Respuesta :
Answer:
[tex]z=\frac{46.5-46.7}{\frac{1.1}{\sqrt{150}}}=-2.23[/tex]
The p value would be given by:
[tex]p_v =2*P(z<-2.23)=0.0257[/tex]
For this case since th p value is lower than the significance level of0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case is significantly different from 46.7 MPG
Step-by-step explanation:
Information given
[tex]\bar X=46.5[/tex] represent the mean
[tex]\sigma=1.1[/tex] represent the population standard deviation
[tex]n=150[/tex] sample size
[tex]\mu_o =46.7[/tex] represent the value to verify
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
z would represent the statistic
[tex]p_v[/tex] represent the p value
Hypothesis to test
We want to test if the true mean for this case is 46.7, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 46.7[/tex]
Alternative hypothesis:[tex]\mu \neq 46.7[/tex]
Since we know the population deviation the statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
Replacing we got:
[tex]z=\frac{46.5-46.7}{\frac{1.1}{\sqrt{150}}}=-2.23[/tex]
The p value would be given by:
[tex]p_v =2*P(z<-2.23)=0.0257[/tex]
For this case since th p value is lower than the significance level of0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case is significantly different from 46.7 MPG