Question

In: Statistics and Probability

Dr. Jones is interested in the effects of caffeine on performance. She randomly splits her statistics...

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

Solutions

Expert Solution

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


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