Respuesta :
Answer:
[tex]\S^2_p =\frac{(18-1)(2.5)^2 +(18 -1)(2.3)^2}{18 +18 -2}=5.77[/tex]
[tex]S_p=2.402[/tex]
[tex]t=\frac{(12 -13)-(0)}{2.402\sqrt{\frac{1}{18}+\frac{1}{18}}}=-1.249[/tex]
[tex]p_v =2*P(t_{34}<-1.249) =0.2201[/tex]
Step-by-step explanation:
When we have two independent samples from two normal distributions with equal variances we are assuming that
[tex]\sigma^2_1 =\sigma^2_2 =\sigma^2[/tex]
And the statistic is given by this formula:
[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}[/tex]
Where t follows a t distribution with [tex]n_1+n_2 -2[/tex] degrees of freedom and the pooled variance [tex]S^2_p[/tex] is given by this formula:
[tex]\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}[/tex]
This last one is an unbiased estimator of the common variance [tex]\sigma^2[/tex]
The system of hypothesis on this case are:
Null hypothesis: [tex]\mu_1 = \mu_2[/tex]
Alternative hypothesis: [tex]\mu_1 \neq \mu_2[/tex]
Or equivalently:
Null hypothesis: [tex]\mu_1 - \mu_2 = 0[/tex]
Alternative hypothesis: [tex]\mu_1 -\mu_2 \neq 0[/tex]
Our notation on this case :
[tex]n_1 =18[/tex] represent the sample size for group 1
[tex]n_2 =18[/tex] represent the sample size for group 2
[tex]\bar X_1 =12[/tex] represent the sample mean for the group 1
[tex]\bar X_2 =13[/tex] represent the sample mean for the group 2
[tex]s_1=2.5[/tex] represent the sample standard deviation for group 1
[tex]s_2=2.3[/tex] represent the sample standard deviation for group 2
First we can begin finding the pooled variance:
[tex]\S^2_p =\frac{(18-1)(2.5)^2 +(18 -1)(2.3)^2}{18 +18 -2}=5.77[/tex]
And the deviation would be just the square root of the variance:
[tex]S_p=2.402[/tex]
And now we can calculate the statistic:
[tex]t=\frac{(12 -13)-(0)}{2.402\sqrt{\frac{1}{18}+\frac{1}{18}}}=-1.249[/tex]
Now we can calculate the degrees of freedom given by:
[tex]df=18+18-2=34[/tex]
And now we can calculate the p value using the altenative hypothesis:
[tex]p_v =2*P(t_{34}<-1.249) =0.2201[/tex]
If we compare the p value obtained and using the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.
