Conclusion :-

Since we do not have enough evidence to reject null hypothesis H0 , we conclude that mean salary (Salary column) of the teams may not be different from $75.0 million

So correct option is

a. p = 19.2%. Do not reject the null. The mean salary could be $75 million.