Question

A study was conducted to see if students’ writing skills continue to improve as the academic year progresses. English 101 students were asked to write essays throughout the year in September, December, March and June. The scores were compared to see if there is any improvement in this one-year course. Was there a significant improvement?

Student	September	December	March	June
1	17	23	21	24
2	25	21	21	24
3	24	16	21	29
4	18	18	25	27
5	25	21	24	24
6	25	25	16	25
7	24	15	23	22
8	24	21	19	23
9	20	17	16	28
10	24	22	21	22
11	16	24	25	29
12	18	24	23	24
13	15	16	22	27
14	23	24	22	22
15	21	24	20	25
16	22	16	25	26
17	16	16	20	23
18	24	22	22	27
19	19	23	23	27
20	19	17	25	25
21	20	17	19	22
22	22	19	25	24
23	15	24	22	26
24	22	21	15	26
25	25	16	24	22
26	15	22	25	23
27	15	17	25	29
28	18	18	19	23
29	17	17	21	29
30	19	17	15	26
31	19	16	18	26
32	22	21	16	29
33	16	19	19	29

a. The appropriate test for this problem is:

a.one-way ANOVA

b. repeated measurements

b. The obtained statistic is:

a. F = 24.17

b. F = 12.71

c. F = 21.47

d. F = 17.42

c. The associated p value is:

a. < .05

b. < .01

c. < .001

d. < .0001

Decision is:

a. reject the null

b. retain the null

Conclusion is:

a. March scores are significantly lower than the rest of the months

b. December scores are significantly higher than September scores

c. March scores are significantly higher than September scores

d. June scores are significantly higher than the rest of the months

e. no conclusion can be drawn

Answer 1

Solution

[NOTE: Answers are given below. Detailed Working and Back-up Theory follow at the end.]

Part (a)

The appropriate test for this problem is:

a.one-way ANOVA

Answer

Part (b)

The obtained statistic is:

a. F = 24.17

Answer

Part (c)

The associated p value is:

d. < .0001

Answer

Part (d)

Decision is:

a. reject the null

Answer

Part (e)

Conclusion is:

d. June scores are significantly higher than the rest of the months

Answer

Back-up Theory and the solution work details,

ONE-WAY CLASSIFICATION - EQUAL NUMBER OF OBSERAVATIONS (33)PER Treatment

Obsn #	Tr 1			Tr 2		Tr 3		Tr 4
1	17			23		21		24
2	25			21		21		24
3	24			16		21		29
4	18			18		25		27
5	25			21		24		24
6	25			25		16		25
7	24			15		23		22
8	24			21		19		23
9	20			17		16		28
10	24			22		21		22
11	16			24		25		29
12	18			24		23		24
13	15			16		22		27
14	23			24		22		22
15	21			24		20		25
16	22			16		25		26
17	16			16		20		23
18	24			22		22		27
19	19			23		23		27
20	19			17		25		25
21	20			17		19		22
22	22			19		25		24
23	15			24		22		26
24	22			21		15		26
25	25			16		24		22
26	15			22		25		23
27	15			17		25		29
28	18			18		19		23
29	17			17		21		29
30	19			17		15		26
31	19			16		18		26
32	22			21		16		29
33	16			19		19		29
Total xi.	664			649		697		837
Grand Total G = ∑xi. = 2847
r	4
n	33
N= rn	132
Correction Factor C = G^2/N						61404.614
∑xij^2	13748			13085		15041		21417
Total SS Raw = Sum∑xij^2						63291
Total SS = Total SS Raw - C						1886.3864
Treatment SS Raw = ∑xi.^2/n						62075
Treatment SS = Treatment SS Raw - C								670.3864
Error SS = Total SS - Treatment SS								1216
Degrees of freedom df
Total: N - 1
Treatment: r - 1
Error: Total - Treatment = N - r
Mean Sum of squares = SS/df
F = MSTr/MSE
Fcrit = Fr-1, N-r,α
p-value = P(Fr-1, N-r > F)
ANOVA TABLE			α		0.05
Source		df	SS		MS		F		Fcrit		p-value
Treatment		3	670.3864		223.46212		23.5223		2.675387		3.4257E-12
Error		128	1216		9.5
Total		131	1886.3864

DONE