Question

a. all Hypothesis Tests must include all four steps, clearly labeled;

b. all Confidence Intervals must include all output as well as the CI itself

c. include which calculator function you used for each problem.

3. At a community college, the mathematics department has been experimenting with four different delivery mechanisms for content in their Statistics courses. One method is traditional lecture (Method I), the second is a hybrid format in which half the time is spent online and half is spent in-class (Method II), the third is online (Method III), and the fourth is an emporium model from which students obtain their lectures and do their work in a lab with an instructor available for assistance (Method IV). To assess the effective of the four methods, students in each approach are given a final exam with the results shown in the following table. Assume an approximate normal distribution for each method. At the 5% significance level, does the data suggest that any method has a different mean score from the others?

Method I	81	81	85	67	88	72	80	63	62	92	82	49	69	66	74	80
Method II	85	53	80	75	64	39	60	61	83	66	75	66	90	93
Method III	81	59	70	70	64	78	75	80	52	45	87	85	79
Method IV	86	90	81	61	84	72	56	68	82	98	79	74	82

Answer 1

The Summary statistics from the data given is as below

	Method I	Method II	Method III	Method IV	Total
Total	1191	990	925	1013	4119
n	16	14	13	13	56
Average	74.44	70.71	71.15	77.92
Sum of Squares	1903.94	2964.86	1993.69	1610.92	8473.41
Standard Deviation	11.2663	15.1019	12.8896	11.5864

The Hypothesis:

H0: There is no difference between the means of the.four Methods

Ha: The mean of at least one method is different from the others..

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The ANOVA table is as below.

Source	SS	DF	Mean Square	F	Fcv	p
Between	448.73	3	149.58	0.92	2.783	0.4387
Within/Error	8473.42	52	162.95
Total	8922.15	55

The p value is calculated for F = 0.92 for df1 = 3 and df2 = 52

The F critical is calculated at = 0.05 for df1 = 3 and df2 = 52

The Decision Rule:

If F test is > F critical, Then Reject H0.

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

The Decision:

Since F test (0.92) is > F critical (2.783), We Fail to Reject H0.

Also since p-value (0.4387) is < (0.05), We Reject H0.

The Conclusion: There isn't sufficient evidence at the 95% level of significance to conclude that the mean of at least one method is different from the others.

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Calculations For the ANOVA Table:

Overall Mean = [4119 / 56] = 73.55

SS treatment = SUM [n* ( - overall mean)²] = 16 * (74.44 - 73.55)² + 14 * (70.71 - 73.55)² + 13 * (71.15 - 73.55)² + 13 * (77.92 - 73.55)² = 448.73

df1 = k - 1 = 4 - 1 = 3

MSTR = SS treatment/df1 = 448.73 / 3 = 149.58

SSerror = SUM (Sum of Squares) = 1903.94 + 2964.86 + 1993.69 + 1610.92 = 8473.42

df2 = N - k = 56 - 4 = 52

Therefore MS error = SSerror/df2 = 8473.42 / 52 = 162.95

F = MSTR/MSE = 149.58 / 162.95 = 0.92

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