Respuesta :
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).
[tex]$H_0: \mu_1 = \mu_2$[/tex]
[tex]$H_a:\mu_1 < \mu_2$[/tex]
Therefore, [tex]$t = \frac{\bar x_1 - \bar x_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{S_2^2}{n_2}}}$[/tex]
[tex]$t = \frac{34.86 - 64.14}{\sqrt{\frac{21.99^2}{7}+\frac{20.46^2}{7}}}$[/tex]
t = -2.579143
Now, [tex]$df = min(n_1 - 1, n_2 - 1)$[/tex]
df = 7 - 1
= 6
Therefore the value of p :
[tex]$=T.DIST(-2.579143,6,TRUE)$[/tex]
= 0.020908803
The p value is 0.0209
[tex]$p< 0.05$[/tex]
So we reject the null hypothesis and conclude that [tex]$\mu_1 < \mu_2$[/tex]
