Question

When only two treatments are involved, ANOVA and the Student’s t test (Chapter 11) result in the same conclusions. Also, for computed test statistics, t^₂ = F. To demonstrate this relationship, use the following example. Fourteen randomly selected students enrolled in a history course were divided into two groups, one consisting of 6 students who took the course in the normal lecture format. The other group of 8 students took the course as a distance course format. At the end of the course, each group was examined with a 50-item test. The following is a list of the number correct for each of the two groups.

Traditional Lecture	Distance
36	43
31	31
35	44
30	36
33	44
37	35
	46
	43

  

   

1. a-1. Complete the ANOVA table. (Round your SS, MS, and F values to 2 decimal places and p value to 4 decimal places.)

1. a-2. Use a α = 0.01 level of significance. (Round your answer to 2 decimal places.)

2. Using the t test from Chapter 11, compute t. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)

3. There is any difference in the mean test scores.

Answer 1

Solution: We can use the excel "ANOVA: Single Factor" to answer the first two parts of the question. The excel output is given below:

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Traditional Lecture	6	202	33.66666667	7.866666667
Distance	8	322	40.25	29.64285714
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	148.5952381	1	148.5952381	7.224076396	0.019751342	4.747225336
Within Groups	246.8333333	12	20.56944444
Total	395.4285714	13

a-1. Complete the ANOVA table. (Round your SS, MS, and F values to 2 decimal places and p value to 4 decimal places.)

Answer:

Source of Variation	SS	df	MS	F	P-value
Between Groups	148.60	1	148.60	7.22	0.0198
Within Groups	246.83	12	20.57
Total	395.43	13

a-2. Use a α = 0.01 level of significance. (Round your answer to 2 decimal places.)

Answer: Since the p-value is 0.0198, which is greater than the significance level 0.01, we, therefore, fail to reject the null hypothesis and conclude that there is no significant difference between the two means.

We can use the Excel "t-Test: Two-Sample Assuming Equal Variances" Data analysis tool to find the answer to the below questions. The excel output is given below:

t-Test: Two-Sample Assuming Equal Variances
	Traditional Lecture	Distance
Mean	33.66667	40.25
Variance	7.866667	29.64286
Observations	6	8
Pooled Variance	20.56944
Hypothesized Mean Difference	0
df	12
t Stat	-2.68776
P(T<=t) one-tail	0.009876
t Critical one-tail	2.680998
P(T<=t) two-tail	0.019751
t Critical two-tail	3.05454

Using the t test from Chapter 11, compute t. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)

Answer:

There is any difference in the mean test scores.

Answer: Since the p-value is 0.0198, which is greater than the significance level 0.01, we, therefore, fail to reject the null hypothesis and conclude that there is no significant difference between the two means.