Question

Kaneko has two groups. She randomly assigns subjects to her two groups. She has one group read a news article about the importance of a public speaking classes to one’s ability to get a job, while the other group reads a news article about a recent college football game that took place at her university. Then she asks both groups to rate how important they believe a public speaking class is to one’s ability to get a job(0-50). She thinks the groups will be different but does not make a specific prediction about the direction of the difference. She wants to use α = .01. Public SpeakingGroupFootball GroupMean4530s64n4035a.What test should Kaneko use?b.Write out Kaneko’s null and alternative hypotheses in formula form.c.What is the calculated value?d.What is the critical value using α = .01?e.Make a statistical and substantive conclusion for the above problem.f. Calculate eta-square and omega-square for this problem. Interpret what these numbers mean.

Answer 1

Claim : The means of the two groups are different.

First we need to run variance test to check the equality between the variances of two groups.

H0 : = vs Ha :   ≠ ​  

Test statistic F =   > here = 6² = 36 and = 4² = 16

F = 36/16 = 2.25

To find p value we can use excel function =2*FDIST(F,n₁,n₂) =2*FDIST( 2.25,39,34) = 0.018

We are given = 0.01 ,

Since p value ( 0.018) is greater than = 0.01 , we fail to reject H0, So the variances of the groups are equal   =   

So Kaneko should use 2 sample t test with pooled variance.

H0 : ₁ = ₂

Ha: ₁≠ ₂

Now we are given   = 45 , = 30 , s₁ = 6 , s₂ = 4 , n₁ = 40 and n₂ = 35

We can perform 2 sample t test with pooled variance using TI-84 calculator.

Press STAT key ---> Scroll to TESTS ---> Scroll down to 2-sample T test and hit enter.

Then scroll to Stats and hit enter , plug the given values accordingly , then select the appropriate sign under Ha

For pooled select Yes and hit enter.

Then scroll to calculate and hit enter.

Test statistics t = 12.55

We are given   = 0.01 , to find critical value we have to use excel function = TINV( probability , d.f)

probability = 0.01

d.f = n₁ = 40 + n₂ - 2 = 73

=TINV(0.01,73) = 2.645

Critical values = -2.645 and 2.645

Conclusion :

Since t statistic (12.55) is greater than critical value (2.645), we reject H0,

There is significant evidence to support the claim that there is difference between the two groups.

eta-squared ( ²) =   =   = 0.68

larger value of  ² represent the magnitude of difference between the groups is large.

Omega-square ² = = = 0.68

So we can conclude that approximately 68% of the variation in the DV can be attributed to the difference between groups.