Respuesta :
Answer:
[tex]t=\frac{51.5-50}{\frac{6}{\sqrt{36}}}=1.5[/tex]
Now the p value for this case would be given by:
[tex]p_v =P(z>1.5)=0.0668[/tex]
Assuming a significance level of 0.05 or 5% we see that the p value is higher than the significance level we can conclude that we FAIL to reject the null hypothesis and we don't have enough evidence to conclude that the true mean is higher than 50 miles per gallon
Step-by-step explanation:
Information given
[tex]\bar X=51.5[/tex] represent the sample mean for the new small cars
[tex]\sigma=6[/tex] represent the population deviation
[tex]n=36[/tex] sample size
[tex]\mu_o =50[/tex] represent the value to check
t would represent the statistic
[tex]p_v[/tex] represent the p value for the test
System of hypothesis
We want to check if the new small cars average more than 50 miles per gallon in highway driving, the system of hypothesis are:
Null hypothesis:[tex]\mu \leq 50[/tex]
Alternative hypothesis:[tex]\mu > 50[/tex]
Since the population deviation is given the the statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
The statistic for this case is given by:
[tex]t=\frac{51.5-50}{\frac{6}{\sqrt{36}}}=1.5[/tex]
Now the p value for this case would be given by:
[tex]p_v =P(z>1.5)=0.0668[/tex]
Assuming a significance level of 0.05 or 5% we see that the p value is higher than the significance level we can conclude that we FAIL to reject the null hypothesis and we don't have enough evidence to conclude that the true mean is higher than 50 miles per gallon
