Question

HW9-Practice: Problem 3

Problem Value: 1 point(s). Problem Score: 50%. Attempts Remaining: Unlimited.

(1 point) 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 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 are not guaranteed. Report all answers to 4 decimal places.

1. 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?

? Yes No , because  ? the data distribution seems approximately normal there are 14 data points in the sample user estimates are reliable user estimates are not reliable .

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? Conduct a hypothesis test. Round numeric answers to 3 decimal places where necessary.

2. What are the correct hypotheses for conducting a hypothesis test to determine if these data provide strong evidence against this estimate for drivers who participate on fueleconomy.gov? (Reminder: check conditions)

A. ?0:?=50H0:μ=50 vs. ??:?≠50HA:μ≠50
B. ?0:?=53.3H0:μ=53.3 vs. ??:?≠53.3HA:μ≠53.3
C. ?0:?=50.3H0:μ=50.3 vs. ??:?<50HA:μ<50
D. ?0:?=50H0:μ=50 vs. ??:?>50.3HA:μ>50.3

3. Calculate the test statistic.

4. Calculate the p-value.

5. How much evidence do we have that the null model is not compatible with our observed results?

A. extremely strong evidence
B. little evidence
C. very strong evidence
D. strong evidence
E. some evidence

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

( ,  )

Answer 1

1)Yes because  the data distribution seems approximately normal

2)

option A

3)

population mean μ=	50
sample mean 'x̄=	53.300
sample size   n=	14.00
sample std deviation s=	5.200
std error 'sx=s/√n=	1.390
test stat t ='(x-μ)*√n/sx=	2.3745

4)

p value      =

0.0337

5)

C. very strong evidence

6)

for 95% CI; and 13 df, value of t=		2.1604
margin of error E=t*std error    =		3.0024
lower bound=sample mean-E =		50.2976
Upper bound=sample mean+E =		56.3024
from above 95% confidence interval for population mean =(50.2976 ,56.3024)