Question

Criterion: Identify IV, DV, and hypotheses and evaluate the null hypothesis for an independent samples t test. Data: Use the information from Problem Set 3.8. Instructions: Complete the following: a. Identify the IV and DV in the study. _____________________________________ b. State the null hypothesis and the directional (one-tailed) alternative hypothesis. ___________________________________________________ c. Can you reject the null hypothesis at α = .05? Explain why or why not. ___________________________________________________

Answer 1

The table is missing in the question. The table is provided here :

Group 1        Group 2

 34.86            64.14      mean

 21.99            20.46      standard deviation

  7                    7                n

Solution :

a). The IV or independent variable = Group 1

    The DV or the dependent variable = Group 2

b).

  $H_0: \mu_1 = \mu_2$

  $H_a:\mu_1 < \mu_2$

  Therefore,   $t = \frac{\bar x_1 - \bar x_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{S_2^2}{n_2}}}$

$t = \frac{34.86 - 64.14}{\sqrt{\frac{21.99^2}{7}+\frac{20.46^2}{7}}}$

t = -2.579143

Now,   $df = min(n_1 - 1, n_2 - 1)$

           df = 7 - 1

               = 6

Therefore the value of p :

  $=T.DIST(-2.579143,6,TRUE)$

 = 0.020908803

The p value is 0.0209

$p< 0.05$

So we reject the null hypothesis and conclude that $\mu_1 < \mu_2$