Question

A researcher wants to see if there are any differences between three types of studying: Studying in a quiet room, studying in a room with classical music playing, and studying in a room with loud lawn equipment going on outside the window. She recruits 9 college students and puts three in each group. She has them read an excerpt from a physics textbook and then gives them a 5-question quiz. Below are their scores. Does study condition have an effect on academic performance?

Quiet Room Classical Music Loud Lawn Equipment

122

243

524

8. State the null and alternative hypotheses for this study, both statistically and in English.
9. Calculate the dfbetween, dfwithin, and dftotal and find Fcritical. 2
10. Calculate the mean of each group (M1, M2, and M3), ∑x and SSwithin
11. Determine G and N and calculate the SStotal and the SSbetween.
12. Find the MSbetween and MSwithin.

13. Calculate F and make your decision (accept or reject the null?) and report your results in APA format (see the “In The Literature” section on page 349 in your textbook to help you with this).

The following questions assess your understanding of correlational data.

14. Draw a graph that demonstrates a positive correlation. Label your axes with two variables that are normally positive correlated in the world.
15. Draw a graph that demonstrates a negative correlation. Label your axes with two variables that are normally negative correlated in the world.
16. Summarize why a correlation cannot imply causation. And give an example that illustrates this fact (that is not already given in your textbook.

Answer 1

Let μ1, μ2, μ3  denote the mean quiz score of the students who prepared for the quiz studying in a quiet room, studying in a room with classical music playing, and studying in a room with loud lawn respectively.

We have to test:

H0: μ1= μ2 = μ3   Vs Ha: Not all means are equal

In words,

H0: Study condition has no effect on academic performance Vs Ha: Study condition has a significant effect on academic performance

The appropriate statistical test to test the above hypothesis would be a One way ANOVA.

Constructing an ANOVA table of the format:

Source of Variation	SS	df	MS	F
Between Groups	SSB	k-1	MSB = SSB / (k-1)	F = MSB / MSW
Within Groups	SSW	N-k	MSW = SSW / (N-k)
Total	SST	N-1

where, K = No. of groups copared and N = No .of observations in the sample.

2.

dfbetween = k - 1 = 3 - 1 = 2

dfwithin = n - k = 9 = 3 = 6

dftotal = n - 1 = 9 - 1 = 8

and F critical is obtained from the F table for (k-1,n-k) = ( 2,6) degrees of freedom for alpha = 0.05:

F2,6 = 5.14

3. Computing the means M1, M2 and M3:

M1 = (1 + 2 + 5) / 3 = 2.67

M2 = (2 + 4 + 2) / 3 = 2.67

M1 = (2 + 3 + 4) / 3 = 3

SS(within) = SSW = Sum [Individual observations in a group - Group mean ]2 for the three groups

= [(1-2.67)^2+(2-2.67)^2+(5-2.67)^2] + [(2-2.67)^2+(4-2.67)^2+(2-2.67)^2] + [(2-3)^2+ (3-3)^2+(4-3)^2]

= 13.34

4. The grand mean is computed using the formula:

G = Sum of all sample observations / Total No. of observations

= (1+2+5+2+4+2+2+3+4) / 9..............(Since, N = 9)

= 2.78

SS(Between) = SSB = Group size *  (Group mean - Grand Mean) for the three groups

= 3 [((2.67-2.78)^2)+((2.67-2.78)^2)+((3-2.78)^2 )]

= 0.222

SSTotal = SSW + SSB

= 13.34 + 0.22

= 13.56

5. MSW = SSW / (N-k)

= 13.34 / (9-3)

= 2.22

MSB = SSB / (k-1)

= 0.22 / (3-1)

= 0.11

13. F = MSB / MSW

= 0.11 / 2.22

= 0.05

Comparing the F value obtained with the critical value,

Since F = 0.05 < 5.14 does not lie in the rejection region, we fail to reject the null hypothesis. We may conclude that the data does not provide sufficient evidence to support the claim that Study condition affects academic performance.

Also, to obtain the p-value, we may use the excel function: FDIST

We get p-value = 0.952 > 0.05, which brings us the same conclusion as above.

In APA style,

A one-way between subjects ANOVA was conducted to compare the effect of Study condition on academic performance.There no significant effect of Study condition on DV academic performance at the p<.05 level for the three conditions [F(2,6) = 0.05, p = 0.952)]