Question

Two samples from the same population both have M = 84 and s2 = 20, but one sample has n = 10 and the other has n = 20 scores. Both samples are used to evaluate a hypothesis stating that μ = 80 and to compute Cohen’s d. How will the outcomes for the two samples compare?

Answer 1

Complete Question

Two samples from the same population both have M = 84 and s2 = 20, but one sample has n = 10 and the other has n = 20 scores. Both samples are used to evaluate a hypothesis stating that μ = 80 and to compute Cohen’s d. How will the outcomes for the two samples compare?

a.

The larger sample is more likely to reject the hypothesis and will produce a larger value for Cohen’s d.

b.

The larger sample is more likely to reject the hypothesis, but the two samples will have the same value for Cohen’s d.

c.

The larger sample is less likely to reject the hypothesis and will produce a larger value for Cohen’s d.

d.

The larger sample is less likely to reject the hypothesis, but the two samples will have the same value for Cohen’s d.

Answer:

The Cohen's d value is  $d = 0.895$

The  correct option is b

Step-by-step explanation:

From the question we are told that

   The  sample mean of  each population is  $M = 84$

    The  variance of each  population is  $s^2 = 20$

    The  first sample size is  $n_1 = 10$

    The  second  sample size is  $n_2 = 20$

 

The  null hypothesis is  $H_o : \mu = 80$

   Generally the standard deviation is mathematically evaluated as

            $s = \sqrt{20 }$

=>         $s = 4.47$

  The first test statistics is evaluated as

            $t_1 = \frac{M - \mu }{ \frac{\sigma }{ \sqrt{n_1} } }$

    =>    $t_1 = \frac{84 - 80 }{ \frac{4.47 }{ \sqrt{10} } }$

   =>     $t_1 = 2.8298$

  The second  test statistics is evaluated as

           $t_2 = \frac{M - \mu }{ \frac{\sigma }{ \sqrt{n_2} } }$

=>       $t_2 = \frac{84 - 80 }{ \frac{4.47 }{ \sqrt{20} } }$

=>       $t_2 = 4.0$

The sample with the larger test statistics (sample  size) will more likely reject the null hypothesis

Generally the Cohen's d value is mathematically evaluated as

          $d = \frac{M - \mu }{s }$

  =>     $d = \frac{ 84 - 80 }{4.47 }$    

=>         $d = 0.895$

Given that the the sample mean and  sample size are the same for both sample the Cohen's d value will be  the same