Answer:
[tex]p_v =P(t_{88}>6.960) =2.91x10^{-10}[/tex]
So with the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group A is significantly lower than the mean for the group B.
Step-by-step explanation:
The statistic to test the hypothesis is given by this formula:
[tex]t=\frac{(\bar X_A-\bar X_B)-(\mu_{A}-\mu_B)}{\sqrt{\frac{s_A^2}{n_A}}+\frac{s_B^2}{n_B}}[/tex]
Where t follows a t distribution with [tex]n_1+n_2 -2[/tex] degrees of freedom
The system of hypothesis on this case are:
Null hypothesis: [tex]\mu_A \leq \mu_B[/tex]
Alternative hypothesis: [tex]\mu_A >\mu_B[/tex]
Or equivalently:
Null hypothesis: [tex]\mu_A - \mu_B \leq 0[/tex]
Alternative hypothesis: [tex]\mu_A -\mu_B>0[/tex]
Our notation on this case :
[tex]n_A =50[/tex] represent the sample size for group A
[tex]n_B =40[/tex] represent the sample size for group B
[tex]\bar X_A =280[/tex] represent the sample mean for the group A
[tex]\bar X_B =250[/tex] represent the sample mean for the group B
[tex]s_A=20[/tex] represent the sample standard deviation for group A
[tex]s_B=23[/tex] represent the sample standard deviation for group B
And now we can calculate the statistic:
[tex]t=\frac{(280 -250)-(0)}{\sqrt{\frac{20^2}{50}}+\frac{23^2}{40}}=6.511[/tex]
Now we can calculate the degrees of freedom given by:
[tex]df=50+40-2=88[/tex]
And now we can calculate the p value using the alternative hypothesis:
[tex]p_v =P(t_{88}>6.960) =2.91x10^{-10}[/tex]
So with the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group A is significantly lower than the mean for the group B.
