Question

2. The same exam was given to all the students in the sixth grade, but they were divided into 4 classrooms. Each classroom had a different distraction taking place except for Room 1 that was the control group and free of noise. Using the scores ranging from 0 to 60 of the students below, state a conclusion about the nature of testing practices used.

Control	60 degrees	Classical Music	Next to Cafeteria
31	53	27	29
16	39	36	18
11	41	32	16
19	39	14	20
43	54	40	7
49	41	60	12
4	9	47	20
11	17	2	45
34	13	27	33
39	8	5	53
60	32	55	23
40	29	53	24
57	35	27	18
59	45	4	49
43	42	50	29

Answer 1

The table below gives the data which has been calculated from the given figures

Total	516	497	479	396
n	15	15	15	15
Mean	34.40	33.13	31.93	26.40
Sum Of Squares	4791.6	3143.7	6179.2	2858.6
Variance	342.26	224.55	441.37	204.19
Standard deviation	18.50	14.98	21.01	14.29

The Hypothesis:

H0: There is no difference between the mean scores due to the 4 different testing methods.

Ha: There is a difference between the mean scores due to the 4 different testing methods.

The ANOVA table is as below. the p value is calculated for F = 0.62 for df1 = 4 and df2 = 56

The Fcritical is calculated at = 0.05 (default) for df1 = 4 and df2 = 56

Source	SS	DF	Mean Square	F	Fcv	p
Between	558.86	3.00	186.29	0.62	2.77	0.6082
Within/Error	16973.18	56.00	303.09
Total	17532.04	59.00

The Decision Rule:

If Ftest is > F critical, Then Reject H0.

Also if p-value is < , Then reject H0.

The Decision:

Since Ftest (0.62) is < F critical (2.77), We Fail to Reject H0.

Also since p-value (0.6082) is > (0.05), We reject H0.

The Conclusion: There isn't sufficient sufficient evidence at the 95% level of significance to conclude that there is a difference between the mean scores due to the 4 different testing methods.

___________________________________________________________________

Calculations For the ANOVA Table:

Overall Mean = (516 + 497 + 479 + 396) / 60 = 31.47

SS treatment = SUM n* ( - overall mean)² = 15 * (34.4 - 31.47)² + 15 * (31.93 - 31.47)² + 15 * (33.13 - 31.47)² + 15* (26.4 - 31.47)² = 558.86

df1 = k - 1 = 3 - 1 = 3

MSTR = SS treatment/df1 = 558.86 / 3 = 186.29

SSerror = SUM (Sum of Squares) = 4791.6 + 3143.7 + 6179.2 + 2858.6 = 16973.18

df2 = N - k = 60 - 4 = 56

Therefore MS error = SSerror/df2 = 16973.18 / 56 = 303.09

F = MSTR/MSE = 186.29 / 303.09 = 0.62

____________________________________________________