Respuesta :
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 = [tex]\overline{\rm x}[/tex] = 6.5
Mean of population = μ[tex]_x[/tex] = 5.8
Sample standard deviation, s = 1.2
The t statistic is given as
[tex]t = \frac{\overline{\rm x} - \mu_x}{\frac{s}{\sqrt{N} } }[/tex]
t = [tex]\frac{6.5 - 5.8}{\frac{1.2}{\sqrt{49} } }[/tex] = 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 = [tex]\frac{M_1-M_2}{s_{pooled}}[/tex] =
Where:
M₁ = Group 1 mean = 6.5
M₂ = Group 2 mean = 5.8
[tex]s_{pooled}[/tex] = [tex]\frac{\sqrt{s_1^2+s_2^2} }{2}[/tex]
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 [tex]s_{pooled}[/tex] = [tex]\sqrt{\frac{2\cdot s_1^2 }{2}[/tex] = [tex]{\sqrt{s_1^2}[/tex] = s₁
Therefore
d = [tex]\frac{M_1-M_2}{s_{pooled}}[/tex] = [tex]\frac{6.5-5.8}{1.2}[/tex] = [tex]\frac{7}{12}[/tex] = 0.583
However since N < 50 we make use of the correction factor as follows.
Correction Factor = [tex](\frac{N-3}{N-2.25} )\times \sqrt{\frac{N-2}{N} }[/tex]
Therefore,
d = [tex]\frac{M_1-M_2}{s_{pooled}}[/tex] ×[tex](\frac{N-3}{N-2.25} )\times \sqrt{\frac{N-2}{N} }[/tex] = [tex]\frac{6.5-5.8}{1.2}[/tex]×[tex](\frac{49-3}{49-2.25} )\times \sqrt{\frac{49-2}{49} }[/tex] = [tex]\frac{7}{12}[/tex] ×0.964 = 0.562
Cohen's d = 0.562.
and
(b) To convert d into the coefficient of correlation, r, we have
r = [tex]\frac{d}{\sqrt{d^2 + 4} }[/tex] = [tex]\frac{0.562 }{\sqrt{0.562 ^2 + 4} }[/tex] = 0.271
r² = 0.07322 or 7.322 × 10⁻².
