Business, 03.01.2020 06:31 kkwolfcityouwc96

8fuel efficiency of prius: fueleconomy. gov, the official us government source for fuel economy information, allows users to share gas mileage information on their vehicles. the histogram below shows the distribution of gas mileage in miles per gallon (mpg) from 14 users who drive a 2012 toyota prius. the sample mean is 53.3 mpg and the sample standard deviation is 5.2 mpg. (note that these data are user estimates and since the source data cannot be verified, the accuracy of these estimates is not guaranteed.) use your ti calculator.

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a) We assume that the plot is given on the figure attached. The total sample size is 1+2+7+2+2= 14 and for this case we can conduct a test since we assume normality and we can use the t distribution in order to work with a sample size less than 30.

b) $p_v =2*P(t_{(13)}2.375)=0.0336$

If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its significant different from 50 at 1% of signficance.

c) The 95% confidence interval would be given by (50.298;56.302)

Explanation:

(a) We would like to use these data to evaluate the average gas mileage of all 2012 Prius drivers. Do you think this is reasonable? Why or why not?

We assume that the plot is given on the figure attached. The total sample size is 1+2+7+2+2= 14 and for this case we can conduct a test since we assume normality and we can use the t distribution in order to work with a sample size less than 30.

(b) The EPA claims that a 2012 Prius gets 50 MPG (city and highway mileage combined). Do these data provide strong evidence against this estimate for drivers who participate on fueleconomy.gov? Note any assumptions you must make as you proceed with the test.

Data given and notation

$\bar X=53.3$ represent the sample mean

s=5.2 represent the sample standard deviation

n=14 sample size

$\mu_o =50$ represent the value that we want to test

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

t 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 need to conduct a hypothesis in order to check if the mean is actually 50 or different, the system of hypothesis would be:

Null hypothesis: $\mu = 50$

Alternative hypothesis: $\mu \neq 50$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-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:

$t=\frac{53.3-50}{\frac{5.2}{\sqrt{14}}}=2.375$

P-value

The first step is calculate the degrees of freedom, on this case:

df=n-1=14-1=13

Since is a one side test the p value would be:

$p_v =2*P(t_{(13)}2.375)=0.0336$

Conclusion

(c) Calculate a 95% confidence interval for the average gas mileage of a 2012 Prius by drivers who participate on fueleconomy.gov.

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

df=n-1=14-1=13

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,13)".And we see that $t_{\alpha/2}=2.16$

Now we have everything in order to replace into formula (1):

$53.3-2.16\frac{5.2}{\sqrt{14}}=50.298$

$53.3+2.16\frac{5.2}{\sqrt{14}}=56.302$

So on this case the 95% confidence interval would be given by (50.298;56.302)

8fuel efficiency of prius: fueleconomy.gov, the official us government source for fuel economy info

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