Question

I ran some test about something truly exciting and found that for Group A, the sample meanis 8.5, the standard deviationis 0.6, for n = 12. For Group B, the sample meanis 7.7, the standard deviationis 0.8, and n = 15. Use these values (which you will need to manipulate to suit your needs!) for #1-3. Yes, this is all the info you need!

1. Calculate the effect size using Cohen’s d.

2. Calculate the amount of variance accounted for using r².

3. Construct a 95% confidence interval to estimate the true difference between the groups. Is there one?

Answer 1

M1=8.5

M2=7.7

SD1=0.6

SD2=0.8

n1=12

n2=15

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a)

Cohen's d = (M₂ - M₁) ⁄ SD_pooled

SD_pooled = √((SD₁² + SD₂²) ⁄ 2)

SD_pooled = √((0.6² + 0.8²) ⁄ 2) = 0.707107

Cohen's d = (7.7 - 8.5) ⁄ 0.707107 = 1.131371.

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df=15+12+-2 = 25

r² = t² / (t²+df) = 0.248

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confidence interval-

μ₁ - μ₂ = (M₁ - M₂) ± ts_{(M1 - M2)}

M₁ & M₂ = sample means
t = t statistic determined by confidence level
s_{(M1 - M2)} = standard error = √((s²_p/n₁) + (s²_p/n₂))

Pooled Variance
s²_p = ((df₁)(s²₁) + (df₂)(s²₂)) / (df₁ + df₂) = 12.92 / 25 = 0.52

Standard Error
s_{(M1 - M2)} = √((s²_p/n₁) + (s²_p/n₂)) = √((0.52/12) + (0.52/15)) = 0.28

Confidence Interval
μ₁ - μ₂ = (M₁ - M₂) ± ts_{(M1 - M2)} = 0.8 ± (2.06 * 0.28) = 0.8 ± 0.573

=[0.227, 1.373]

You can be 95% confident that the difference between your two population means (μ₁ - μ₂) lies between 0.227 and 1.373