Assume you are working at the consumer protection agency, recently you have been

Question

Assume you are working at the consumer protection agency, recently you have been getting complaints about the high gas mileage of a new minivan. The car company agrees to allow you to select randomly 41 of its new minivans to test their highway mileage. The company claims that it's minivans get 28 miles per gallon on the highway.   Your test results show a sample mean of 26.7 and a sample deviation of 4.2
Part one
Calculate a 95% confidence interval around your sample mean
Is the claim mean inside your confidence interval?
What does your result mean, in terms of the company's claim?
Part two (two-tail test)
List the null and alternative hypotheses for the appropriate test.
Use alpha = 0.05. Find the critical value(s) and calculate the observed value of the test statistic.
Is the observed test statistic in the critical (rejection) region?
Will the p-value be higher or lower than your alpha? What does this result mean, in terms of the company's claim?
Part three(one-tail test)
List the null and alternative hypotheses for the appropriate test.
Use alpha = 0.05. Find the critical value(s)and calculate the observed value of the test statistic
Is the observed test statistic in the critical region?
Will the p-value be higher or lower than your alpha?
What does this result mean, in terms of the company's claim?

Explanation / Answer

PART 1.
given that,
sample mean, x =26.7
standard deviation, s =4.2
sample size, n =41
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 40 d.f is 2.021
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 = [ 26.7 ± t a/2 ( 4.2/ Sqrt ( 41) ]
= [ 26.7-(2.021 * 0.656) , 26.7+(2.021 * 0.656) ]
= [ 25.374 , 28.026 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 25.374 , 28.026 ] 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

PART 2.
null, Ho: =28
alternate, H1: !=28
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.021
since our test is two-tailed
reject Ho, if to < -2.021 OR if to > 2.021
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =26.7-28/(4.2/sqrt(41))
to =-1.9819
| to | =1.9819
critical value
the value of |t | with n-1 = 40 d.f is 2.021
we got |to| =1.9819 & | t | =2.021
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -1.9819 ) = 0.0544
hence value of p0.05 < 0.0544,here we do not reject Ho
ANSWERS
---------------
null, Ho: =28
alternate, H1: !=28
test statistic: -1.9819
critical value: -2.021 , 2.021
decision: do not reject Ho
p-value: 0.0544

PART 3
Given that,
population mean(u)=28
sample mean, x =26.7
standard deviation, s =4.2
number (n)=41
null, Ho: >28
alternate, H1: <28
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.684
since our test is left-tailed
reject Ho, if to < -1.684
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =26.7-28/(4.2/sqrt(41))
to =-1.9819
| to | =1.9819
critical value
the value of |t | with n-1 = 40 d.f is 1.684
we got |to| =1.9819 & | t | =1.684
make decision
hence value of | to | > | t | and here we reject Ho
p-value :left tail - Ha : ( p < -1.9819 ) = 0.02719
hence value of p0.05 > 0.02719,here we reject Ho
ANSWERS
---------------
null, Ho: >28
alternate, H1: <28
test statistic: -1.9819
critical value: -1.684
decision: reject Ho
p-value: 0.02719

we have evidence that minivans get 28 miles per gallon on the highway

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