Question

In: Statistics and Probability

1. A researcher wants to know what kind of exercising is most likely to lead to...

1. A researcher wants to know what kind of exercising is most likely to lead to improve people’s moods. He has participants engage in one month of regular workouts (cardio, weight training, or yoga) and then has them complete a mood measure (1-10, higher scores = better mood). Is there a significant difference in groups? Hand calculate the ANOVA using a significance level of .05 and then check your work using SPSS.

Cardio	Weights	Yoga
8	9	7
7	5	9
8	7	9
4	6	10

Hypothesis:

F_crit /threshold:

Source of Variations	SS	df	MS	F_obt
Between groups
within groups
Total

Conclusion:

Solutions

Expert Solution

From the question, the data values are

Cardio	Weights	Yoga
8	9	7
7	5	9
8	7	9
4	6	10

	Cardio	Weights	Yoga	Total
Sum X	27	36	13	76
Sum X^2	193	191	311	695
N	4	4	4	12
Mean	6.75	6.75	8.75

The grand mean is obtained by taking mean of the treatments,

Sum of Squares

Between Groups

The sum of square of treatment is obtained by using the formula,

Where,

j is the column (j=1,2,3,4)

i is the row, (i=1,2,3)

is the average of jth column

is the grand mean

Total Sum of Square


Cardio	8	7.416667	0.583333	0.340278
	7	7.416667	-0.41667	0.173611
	8	7.416667	0.583333	0.340278
	4	7.416667	-3.41667	11.67361
Weights	9	7.416667	1.583333	2.506944
	5	7.416667	-2.41667	5.840278
	7	7.416667	-0.41667	0.173611
	6	7.416667	-1.41667	2.006944
Yoga	7	7.416667	-0.41667	0.173611
	9	7.416667	1.583333	2.506944
	9	7.416667	1.583333	2.506944
	10	7.416667	2.583333	6.673611
Sum				34.91667

Within Groups Sum of Square

Degree of freedoms

Mean Squares

F-statistic

F- critical value

The F-critical value is obtained from F-distribution table for significance level = 0.05, numerator degree of freedom = 2 and denominator degree of freedom = 9

The complete ANOVA table is,

ANOVA
Source of Variation	SS	df	MS	F	F crit
Between Groups	10.66667	2	5.333333	1.979381	4.256495
Within Groups	24.25	9	2.694444
Total	34.91667	11

The values are matched with the results are statistical software.

Conclusion:

Since F-statistic value is less than F-critical value, the null hypothesis is not rejected. There is no significant difference in means.

orchestra answered 11 months ago

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