Answer:
[tex]z=\frac{34-33.7}{\frac{1.6}{\sqrt{120}}}=2.05[/tex]
And the p value would be given by:
[tex] p_v= 2*P(z>2.05) = 0.0404[/tex]
And since the p value is higher than the significance level we have enough evidence to fail to reject the null hypothesis.
Step-by-step explanation:
Data given and notation
[tex]\bar X=34[/tex] represent the sample mean
[tex]\sigma=\sqrt{2.56}= 1.6[/tex] represent the population standard deviation
[tex]n=120[/tex] sample size
[tex]\mu_o =33.7[/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 (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
System of hypotheses.
We need to conduct a hypothesis in order to check if the true mean is 33.7 mpg or no, the system of hypothesis would be:
Null hypothesis:[tex]\mu =33.7[/tex]
Alternative hypothesis:[tex]\mu \neq 33.7[/tex]
Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]z=\frac{34-33.7}{\frac{1.6}{\sqrt{120}}}=2.05[/tex]
And the p value would be given by:
[tex] p_v= 2*P(z>2.05) = 0.0404[/tex]
And since the p value is higher than the significance level we have enough evidence to fail to reject the null hypothesis.