Question

An automobile manufacturer claims that its car has a 33.7 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the MPG for this car since it is believed that the car has an incorrect manufacturer's MPG rating. After testing 120 cars, they found a mean MPG of 34.0. Assume the variance is known to be 2.56. A level of significance of 0.02 will be used. Find the value of the test statistic. Round your answer to 2 decimal places.Enter the value of the test statistic.

Answer 1

Answer:

$z=\frac{34-33.7}{\frac{1.6}{\sqrt{120}}}=2.05$  

And the p value would be given by:

$p_v= 2*P(z>2.05) = 0.0404$

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  

$\bar X=34$ represent the sample mean

$\sigma=\sqrt{2.56}= 1.6$ represent the population standard deviation

$n=120$ sample size  

$\mu_o =33.7$ represent the value that we want to test  

$\alpha=0.02$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

$p_v$ 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:$\mu =33.7$  

Alternative hypothesis:$\mu \neq 33.7$  

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:  

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

$z=\frac{34-33.7}{\frac{1.6}{\sqrt{120}}}=2.05$  

And the p value would be given by:

$p_v= 2*P(z>2.05) = 0.0404$

And since the p value is higher than the significance level we have enough evidence to fail to reject the null hypothesis.