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Statistics for The Behavioral Scie...

10th Edition
Frederick J Gravetter + 1 other
ISBN: 9781305504912

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Section
BuyFindarrow_forward

Statistics for The Behavioral Scie...

10th Edition
Frederick J Gravetter + 1 other
ISBN: 9781305504912
Textbook Problem

It appears that there is some truth to the old adage ‘That which doesn’t kill us makes us stronger.” Seery, Holman, and Silver (2010) found that individuals with some history of adversity report better mental health and higher well-being compared to people with little or no history of adversity. In an attempt to examine this phenomenon, a researcher surveys a group of college students to determine the negative life events that they experienced in the past 5 years and their current feeling of well-being. For n = 18 participants with 2 or fewer negative experiences, the average well-being score is M = 42 with S S = 398 , and for n = 16 participants with 5 to 10 negative experiences the average score is M = 48.6 with S S = 370 .

a. Is there a significant difference between the two populations represented by these two samples? Use a two-tailed test with α = .01 .

b. Compute Cohen’s d to measure the size of the effect.

c. Write a sentence demonstrating how the outcome of the hypothesis test and the measure of effect size would appear in a research report.

To determine

We need to find there is any significant difference in the two populations or not also find the Cohen`s d effect size measure.

Explanation

We have provided with two samples with along with summary statistic of each sample. For the significance difference of the two sample mean will follow two sample t-test on the basis of the sample will construct critical point. Then will calculate Cohen's statistic that will give us the effect size measure. Then will conclude the measures will imply what.

Given:

We have provided two samples with summary statistic

For first sample,

n1=18M1=42SS1=398

For second sample,

n1=16M1=48.6SS1=370

Formula used:

For one sample of size n, we have below formulas

Mean= x¯ = 1ni=1nxi

SS= i=1n(xix¯)2

Variance= 1n1i=1n(xix¯)2

If there are two samples of sizes n1and n2 we have the following formula for pooled SD,

Sp=Pooled Variance

SP2=(n11)×s12+(n21)×s22(n1+n22)

S12=sample variance of first sample

S22=sample variance of second sample

Denote,

Se=standard error of the mean,

Se=Sp1n1+1n2

Calculation:

a)

Here will proceed by the steps as,

S12=1n1i=1n(xix¯)2=398181=23.41

S22=1n21i=1n(xix¯)2=370161=24

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