4. A consumer advocate group would like to disprove a car manufacturer\'s claim

Question

4. A consumer advocate group would like to disprove a car manufacturer's claim that a specific model will average 24 miles per gallon of gasoline. Specifically, the group would like to show that the mean miles per gallon is considerably lower than 24 miles per gallon. The group took a random sample of 30 cars and obtaineda sample mean of 22.5 mpg . Assume -29. Let alpha-0.05. a.) State the null an alternative hypotheses Ho: Ha: b.) Perform the test of significance. c.) Graph. d.) State your conclusion. E) Construct a 95% C.1, for the mean mpg.

Explanation / Answer

4.

Given that,
population mean(u)=24
standard deviation, =2.9
sample mean, x =22.5
number (n)=30
null, Ho: =24
alternate, H1: <24
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 22.5-24/(2.9/sqrt(30)
zo = -2.83305
| zo | = 2.83305
critical value
the value of |z | at los 5% is 1.645
we got |zo| =2.83305 & | z | = 1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : left tail - ha : ( p < -2.83305 ) = 0.00231
hence value of p0.05 > 0.00231, here we reject Ho
ANSWERS
---------------
a.
null, Ho: =24
alternate, H1: <24
b.
test statistic: -2.83305
critical value: -1.645
d.
decision: reject Ho
p-value: 0.00231
we have enough evidence to support the claim that mean miles per gallon is considerable lower than 24 miles per gallon

e.
TRADITIONAL METHOD
given that,
standard deviation, =2.9
sample mean, x =22.5
population size (n)=30
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 2.9/ sqrt ( 30) )
= 0.529
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.645
since our test is left-tailed
value of z table is 1.645
margin of error = 1.645 * 0.529
= 0.871
III.
CI = x ± margin of error
confidence interval = [ 22.5 ± 0.871 ]
= [ 21.629,23.371 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =2.9
sample mean, x =22.5
population size (n)=30
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.645
since our test is left-tailed
value of z table is 1.645
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 22.5 ± Z a/2 ( 2.9/ Sqrt ( 30) ) ]
= [ 22.5 - 1.645 * (0.529) , 22.5 + 1.645 * (0.529) ]
= [ 21.629,23.371 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [21.629 , 23.371 ] 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
[ANSWERS]
best point of estimate = mean = 22.5
standard error =0.529
z table value = 1.645
margin of error = 0.871
confidence interval = [ 21.629 , 23.371 ]

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