Question

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? , because . 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. H0:μ=50 vs. HA:μ>50.3 B. H0:μ=50 vs. HA:μ≠50 C. H0:μ=50.3 vs. HA:μ<50 D. H0:μ=53.3 vs. HA:μ≠53.3 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. little evidence B. very strong evidence C. strong evidence D. extremely 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

Given that,
population mean(u)=50
sample mean, x =53.3
standard deviation, s =5.2
number (n)=14
null, Ho: μ=50
alternate, H1: μ!=50
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.16
since our test is two-tailed
reject Ho, if to < -2.16 OR if to > 2.16
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =53.3-50/(5.2/sqrt(14))
to =2.3745
| to | =2.3745
critical value
the value of |t α| with n-1 = 13 d.f is 2.16
we got |to| =2.3745 & | t α | =2.16
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 2.3745 ) = 0.0337
hence value of p0.05 > 0.0337,here we reject Ho
ANSWERS
---------------
1.
yes,
reasonable
these data to evaluate the average gas mileage of all 2012 Prius drivers
because . The EPA claims that a 2012 Prius gets 50 MPG (city and highway mileage combined)
2.
t test for single mean
3.
null, Ho: μ=50
alternate, H1: μ!=50
test statistic: 2.3745
critical value: -2.16 , 2.16
4.
p-value: 0.0337
5.
decision: reject Ho
option:B
strong evidence
we have enough evidence to support the claim that a 2012 Prius gets 50 MPG (city and highway mileage combined).
6.
TRADITIONAL METHOD
given that,
sample mean, x =53.3
standard deviation, s =5.2
sample size, n =14
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 5.2/ sqrt ( 14) )
= 1.39
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 13 d.f is 2.16
margin of error = 2.16 * 1.39
= 3.002
III.
CI = x ± margin of error
confidence interval = [ 53.3 ± 3.002 ]
= [ 50.298 , 56.302 ]
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DIRECT METHOD
given that,
sample mean, x =53.3
standard deviation, s =5.2
sample size, n =14
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 13 d.f is 2.16
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 53.3 ± t a/2 ( 5.2/ Sqrt ( 14) ]
= [ 53.3-(2.16 * 1.39) , 53.3+(2.16 * 1.39) ]
= [ 50.298 , 56.302 ]
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interpretations:
1) we are 95% sure that the interval [ 50.298 , 56.302 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean