Respuesta :
Answer:
[tex]z=\frac{31.1-31.3}{\frac{1.3}{\sqrt{140}}}=-1.82[/tex]
The p value for this case would be given by:
[tex]p_v =2*P(z<-1.82)=0.0688[/tex]
For this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly different from 31.3 MPG
Step-by-step explanation:
Information given
[tex]\bar X=31.1[/tex] represent the sample mean
[tex]\sigma=1.3[/tex] represent the population standard deviation
[tex]n=140[/tex] sample size
[tex]\mu_o =31.3[/tex] represent the value that we want to test
[tex]\alpha=0.02[/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 is equal to 31.3 MPG, the system of hypothesis would be:
Null hypothesis:[tex]\mu =31.3[/tex]
Alternative hypothesis:[tex]\mu \neq 31.3[/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{31.1-31.3}{\frac{1.3}{\sqrt{140}}}=-1.82[/tex]
The p value for this case would be given by:
[tex]p_v =2*P(z<-1.82)=0.0688[/tex]
For this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly different from 31.3 MPG