Question

An industrial/organizational psychologist has been consulting with a company that runs weekend job-seeking workshops for the unemployed. She collected data on several issues related to these workshops and, after conducting statistical tests, obtained statistically significant findings. She needs to find a way to evaluate effect size so that she can make recommendations to the company.
One of the psychologistâs findings is that immediately after the workshop, a sample of 49 job seekers who received training on resume writing scored an average of 6.5 as measured on a 9-point job-search satisfaction scale, with a standard deviation of 1.2. The typical job seeker scores 5.8 points.
(a) The psychologist finds that the estimated Cohenâs d is _________ .
(b) The t statistic is 4.12, and r² is __________.

Answer 1

Answer:

The answers to the question are;

(a) The psychologist finds that the estimated Cohen's d is __0.562.

(b) The t statistic is 4.12, and r² is ___ 7.322 × 10⁻² .

Step-by-step explanation:

To solve the question, we note the given variables as follows

Sample count N = 49

Mean of sample statistics = $\overline{\rm x}$ = 6.5

Mean of population = μ$_x$ = 5.8

Sample standard deviation, s = 1.2

The t statistic is given as

$t = \frac{\overline{\rm x} - \mu_x}{\frac{s}{\sqrt{N} } }$

t = $\frac{6.5 - 5.8}{\frac{1.2}{\sqrt{49} } }$ = 4.08

(a) The Cohen's D is used as an indication of the effect of the size or magnitude of an object.

d is given as

d = $\frac{M_1-M_2}{s_{pooled}}$ =  

Where:

M₁ = Group 1 mean = 6.5

M₂ = Group 2 mean = 5.8

$s_{pooled}$ =  $\frac{\sqrt{s_1^2+s_2^2} }{2}$

Where:

s₁ = Standard deviation of group 1 sample = 1.2

s₂ = Standard deviation of group 2 sample

Since the statistics is pooled we can take s₁² = s₂²

Therefore $s_{pooled}$  = $\sqrt{\frac{2\cdot s_1^2 }{2}$ = ${\sqrt{s_1^2}$ = s₁

Therefore

d = $\frac{M_1-M_2}{s_{pooled}}$ =  $\frac{6.5-5.8}{1.2}$ = $\frac{7}{12}$ = 0.583

However since N < 50 we make use of the correction factor as follows.

Correction Factor = $(\frac{N-3}{N-2.25} )\times \sqrt{\frac{N-2}{N} }$

Therefore,

d =  $\frac{M_1-M_2}{s_{pooled}}$ ×$(\frac{N-3}{N-2.25} )\times \sqrt{\frac{N-2}{N} }$  =  $\frac{6.5-5.8}{1.2}$×$(\frac{49-3}{49-2.25} )\times \sqrt{\frac{49-2}{49} }$ =  $\frac{7}{12}$ ×0.964 = 0.562

Cohen's d = 0.562.

and

(b) To convert d into the coefficient of correlation, r, we have

r = $\frac{d}{\sqrt{d^2 + 4} }$ =  $\frac{0.562 }{\sqrt{0.562 ^2 + 4} }$ = 0.271

r² =  0.07322 or 7.322 × 10⁻².