Question

An automobile manufacturer has given its jeep a 50.5 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the actual MPG for this jeep since it is believed that the jeep has an incorrect manufacturer's MPG rating. After testing 280 jeeps, they found a mean MPG of 50.7. Assume the population standard deviation is known to be 1.3. A level of significance of 0.05 will be used. State the null and alternative hypotheses.

Answer 1

Answer:

$z=\frac{50.7-50.5}{\frac{1.3}{\sqrt{280}}}=2.574$    

$p_v =2*P(Z>2.574)=0.010$  

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is different from 50.5 mpg

Step-by-step explanation:

Data given and notation  

$\bar X=50.7$ represent the sample  mean

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

$n=280$ sample size  

$\mu_o =50.5$ represent the value that we want to test

$\alpha=0.05$ 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)  

Sytem of hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 50.5, the system of hypothesis would be:  

Null hypothesis:$\mu = 50.5$  

Alternative hypothesis:$\mu \neq 50.5$  

The statistic is given by:

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

Replacing we got:  

$z=\frac{50.7-50.5}{\frac{1.3}{\sqrt{280}}}=2.574$    

P-value

Since is a two tailed test the p value would be:  

$p_v =2*P(Z>2.574)=0.010$  

Conclusion  

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is different from 50.5 mpg