Chapter 9, Problem 22P

Oishi and Schimmk (2010) report that people who move from home to home frequently as children tend to have lower than average levels of well-being as adults. To further examine this relationship, a psychologist obtains a sample of $n=12$ young adults who each experienced 5 or more different homes before they were 16 years old. These participants were given a standardized well-being questionnaire for which the

general population has an average score of $\mu =40$ .

The well-being scores for this sample are as follows:

, 37, 41, 35, 42, 40, 33, 33, 36, 38, 32, 39.

a. On the basis of this sample, is well-being for frequent movers significantly different from wellbeing in the general population? Use a two-tailed test with $\alpha =.05$ .

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

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

To determine

1. To test whether wellbeing for frequent mover is significantly different from wellbeing in general population using two tail t-test.
2. Estimated Cohen's d.

Answer to Problem 22P

Solution:

1. Based on two tailed t test we conclude that well-being for frequent mover is significantly different from well-being in general population.
2. d = -0.9174311
3. The t score of -3.18 is outside the rejection area. The 2 critical values (cutoff points) are -3.25 and 3.25. Since the t score, 4.09, is outside this interval, we reject the null hypothesis. We accept the alternative hypothesis and on the basis of given sample we conclude that it is significant to say that wellbeing for frequent mover is significantly different from wellbeing in general population.

Explanation of Solution

Given:

1. Scores: 38, 37, 41, 35, 42, 40, 33, 33, 36, 38, 32, 39
2. n =12, μ = 40, M = 37, $\sigma$ =3.27
3. n =12, μ = 40, M = 37

Formula used:

1. Mean ( $\overline{X}$ ) = $\frac{1}{n}{\displaystyle \sum _{i=1}^n{X}_i}$, Standard deviation (s. d.)= $\sqrt{\frac{1}{n-1}{\displaystyle \sum _{i=1}^n{(X_i-\overline{X})}^2}}$ Standard error = $\frac{(s.d.)}{\sqrt{n}}$

2. For testing the significant effect, we use t test for single mean. Firstly, we define null and alternative hypothesis, which is,

   Null hypothesis: H_o= well-being for frequent mover is significantly not different from wellbeing in general population.

   Alternative hypothesis: H₁= well-being for frequent mover is significantly different from well-being in general population.

   OR

   Null hypothesis: H_oμ = 40

   Alternative hypothesis: H₁: μ = 40

   For two tailed t test

   $t_{(calculated)}=\frac{M-\mu }{(\frac{\sigma }{\sqrt{n}})}\sim t_{(n-1)}$

3. Cohen's d is given as:$d=\frac{M-\mu }{\sigma }$

Calculations:

1. $\begin{array}{c}{\displaystyle \sum _{i=1}^6{X}_i}=\\ =444\end{array}$

   38 + 37 + 41 + 35 + 42 +40+ 33 + 33 + 36 + 38+ 32+ 39

   Here, n=12

   Mean,

   $\frac{1}{12}{\displaystyle \sum _{i=1}^{12}{X}_i}$ = $\frac{{444}^{}}{12}$ = 37

   ${\displaystyle \sum _{i=1}^{12}{(X_i-\overline{X})}^2}$$\left(\begin{array}{l}{\left(38-37\right)}^2+{\left(37-37\right)}^2+{\left(41-37\right)}^2+{\left(35-37\right)}^2+{(}^{42}+{(}^{40}\\ +{(}^{33}+{(}^{33}+{(}^{36}+{(}^{38}+{(}^{32}+{(}^{39}\end{array}\right)$ =118

   Standard deviation,

   $\sqrt{\frac{1}{n-1}{\displaystyle \sum _{i=1}^n{(X_i-\overline{X})}^2}}$ = $\begin{array}{l}\sqrt{\frac{118}{11}}\\ =3.27\end{array}$

2. $t_{(calculated)}=\frac{37-40}{(\frac{3.27}{\sqrt{12}})}$ = -3.18

   The 2 critical values (cutoff points) are -2.201 and 2.201. Since the t score, -3.18, is outside this interval, we reject the null hypothesis. We accept the alternative hypothesis.

3. $d=\frac{37-40}{3.27}$= -0.9174311

Conclusion:

Based on two tailed t test we conclude that well-being for frequent mover is significantly different from well-being in general population.