Respuesta :
Answer:
[tex]t=\frac{(10 -7)-(0)}{\sqrt{\frac{2.268^2}{8}+\frac{1.414^2}{6}}}=3.036[/tex]
[tex]p_v =P(t_{12}>3.036) =0.0051[/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 1 (Orno) is significantly higher than the mean for the group 2 (Edne).
Step-by-step explanation:
The system of hypothesis on this case are:
Null hypothesis: [tex]\mu_1 \leq \mu_2[/tex]
Alternative hypothesis: [tex]\mu_1 > \mu_2[/tex]
Or equivalently:
Null hypothesis: [tex]\mu_1 - \mu_2 \leq 0[/tex]
Alternative hypothesis: [tex]\mu_1 -\mu_2 > 0[/tex]
Our notation on this case :
[tex]n_1 =8[/tex] represent the sample size for group 1 (Orno)
[tex]n_2 =6[/tex] represent the sample size for group 2 (Edne)
We can calculate the sampel means and deviations with the following formulas:
[tex]\bar X =\frac{\sum_{i=1}^n X_i}{n}[/tex]
[tex]s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
[tex]\bar X_1 =10[/tex] represent the sample mean for the group 1
[tex]\bar X_2 =7[/tex] represent the sample mean for the group 2
[tex]s_1=2.268[/tex] represent the sample standard deviation for group 1
[tex]s_2=1.414[/tex] represent the sample standard deviation for group 2
If we see the alternative hypothesis we see that we are conducting a right tailed test.
On this case since the significance assumed is 0.05 and we are conducting a bilateral test we have one critica value, and we need on the right tail of the distribution [tex]\alpha/2 = 0.05[/tex] of the area.
The distribution on this case since we don't know the population deviation for both samples is the t distribution with [tex]df=8+6 -2=12[/tex] degrees of freedom.
We can use the following excel code in order to find the critical value:
"=T.INV(1-0.05,12)"
And the rejection zone is: (1.78,infinity)
The statistic is given by this formula:
[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}+\frac{S^2_2}{n_2}}}[/tex]
And now we can calculate the statistic:
[tex]t=\frac{(10 -7)-(0)}{\sqrt{\frac{2.268^2}{8}+\frac{1.414^2}{6}}}=3.036[/tex]
The degrees of freedom are given by:
[tex]df=8+6-2=12[/tex]
And now we can calculate the p value using the altenative hypothesis:
[tex]p_v =P(t_{12}>3.036) =0.0051[/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 1 (Orno) is significantly higher than the mean for the group 2 (Edne).
