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1. (25 points) 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.Use the five steps of hypothesis testing (report results in APA format).Figure the effect size of this study.Conduct a planned contrast for Baseball versus Football players (using Tukey’s HSD).
1. 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. 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.Make a graph with the regression line.Figure the standardized regression coefficient.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?
1. Use the five steps of hypothesis testing.Figure a measure of effect size and indicate whether it is small, medium, or large.

## 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

HA: 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.

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250 words