In: Statistics and Probability
Dr. Jones is interested in the effects of caffeine on performance. She randomly splits her statistics class of 15 students into three groups. Prior to the final exam, she has one group take no caffeine prior to taking the exam, students in another group take 100 mg of caffeine, and students in the third group take 200 mg of caffeine. She records the grades in each condition.
0 mg |
100 mg |
200 mg |
65 |
75 |
72 |
62 |
80 |
78 |
70 |
85 |
87 |
80 |
94 |
92 |
78 |
86 |
91 |
a. State the null hypothesis.
b. Identify the appropriate statistical test
c. Calculate the appropriate test statistic. Make sure to report the degrees of freedom for the statistical test (if appropriate). You must show your work to receive full credit.
d. State your conclusions (use a two-tailed test with = .05 for all tests).
e. Compute an effect size
a. state the null hypothesis:
H0: There is no significance difference between three group (i.e. 0, 100, 200 mg)
Ha: There is significance difference between three group (i.e. 0, 100, 200 mg)
b. Identify the appropriate statistical tet:
In statistics, one-way analysis of variance is a technique that can be used to compare means of two or more samples.
So, to check significance between there groups we should be use one way anova.
c. Calculate the appropriate test statistic.
By using excel data analysis tool we have to calculate one way ANOVA,
First arrange data in excel as,
0 mg | 100 mg | 200 mg |
65 | 75 | 72 |
62 | 80 | 78 |
70 | 85 | 87 |
80 | 94 | 92 |
78 | 86 | 91 |
Then use data analysis tool as,
The output of ANOVA is,
Anova: Single Factor | ||||||
SUMMARY | ||||||
Groups | Count | Sum | Average | Variance | ||
0 mg | 5 | 355 | 71 | 62 | ||
100 mg | 5 | 420 | 84 | 50.5 | ||
200 mg | 5 | 420 | 84 | 75.5 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 563.33 | 2 | 281.67 | 4.49 | 0.03 | 3.89 |
Within Groups | 752.00 | 12 | 62.67 | |||
Total | 1315.33 | 14 |
d. State your conclusions (use a two-tailed test with alpha = .05 for all tests).
We can see in above output,
P-vlaue = 0.03
Decision: the p-value (0.03) is less than 0.05 then we Reject H0
Conclusion: There is significance difference between three group (i.e. 0, 100, 200 mg)
e. Compute an effect size
Overall effect size = Sqrt(MSbetween / MSwithin)
= sqrt(281.67/62.67)
= sqrt(4.49)
= 2.1200
The effect size is 2.1200