Question

A researcher randomly assigns 33 subjects to one of three groups. Group 1 receives technical dietary information interactively from an on-line website. Group 2 receives the same information from a nurse practitioner, while Group 3 receives the information from a video made by the same nurse practitioner.

The researcher looked at three different ratings of the presentation: difficulty, usefulness, and importance to determine if there is a difference in the modes of presentation. In particular, the researcher is interested in whether the interactive website is superior because that is the most cost-effective way of delivering the information.

Group	Usefulness	Difficulty	Importance
1	20	5	18
1	25	9	8
1	23	15	20
1	16	9	22
1	20	6	22
1	28	14	8
1	20	6	13
1	25	8	13
1	24	10	24
1	18	10	20
1	17	9	4
2	28	7	14
2	25	14	5
2	26	9	20
2	19	15	22
2	29	14	12
2	15	6	2
2	29	10	5
2	26	11	1
2	22	5	2
2	15	15	14
2	29	6	4
2	15	6	3
3	22	8	12
3	27	9	14
3	21	10	7
3	17	9	1
3	16	7	12
3	19	9	7
3	23	10	1
3	27	9	5
3	23	9	6
3	16	14	22

1. Run the appropriate analysis of the data and interpret the results.
2. How could this study have been done differently? Why would this approach be better or why would it not be?

Answer 1

Answer:

Given Data

A researcher randomly assigns 33 subjects to one of three groups. Group 1 receives technical dietary information interactively from an on-line website. Group 2 receives the same information from a nurse practitioner, while Group 3 receives the information from a video made by the same nurse practitioner.

We can conduct individual tests for each of the aspect for the methods:

comparing the usefulness of the methods

We can enter the data in excel as follows:

K	L	M
1	2	3
20	28	22
25	25	27
23	26	21
16	19	17
20	29	16
28	15	19
20	29	23
25	26	27
24	22	23
18	15	16
17	29
	15

Now click on data tab > data analysis > ANOVA single factor, the following window pops up
Enter the input range as the cell addresses of all the data cells and click ok to get the output as shown below

Since the P value is 0.53966 which is greater than the significance level 0.05, we fail to reject the null hypothesis and conclude that all the three methods provide same level of understanding

.

Next enter the data for difficulty level as follows:

O	P	Q
1	2	3
5	7	8
9	14	9
15	9	10
9	15	9
6	14	7
14	6	9
6	10	10
8	11	9
10	5	9
10	15	14
9	6
	6

Now again click on data tab > data analysis > ANOVA single factor

enter the range of cells and hit ok to get the output as shown below:

Again P vlaue 0.8788 is greater than the significance level thus all the modes have same dififculty level

.

Next we conduct the test for inportance factor in the same manner and get the output as follows:

Since the P value is 0.036425 which is less than the significance level 0.05 we conclude that the "importance" factor varies in all the three methods. Thsu the three methods are different in terms of importance comparison

.

This method of testing "ANOVA: Single factor" helps to identify the difference if any among the factors, But fails to help us identify which factor or method is better and which is not.

A better way to conduct such a study is: compare the factors in pairs, that helps us knwo which method is better and which is not

we can conduct a 2 sample t test for the same, But we need to conduct the test 3 times and then get a combined conclusion about it.

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