Respuesta :
Answer:
[tex]z=\frac{(3.2-3.4)-0}{\sqrt{\frac{0.141^2}{60}+\frac{0.224^2}{60}}}}=-5.85[/tex]
The p value can be calculated with this probability:
[tex]p_v =2*P(z<-5.85)=4.91x10^{-9}[/tex]
The p value for this case is a value very low and near to 0 so then we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different
Step-by-step explanation:
Information provided
[tex]\bar X_{1}=3.2[/tex] represent the mean for sample A
[tex]\bar X_{2}=3.4[/tex] represent the mean for sample B
[tex]\sigma_{1}=\sqrt{0.02}= 0.141[/tex] represent the sample standard deviation for A
[tex]s_{2}=\sqrt{0.05}= 0.224[/tex] represent the sample standard deviation for B
[tex]n_{1}=60[/tex] sample size for the group A
[tex]n_{2}=60[/tex] sample size for the group B
[tex]\alpha=0.05[/tex] Significance level provided
z would represent the statistic
Hypothesis to test
We want to verify if that there is a difference in the interest rates paid by the two states, 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 for this case since we know the population deviations is given by:
[tex]z=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}[/tex] (1)
Replacing the info given we got:
[tex]z=\frac{(3.2-3.4)-0}{\sqrt{\frac{0.141^2}{60}+\frac{0.224^2}{60}}}}=-5.85[/tex]
The p value can be calculated with this probability:
[tex]p_v =2*P(z<-5.85)=4.91x10^{-9}[/tex]
The p value for this case is a value very low and near to 0 so then we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different
