Respuesta :
Answer:
[tex]t=\frac{(350-342)-0}{\sqrt{\frac{12^2}{15}+\frac{15^2}{17}}}}=1.674[/tex]
The degreess of freedom are given by:
[tex] df = 15+12-2 =25[/tex]
Now we can calculate the p value with the following probability:
[tex]p_v =2*P(t_{25}>1.674)=0.107[/tex]
For this case since the p value is higher than the significance level we can FAIL to reject the null hypothesis and we can conclude that the true means are NOT significantly different at 10% of significance.
Step-by-step explanation:
Information provided
[tex]\bar X_{1}=350[/tex] represent the mean for sample 1
[tex]\bar X_{2}=342[/tex] represent the mean for sample 2
[tex]s_{1}=12[/tex] represent the sample standard deviation for 1
[tex]s_{2}=15[/tex] represent the sample standard deviation for 2
[tex]n_{1}=15[/tex] sample size for the group 2
[tex]n_{2}=17[/tex] sample size for the group 2
[tex]\alpha=0.1[/tex] Significance level provided
t would represent the statistic
Hypothesis to test
We want to verify if the true means for this case are significantly different, the system of hypothesis would be:
Null hypothesis:[tex]\mu_{1}-\mu_{2}=0[/tex]
Alternative hypothesis:[tex]\mu_{1} - \mu_{2}\neq 0[/tex]
The statistic is given by:
[tex]t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}[/tex] (1)
And the degrees of freedom are given by [tex]df=n_1 +n_2 -2=15+17-2=30[/tex]
Replacing the info given we got:
[tex]t=\frac{(350-342)-0}{\sqrt{\frac{12^2}{15}+\frac{15^2}{17}}}}=1.674[/tex]
The degreess of freedom are given by:
[tex] df = 15+12-2 =25[/tex]
Now we can calculate the p value with the following probability:
[tex]p_v =2*P(t_{25}>1.674)=0.107[/tex]
For this case since the p value is higher than the significance level we can FAIL to reject the null hypothesis and we can conclude that the true means are NOT significantly different at 10% of significance.
