Question

. An independent-measures research study was used to compare two treatment conditions with n=12 participants for each treatment. The first treatment condition had M=55 with a variance of s2=8, and the second treatment condition had M= 52 and s2=4. Look up the appropriate critical value based on your computed degrees of freedom. Do these data indicate a significant difference between the two treatments

Answer 1

Answer:

Decision rule  is  

Fail to reject the null hypothesis

 The conclusion is  

There is sufficient evidence that indicate that there is a significant difference between the two treatments

Step-by-step explanation:

From the question we are told that  

    The sample size is  n =  12

    The first  mean is  $M_1 = 55$

    The variance is  $s^2_1 = 8$

    The second mean is  $M_2 = 52$

     The second variance is $s^2_2 = 4$

Let the level of significance be  $\alpha = 0.05$

The null hypothesis is  $H_o : M_1 = M_2$

The alternative hypothesis is $H_a : M_1 \ne M_2$

  Generally the test statistics is mathematically

     $t = \frac{M_1 - M_2 }{ \sqrt{\frac{s_1^2}{n_1} +\frac{s_2^2}{n_2}} }$

=>   $t = \frac{55 - 52 }{ \sqrt{\frac{8}{12} +\frac{4}{12 }} }$

=>   $t = 1.1619$

Generally the degree of freedom is mathematically represented as

      $df = 12 -1$

=>    $df = 11$

Generally  from the t- distriibution table the critical  value  of  $\alpha$ at a degree of freedom of $df = 11$  is

    $t_{0.05 , 11} = 2.20098516$

Here given that the critical value is greater than the t statistics value the

Decision rule  is  

Fail to reject the null hypothesis

 The conclusion is  

There is sufficient evidence that indicate that there is a significant difference between the two treatments