Question

An automobile manufacturer has given its car a 31.131.1 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the actual MPG for this car since it is believed that the car performs over the manufacturer's MPG rating. After testing 110110 cars, they found a mean MPG of 31.331.3. Assume the population variance is known to be 3.613.61. Is there sufficient evidence at the 0.10.1 level to support the testing firm's claim?

Answer 1

Answer:

$z=\frac{31.3-31.1}{\frac{1.9}{\sqrt{110}}}=1.104$    

Now we can calculate the p value with this probability:

$p_v =P(z>1.104)=0.135$  

Since the p value is higher than the significance level provided of 0.1 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean for this case is higher than 31.1MPG using a significance level of 10%. So then the cliam makes sense

Step-by-step explanation:

Information given

$\bar X=31.3$ represent the sample mean

$\sigma=\sqrt{3.61}= 1.9$ represent the population deviation

$n=110$ sample size  

$\mu_o =31.1$ represent the value to verify

$\alpha=0.1$ represent the significance level

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to check if the true mean for this case is higher than 31.1 MPG, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 31.1$  

Alternative hypothesis:$\mu > 31.1$  

Since we know the population deviation the statistic would be given by:

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

Replacing the info given we got:

$z=\frac{31.3-31.1}{\frac{1.9}{\sqrt{110}}}=1.104$    

Now we can calculate the p value with this probability:

$p_v =P(z>1.104)=0.135$  

Since the p value is higher than the significance level provided of 0.1 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean for this case is higher than 31.1MPG using a significance level of 10%. So then the cliam makes sense