An advertised average fuel consumption for a 2017 Land Rover LWB is 14 MPG (City

Question

An advertised average fuel consumption for a 2017 Land Rover LWB is 14 MPG (City). You choose a 0.05 significance level to perform a hypothesis test to verify this claim using your brand new Land Rover. You record your fuel consumption and record the averages for every month. After 16 month you have a sample consisting of 16 data points: 14.61, 8.60, 11.85, 10.13, 15.26, 17.16, 18.77, 15.01, 9.21, 17.20, 16.22, 15.54, 11.8, 13.94, 9.36, 10.71 (a) Set the significance level to . 0.05 and use the null hypothesis significance test to verify that the actual average fuel consumption is 14 MPG. (b) Find the p-value corresponding to the given sample. (c) Using the p-value from (b), find the smallest significance level that would lead to the rejection of the null hypothesis.

Explanation / Answer

Given that,
population mean(u)=14
sample mean, x =13.4613
standard deviation, s =3.2401
number (n)=16
null, Ho: =14
alternate, H1: !=14
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.131
since our test is two-tailed
reject Ho, if to < -2.131 OR if to > 2.131
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =13.4613-14/(3.2401/sqrt(16))
to =-0.665
| to | =0.665
critical value
the value of |t | with n-1 = 15 d.f is 2.131
we got |to| =0.665 & | t | =2.131
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -0.665 ) = 0.5161
hence value of p0.05 < 0.5161,here we do not reject Ho
ANSWERS
---------------
null, Ho: =14
alternate, H1: !=14
test statistic: -0.665
critical value: -2.131 , 2.131
decision: do not reject Ho
p-value: 0.5161

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