A plant manager randomly divides her employees into two grou

A plant manager randomly divides her employees into two groups. One group (Group A) includes 32 employees and the second one (Group B) includes 30 employees. After dividing the employees, the manager wants to confirm that the two groups are indeed similar in performance. He hypothesizes that there is no statistically significant difference between the two groups. To compare the two groups, the plant manager use ratings given by the employees\' front-line supervisors at the end of the previous year. The rating scale ranges from 5 (excellent employee) to 1 (on the brink of termination). Using these ratings, the manager conducts a t test to determine whether the two groups are similar. The results are as follows:

Group

n

Mean

SD

t

A

32

3.66

1.31

2.008

B

30

3.00

1.26

t crit(.05,df)=2.0;

  t crit(.02,df)=2.390;

t crit(.01 ,df)=2.660

< >which t test was used and why?

< >what were the degrees of freedom?

Solution

A plant manager randomly divides her employees into two groups. One group (Group A) includes 32 employees and the second one (Group B) includes 30 employees. After dividing the employees, the manager wants to confirm that the two groups are indeed similar in performance. He hypothesizes that there is no statistically significant difference between the two groups. To compare the two groups, the plant manager use ratings given by the employees\' front-line supervisors at the end of the previous year. The rating scale ranges from 5 (excellent employee) to 1 (on the brink of termination). Using these ratings, the manager conducts a t test to determine whether the two groups are similar. The results are as follows:

Group

n

Mean

SD

t

A

32

3.66

1.31

2.008

B

30

3.00

1.26

t crit(.05,df)=2.0;

  t crit(.02,df)=2.390;

t crit(.01 ,df)=2.660

< >which t test was used and why?

Two group independent sample t test was used because two different groups A and B are compared.

< >what were the degrees of freedom?

degrees of freedom = n1+n2-2 = 32+30-2 = 60

The null hypothesis is rejected at 0.05 level.