 # It appears that there is some truth to the old adage “That which doesn’t kill us Zoe Oneal 2021-10-18 Answered
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 $n=16$ participants with 2 or fewer negative experiences, the average well-being score is $M=42$ with $SS=398$, and for $n=16$ participants with 5 to 10 negative experiences the average score is $M=48.6$ with $SS=370$.
1. Is there a significant difference between the two populations represented by these two samples? Use a two-tailed test with $\alpha =.01$ [use the 4-step procedure]. [Remember to assess whether the assumption of homogeneity of variances is satisfied or not, $\alpha =.01$ for homogeneity test].
2. Compute Cohen’s d to measure the size of the effect.
3. Report your hypothesis test results and cohen’s d in APA format.
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Step 1
Given Information :
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.
Participants with 2 or fewer negative experiences:
${\overline{X}}_{1}=42$
$SS=398$
$n=16$
${s}_{1}=\sqrt{\frac{SS}{n-1}}$
$=\sqrt{\frac{398}{16-1}}$
${s}_{1}=5.151$
Participants with 5 to 10 negative experiences:
${\overline{X}}_{2}=48.6$
$SS=370$
$n=16$
${s}_{2}=\sqrt{\frac{SS}{n-1}}$
$=\sqrt{\frac{370}{16-1}}$
${s}_{2}=4.967$
Step 2
Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
Ho: there is no significant difference between the two populations represented by these two samples
i.e. ${\mu }_{1}={\mu }_{2}$
Ha: there is a significant difference between the two populations represented by these two samples
i.e ${\mu }_{1}\ne {\mu }_{2}$
This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.
Rejection Region
Based on the information provided, the significance level is $\alpha =0.01$, and the degrees of freedom are $df=30$. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:
Hence, it is found that the critical value for this two-tailed test is $tc=2.75$, for $\alpha =0.01$ and $df=30$.
The rejection region for this two-tailed test is $R=\left\{t:\mid t\mid >2.75\right\}$.
Test Statistics
Since it is assumed that the population variances are equal, the t-statistic is computed as follows:
$t=\frac{{\overline{X}}_{1}-{\overline{X}}_{2}}{\sqrt{\frac{\left({n}_{1}-1\right){s}_{1}^{2}+\left({n}_{2}-1\right){s}_{2}^{2}}{{n}_{1}+{n}_{2}-2}\left(\frac{1}{{n}_{1}}+\frac{1}{{n}_{2}}\right)}}$
$t=\frac{42-48.6}{\sqrt{\frac{\left(16-1\right){5.151}^{2}+\left(16-1\right){4.967}^{2}}{16+16-2}\left(\frac{1}{16}+\frac{1}{16}\right)}}=-3.689$
Since it is observed that $\mid t\mid =3.689>tc=2.75$, it is then concluded that the null hypothesis is rejected.
Using the P-value approach: The p-value is $p=0.0009$, and since $p=0.0009<0.01$, it is concluded that the null hypothesis is rejected.
Conclusion
It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that there is a significant difference between the two populations represented by these two samples​, at the 0.01 significance level.
Step 3
Cohen's d using t-statistic:
For the independent samples T-test, Cohen's d is determined by calculating the mean difference between your two groups, and then dividing the result by the pooled standard deviation.
Cohen's $d=\frac{M2-M1}{S}{D}_{pooled}$
where:
$S{D}_{pooled}=\sqrt{\left(\left(S{D}_{1}^{2}+S{D}_{2}^{2}\right)⁄2\right)}$
Pluggin all the values in stated formula ,
Cohen's $d=\left(48.6-42\right)/5.059836=1.30439.$
Since , the cohen's d is equal to 1.304 . It shows a large effect size.