Factorial Analysis Of Variance And Hypothesis Testing In Relation To Professional Athletes’ Healthy Eating Habits

Factorial Analysis of Variance in Relation to Professional Athletes’ Healthy Eating Habits

A sports psychologist gave a questionnaire about healthy eating habits to randomly selected professional athletes. The results are displayed below. Using the .05 significance level, is there a difference in healthy eating habits among professionals in the three sports?

Baseball Players              Basketball Players              Football Players

            32                                    27                                        27

            27                                    36                                        23

            26                                    25                                        26

            35                                    30                                        20

1. Make a graph for the data set.

Answer 

Use the five steps of hypothesis testing (report results in APA format).

Step1: Null and Alternative hypothesis

At 5% level of significance

Step 2: Test Statistics

To test the hypothesis analysis of variance (ANOVA) test will be performed.

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Baseball Players	4	120	30	18
Basketball Players	4	118	29.5	23
Football Players	4	118	29.5	23

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	0.666667	2	0.333333	0.015625	0.984523	4.256495
Within Groups	192	9	21.33333
Total	192.6667	11

Step 3: Critical value

We find the critical value at 5% level

Step 4: Decision

The null hypothesis is rejected if the computed F value is greater than the F-critical value otherwise the null hypothesis is accepted.

Step 5: Conclusion

The computed F-value (0.0156) is less than the F-critical (4.2565) hence the null hypothesis is not rejected as such we conclude that there is no significant difference in healthy eating habits among professionals in the three sports.

Effect size output

The overall effect size f = 0.0510
The effect size for Group 1 vs Group 2 is f = 0.0442
The effect size for Group 1 vs Group 3 is f = 0.0442
The effect size for Group 2 vs Group 3 is f = 0.0000

1. Conduct a planned contrast for Baseball versus Football players (using Tukey’s HSD).

Answer

Multiple Comparisons
Dependent Variable:   VAR00001
Tukey HSD
(I) VAR00002	(J) VAR00002	Mean Difference (I-J)	Std. Error	Sig.	95% Confidence Interval
Lower Bound	Upper Bound
Baseball	Basketball	.50000	3.26599	.987	-8.6187	9.6187
Football	.50000	3.26599	.987	-8.6187	9.6187
Basketball	Baseball	-.50000	3.26599	.987	-9.6187	8.6187
Football	.00000	3.26599	1.000	-9.1187	9.1187
Football	Baseball	-.50000	3.26599	.987	-9.6187	8.6187
Basketball	.00000	3.26599	1.000	-9.1187	9.1187

Results of the post-hoc Tukey HSD showed that there is no significant difference in healthy eating habits between the Baseball and football players.

1. (25 points) A researcher is interested in the effects of sleep deprivation and caffeine intake on mood. Participants were randomly assigned to a sleep condition (normal or deprived) and a caffeine condition (0 cups, 2 cups, or 4 cups). After the manipulations, mood was measured (such that higher numbers indicated better mood). The results were as follows:

               Normal Condition                            Deprived Condition

            0 cups  2 cups  4 cups                          0 cups  2 cups  4 cups

                16       18         18                                0          5          6

                17       20         17                                6          4          8

                20       20         17                                3          4          6

                19       19         17                                2          2          7

                18       18         16                                4          5          8

Analyze these data using a factorial analysis of variance and including R² for each effect.

Answer

Tests of Between-Subjects Effects
Dependent Variable:   mood score
Source	Type III Sum of Squares	df	Mean Square	F	Sig.	Partial Eta Squared
Corrected Model	1386.667a	5	277.333	144.696	.000	.968
Intercept	3853.333	1	3853.333	2010.435	.000	.988
Caffeine	11.667	2	5.833	3.043	.066	.202
Sleep	1333.333	1	1333.333	695.652	.000	.967
Caffeine * Sleep	41.667	2	20.833	10.870	.000	.475
Error	46.000	24	1.917
Total	5286.000	30
Corrected Total	1432.667	29
a. R Squared = .968 (Adjusted R Squared = .961)

Results shows that there was a statistically significant two-way interaction between Caffeine and Sleep, F(2, 24) = 10.87, p = .000.

The value of R-Squared is 0.968; this implies that the two factors (caffeine and sleep conditions) together with their interaction explain 96.8% of the variation in the mood score.

1. (25 points) A researcher was interested in whether college GPA (X) would predict starting salary after college (Y). (For simplicity, salary was converted to a 100-point scale.) The participants’ scores were:

            X                      Y         

            2.25                 55

            1.75                 23

            3.25                 80

            3.75                 42       

            M = 2.75          M = 50 

1. Report the correlation and linear prediction equation.

Answer

Correlation

	X	Y
X	1
Y	0.48065	1

The correlation coefficient between x and y is 0.4807; this implies a weak positive correlation between the two variables (X and Y).

Linear prediction equation;

1. Make a graph with the regression line.

Answer

Coefficientsa
Model	Unstandardized Coefficients	Standardized Coefficients	t	Sig.
B	Std. Error	Beta
1	(Constant)	15.350	46.511		.330	.773
x	12.600	16.255	.481	.775	.519
a. Dependent Variable: y

The standardized regression coefficient is 0.481.

1. (25 points) An advertising firm wanting to target people with strong desires for success conducted a study to see if such people differed in the types of television shows they watched. Randomly selected participants recorded the shows they watched for a week, then their desire for success was assessed, and finally they were divided into two groups. Low Success seekers watched 8 comedies, 15 romances, 6 documentaries, 13 dramas, and 3 news shows. High Success seekers watched 3 comedies, 3 romances, 9 documentaries, 7 dramas, and 8 news shows. Using the .05 significance level, is the distribution of type of shows watched different for participants having high and low desires for success?

	Low Success seekers	High Success seekers	Total
Comedies	8	3	11
Romances	15	3	18
Documentaries	6	9	15
Dramas	13	7	20
News shows	3	8	11
Total	45	30	75

Use the five steps of hypothesis testing.

Answer

Step1: Null and Alternative hypothesis

H₀: There is no significant association between desire for success and type of television shows watched

H_A: There is significant association between desire for success and type of television shows watched

At 5% level of significance

Step 2: Test Statistics

To test the hypothesis Chi-Square test of association will be performed.

Results
	Low Success seekers	High Success seekers	Row Totals
Comedies	8  (6.60)  [0.30]	3  (4.40)  [0.45]	11
Romances	15  (10.80)  [1.63]	3  (7.20)  [2.45]	18
Documentaries	6  (9.00)  [1.00]	9  (6.00)  [1.50]	15
Dramas	13  (12.00)  [0.08]	7  (8.00)  [0.12]	20
News shows	3  (6.60)  [1.96]	8  (4.40)  [2.95]	11
Column Totals	45	30	75  (Grand Total)

The chi-square statistic is 12.4432. The p-value is .014343.

Step 3: Critical value

We find the critical value at 5% level

Step 4: Decision

The null hypothesis is rejected if the computed Chi-Square value is greater than the critical Chi-Square value otherwise the null hypothesis is accepted.

Step 5: Conclusion

The computed Chi-Square value (12.4432) is greater than the critical Chi-Square value (9.4877) hence the null hypothesis is rejected as such we conclude that there is significant association between desire for success and type of television shows watched.