Question

An automobile manufacturer claims that its van has a 31.3 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the MPG for this van since it is believed that the van has an incorrect manufacturer's MPG rating. After testing 140 vans, they found a mean MPG of 31.1. Assume the standard deviation is known to be 1.3. A level of significance of 0.02 will be used. Find the value of the test statistic. Round your answer to 2 decimal places.

Answer 1

Answer:

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

$z=\frac{31.1-31.3}{\frac{1.3}{\sqrt{140}}}=-1.82$  

$p_v =2*P(z$  

If we compare the p value and the significance level given $\alpha=0.02$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is NOT significantly different from 31.3 at 2% of signficance.  

Step-by-step explanation:

Data given and notation  

$\bar X=31.1$ represent the sample mean

$\sigma=1.3$ represent the population standard deviation for the sample  

$n=140$ sample size  

$\mu_o =31.3$ 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)  

State the null and alternative hypotheses.  

We can assume that we conduct a hypothesis in order to check if the true mean is equal to 31.3 , the system of hypothesis would be:  

Null hypothesis:$\mu =31.3$  

Alternative hypothesis:$\mu \neq 31.3$  

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)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

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

$z=\frac{31.1-31.3}{\frac{1.3}{\sqrt{140}}}=-1.82$  

P-value  

Since is a two-sided test the p value would be:  

$p_v =2*P(z$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.02$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is NOT significantly different from 31.3 at 2% of signficance.