NT16. 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 withsome history of adversity report better mental healthand higher well-being compared to people with littleor no history of adversity. In an attempt to examinethis phenomenon, a researcher surveys a group of col-lege students to determine the negative life events thatthey experienced in the past 5 years and their currentfeeling of well-being. For n = 18 participants with 2or fewer negative experiences, the average well-beingscore is M = 42 with SS = 398, and for n =ticipants with 5 to 10 negative experiences the averagescore is M = 48.6 with SS = 370.a. Is there a significant difference between the twopopulations represented by these two samples?Use a two-tailed test with a = .01.16 par-%3D%3D%3D%3Db. Compute Cohen's d to measure the size of the effect.c. Write a sentence demonstrating how the outcomeof the hypothesis test and thS Exeasdre of effectsize would appear in a research report.

For 2 or fewer negative experiences : n1 = 18 sample mean, x1 = 42 Standard deviation , s1 = SSn-1=39817=4.838 For 5 to 10 negative experiences : n2 = 16 sample mean, x2 = 48.6 Standard deviation , s2 = SSn-1=37015=4.966Now, Null and Alternative Hypotheses : H0:μ1=μ2Ha:μ1≠μ2 (There is a significant difference between the two populations) It is a two-tailed test Decision Rule : Significance level =α=0.01\alpha = 0.01 Degrees of freedom = dfTotal=df1+df2=17+15=32df_{Total} = df_1 + df_2 = 17 + 15 = 32 Critical value =tc=2.738t_c = 2.738 Rejection region =R={t:∣t∣>2.738}R = \{t: |t| > 2.738\} Reject H0 : if R={t:∣t∣>2.738}R = \{t: |t| > 2.738\} Test Statistics : t=Xˉ1−Xˉ2(n1−1)s12+(n2−1)s22n1+n2−2(1n1+1n2)t = \displaystyle \frac{\bar X_1 - \bar X_2}{\sqrt{ \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}(\frac{1}{n_1}+\frac{1}{n_2}) } } =42−48.6(18−1)4.8382+(16−1)4.966218+16−2(118+116)=−3.921 = \displaystyle \frac{ 42 - 48.6}{\sqrt{ \frac{(18-1)4.838^2 + (16-1)4.966^2}{ 18+16-2}(\frac{1}{ 18}+\frac{1}{ 16}) } } = -3.921 Decision : We noticed that ∣t∣=3.921>tc=2.738|t| = 3.921 > t_c = 2.738 hence null hypothesis is rejected Conclusion : There is enough evidence to claim that There is a significant difference between the two populations, at the α=0.01\alpha = 0.01 significance level The formula for the calculation of Cohen d : Cohen's d = (M2 - M1) ⁄ SDpooled SDpooled = √((SD12 + SD22) ⁄ 2) = 4.9024 Cohen's d = (48.6 - 42) ⁄ 4.9024 = 1.346 According to the result of Hypothesis test , There is enough evidence to claim that There is a significant difference between the two populations, at the α=0.01\alpha = 0.01 significance level The effect size is found to be 1.346 . According to the guidelines by Cohen , an effect size of 1.346 is considered to be large effect size. The difference between the two groups is significant