The following are the miles-per-gallon figures for a sample of cars of the same

Question

The following are the miles-per-gallon figures for a sample of cars of the same model tested under identical conditions: 25 30 28 29 24 25 27 28 26 24 27 31 33 a. The consumers' group that conducted this test is skeptical of the manufacturer's claim that its cars average 28 mpg. The group will only publish a rejection of the manufacturer's claim if it can be 99% sure that it is correct in doing so. Specify the null and alternative hypotheses for this test, determine the critical region, and decide whether to reject the manufacturer's claim. b. Find the 95% confidence interval for the mean gas mileage of all cars of this model. c. Interpret the confidence interval you found in (b). d. Can you conclude that 95% of all cars of this model will have mpg values within this interval? Explain.

Explanation / Answer

Q1.
Given that,
population mean(u)=28
sample mean, x =27.4615
standard deviation, s =2.7573
number (n)=13
null, Ho: =28
alternate, H1: !=28
level of significance, = 0.01
from standard normal table, two tailed t /2 =3.055
since our test is two-tailed
reject Ho, if to < -3.055 OR if to > 3.055
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =27.4615-28/(2.7573/sqrt(13))
to =-0.704
| to | =0.704
critical value
the value of |t | with n-1 = 12 d.f is 3.055
we got |to| =0.704 & | t | =3.055
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -0.7042 ) = 0.4948
hence value of p0.01 < 0.4948,here we do not reject Ho
ANSWERS
---------------
null, Ho: =28
alternate, H1: !=28
test statistic: -0.704
critical value: -3.055 , 3.055
decision: do not reject Ho
p-value: 0.4948
we have evidence to support the claim
Q2.
TRADITIONAL METHOD
given that,
sample mean, x =27.4615
standard deviation, s =2.7573
sample size, n =13
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 2.7573/ sqrt ( 13) )
= 0.765
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 12 d.f is 2.179
margin of error = 2.179 * 0.765
= 1.666
III.
CI = x ± margin of error
confidence interval = [ 27.4615 ± 1.666 ]
= [ 25.795 , 29.128 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =27.4615
standard deviation, s =2.7573
sample size, n =13
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 12 d.f is 2.179
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 = [ 27.4615 ± Z a/2 ( 2.7573/ Sqrt ( 13) ]
= [ 27.4615-(2.179 * 0.765) , 27.4615+(2.179 * 0.765) ]
= [ 25.795 , 29.128 ]
-----------------------------------------------------------------------------------------------
Q3.
interpretations:
1) we are 95% sure that the interval [ 25.795 , 29.128 ] 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

Q4.
no, we conclude that at 95% confidence with mean value lies in b/w this interval

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