Question

A sleep researcher conducted a study where 4 participants were instructed to count sheep and 4 were told to concentrate on their breathing and 4 were not given any special instruction. The average number of minutes taken for each participants to fall asleep over the next 7 days were 14, 18, 27, 31 minutes for the sheep counters. 25, 22, 17, 14 minutes for the breathing ones, and 45, 33, 30, and 41 for the control condition. using the .05 significance level what is the value of the Grand mean? What is the F ratio? What is the final F value? Is the F value extreme enough to reject the null hypothesis? How do I input this into SPSS?

Answer 1

Solution:

Here, we have to use one way ANOVA test for testing the given claim. The null and alternative hypotheses for this test are given as below:

Null hypothesis: H₀: There is no significant effect of instructions on the average number of minutes for fall asleep.

Alternative hypothesis: H_a: There is a significant effect of instructions on the average number of minutes for fall asleep.

We are given level of significance = α = 0.05

First of all we have to input the given data in SPSS. Organize the given data as below:

Codes for variable instruction:

1: Count Sheep

2: Concentrate on breathe

3: None

Data:

Minutes	Instruction
14	1
18	1
27	1
31	1
25	2
22	2
17	2
14	2
45	3
33	3
30	3
41	3

Use proper label in SPSS in variable view.

Click on Analyze > Drop down and Select compare means > Drop down and select One-Way ANOVA

Move the variable Minutes in dependent list field.

Move the variable Instruction in Factor field.

Click on Options and check box for descriptive.

Press OK.

We will get following SPSS outputs:

Descriptives
Minutes
	N	Mean	Std. Deviation	Std. Error	95% Confidence Interval for Mean		Minimum	Maximum
					Lower Bound	Upper Bound
Count Sheep	4	22.5000	7.85281	3.92641	10.0044	34.9956	14.00	31.00
Concentrate on breathe	4	19.5000	4.93288	2.46644	11.6507	27.3493	14.00	25.00
None	4	37.2500	6.94622	3.47311	26.1970	48.3030	30.00	45.00
Total	12	26.4167	10.11262	2.91926	19.9914	32.8419	14.00	45.00

ANOVA
Minutes
	Sum of Squares	df	Mean Square	F	Sig.
Between Groups	722.167	2	361.083	8.069	.010
Within Groups	402.750	9	44.750
Total	1124.917	11

The value of grand mean is given as 26.4167.

The value of F test statistic is given as 8.069.

df1 = 2

df2 = 9

The value of F critical is given as 4.256495. (by using F-table)

The P-value is given as 0.01.

Level of significance = α = 0.05

P-value = 0.01 < α = 0.05

The F value is extreme enough to reject the null hypothesis

So, we reject the null hypothesis that there is no significant effect of instructions on the average number of minutes for fall asleep.

There is sufficient evidence to conclude that there is a significant effect of instructions on the average number of minutes for fall asleep.