Respuesta :
Answer:
[tex]t=\frac{(4-3.5)-0}{\sqrt{\frac{1.5^2}{11}+\frac{1^2}{9}}}}=0.890[/tex]
The p value for this case would be:
[tex]p_v =P(t_{18}>0.890)=0.193[/tex]
The p value is higher than the significance level so then we can conclude that we can FAIL to reject the null hypothesis and then the true mean for group A is not significantly higher than the mean for B
Step-by-step explanation:
Information given
[tex]\bar X_{1}=4[/tex] represent the mean for sample A
[tex]\bar X_{2}=3.5[/tex] represent the mean for sample B
[tex]s_{1}=1.5[/tex] represent the sample standard deviation for A
[tex]s_{2}=1[/tex] represent the sample standard deviation for B
[tex]n_{1}=11[/tex] sample size for the group A
[tex]n_{2}=9[/tex] sample size for the group B
[tex]\alpha=0.1[/tex] Significance level provided
t would represent the statistic
Hypothesis to test
We want to verify if the student who graduates from college A has taken more math classes, on the average, the system of hypothesis would be:
Null hypothesis:[tex]\mu_{1}-\mu_{2} \leq 0[/tex]
Alternative hypothesis:[tex]\mu_{1} - \mu_{2}> 0[/tex]
The statistic is given by:
[tex]t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}[/tex] (1)
And the degrees of freedom are given by [tex]df=n_1 +n_2 -2=11+9-2=18[/tex]
Replacing we got:
[tex]t=\frac{(4-3.5)-0}{\sqrt{\frac{1.5^2}{11}+\frac{1^2}{9}}}}=0.890[/tex]
The p value for this case would be:
[tex]p_v =P(t_{18}>0.890)=0.193[/tex]
The p value is higher than the significance level so then we can conclude that we can FAIL to reject the null hypothesis and then the true mean for group A is not significantly higher than the mean for B
