Question

College students were randomly sorted into one of two groups. Members of each group performed a series of mental tasks while music was playing in the background. One group listened to pop music and the other to country music. It is hypothesized that more tasks will be completed with pop than with country music. What can be concluded with α = 0.05? The results are below:

pop	country
mean 1 = 47.14 sigma hat 1 =3.77 n1 = 15	mean 2 = 41.09 sigma hat 2 =6.36 n2 = 9

a) What is the appropriate test statistic?
---Select--- na z-test One-Sample t-test Independent-Samples t-test Related-Samples t-test

b)
Condition 1:
---Select--- mental task music school country music pop music
Condition 2:
---Select--- mental task music school country music pop music

c) Obtain/compute the appropriate values to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
critical value =  ; test statistic =  
Decision:  ---Select--- Reject H0 Fail to reject H0

d) If appropriate, compute the CI. If not appropriate, input "na" for both spaces below.
[  ,  ]

e) Compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and/or select "na" below.
d =  ;  ---Select--- na trivial effect small effect medium effect large effect
r² =  ;  ---Select--- na trivial effect small effect medium effect large effect

f) Make an interpretation based on the results.

a)Students that listened to pop music significantly completed more tasks.

b)Students that listened to country music significantly completed more tasks.   

c)Music had no significant effect on the number of tasks completed.

Answer 1

a.

Independent-Samples t-test

b.

Can you please provide options properly?

c.

Hypothesis:

I used Minitab to solve this question.

Go to 'Stat' ---> 'Basic Statistic' ---> '2 sample t test'. New window will pop-up on screen. Refer following screen shot and enter information accordingly.

Minitab output:

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean
1 15 47.14 3.77 0.97
2 9 41.09 6.36 2.1

Difference = μ (1) - μ (2)
Estimate for difference: 6.05
95% lower bound for difference: 2.52
T-Test of difference = 0 (vs >): T-Value = 2.94 P-Value = 0.004 DF = 22
Both use Pooled StDev = 4.8738

Test statistic t = 2.94

Degrees of freedom = 22

Critical value = 1.717

Since calculated value is greater than critical value we reject null hypothesis

d.

since alternative hypothesis is greater than type, we need to calculated lower bound only, because upper bound is infinity.

Thus 95% confidence interval is ( 2.52, infinity )

e.

Effect size:

This is large effect.

f.

Students that listened to pop music significantly completed more tasks.