Question

Ms. Smith teaches three Algebra I classes that cover exactly the same content. She is wondering if changing the order of some of the lessons would be beneficial to students. In class A, she teaches everything in the traditional order. In class B, she decides to skip Chapter 1 as it is preliminary information that students may already know. In class C, she decides to do Chapter 5 prior to doing Chapter 3. She would like to know if there are any significant differences in the average scores of the three classes.

1. Which test will be used?
2. What are H0 (null hypothesis) and Ha (alternative hypothesis)?
3. Calculate thes tatistic.
4. What decision does this inform to make and why? (may use the critical value in the chart or the calculated p-value, but must indicate what are doing.)
5. a one to two sentence conclusion that correctly responds to the question. 2 or 3 sentences to report your statistic ex. [χ^2(df)=1.1, p<.05]

Class A	Class B	Class C
50.90	91.04	58.93
28.55	81.51	40.52
46.33	76.55	43.89
28.42	85.56	47.99
36.19	36.49	22.14
93.61	70.91	40.42
23.73	92.23	33.73
77.83	50.48	60.29
54.34	40.52	75.05
43.32	38.47	37.40
63.96	45.63	70.01
56.19	89.13	31.17
17.70	65.04	13.41
81.86	83.31	10.12
86.10	74.89	62.96
25.61	41.79	89.27
52.89	42.04	44.58
21.31	98.38	4.75
41.44	76.79	34.15
99.71	20.96	5.06
74.50	18.66	85.20
92.45	82.75	20.23
98.38	55.04	56.49
4.10	13.89	31.12
34.47	49.37	17.87

Answer 1

This problem has been solved using Excel. Go to Data, select Data Analysis, choose Anova: Single Factor. Input values in the following way:

a) ANOVA for single factor is the test to be used.

b) H0: µ1 = µ2= µ3: Mean scores of classes A, B and C are the same

H1: Mean score of at least one of the classes A, B or C is different

c) Excel output

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Class A	25	1333.89	53.36	788.98
Class B	25	1521.43	60.86	635.34
Class C	25	1036.75	41.47	571.46
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	4778.38	2	2389.19	3.59	0.03	3.12
Within Groups	47898.78	72	665.26
Total	52677.16	74

Test statistic: F = 3.59

d) p-value = 0.03

Since p-value is less than 0.05, we reject the null hypothesis and conclude that mean score of at least one of the classes A, B or C is different.

e)

p-value = 0.03

Since p-value(0.03) < 0.05, we reject the null hypothesis and conclude that average score of the three classes A, B and C are different.