The manager of a paint supply store wants to determine whether the mean amount o
ID: 3364964 • Letter: T
Question
The manager of a paint supply store wants to determine whether the mean amount of paint contained in 1-gallon cans purchased from a nationally known manufacturer is actually 1 gallon. You know from the manufacturer's specifications that the standard deviation of the amount of paint is
0.0080.008
gallon. You select a random sample of
4545
cans, and the mean amount of paint per 1-gallon can is
0.9950.995
gallon. Complete parts (a) through (e) below.
Is there evidence that the mean amount is different from
11.0
gallon? (Use
alphaequals=0.050.05.)
Let
mu
be the population mean. Determine the null hypothesis,
Upper H 0H0,
and the alternative hypothesis,
Upper H 1H1.
Upper H 0H0:
mu
greater than or equals
not equals
less than or equals
equals=
11.0
Upper H 1H1:
mu
equals=
not equals
less than<
greater than>
11.0
What is the test statistic?
Upper Z Subscript STATZSTATequals=nothing
(Round to two decimal places as needed.)What is/are the critical value(s)? (Use
alphaequals=0.050.05.)
nothing
(Round to two decimal places as needed. Use a comma to separate answers as needed.)
What is the final conclusion?
A.
Fail to rejectFail to reject
Upper H 0H0.
There is
sufficientsufficient
evidence that the mean amount is different from
11.0
gallon.
B.
Fail to rejectFail to reject
Upper H 0H0.
There is
insufficientinsufficient
evidence that the mean amount is different from
11.0
gallon.
C.
RejectReject
Upper H 0H0.
There is
sufficientsufficient
evidence that the mean amount is different from
11.0
gallon.
D.
RejectReject
Upper H 0H0.
There is
insufficientinsufficient
evidence that the mean amount is different from
11.0
gallon.
b. Compute the p-value and interpret its meaning.
What is the p-value?
nothing
(Round to three decimal places as needed.)
Interpret the meaning of the p-value. Choose the correct answer below.
A.
Fail to rejectFail to reject
Upper H 0H0.
There is
sufficientsufficient
evidence that the mean amount is different from
11.0
gallon.
B.
RejectReject
Upper H 0H0.
There is
insufficientinsufficient
evidence that the mean amount is different from
11.0
gallon.
C.
Fail to rejectFail to reject
Upper H 0H0.
There is
insufficientinsufficient
evidence that the mean amount is different from
11.0
gallon.
D.
RejectReject
Upper H 0H0.
There is
sufficientsufficient
evidence that the mean amount is different from
11.0
gallon.c. Construct a
9595%
confidence interval estimate of the population mean amount of paint.
nothingless than or equalsmuless than or equalsnothing
(Round to four decimal places as needed.)
Draw an appropriate conclusion. Choose the correct answer below.
A.
RejectReject
Upper H 0H0.
The value
11.0
is
withinwithin
the confidence interval.
B.
Fail to rejectFail to reject
Upper H 0H0.
The value
11.0
is
outside ofoutside of
the confidence interval.
C.
Fail to rejectFail to reject
Upper H 0H0.
The value
11.0
is
withinwithin
the confidence interval.
D.
RejectReject
Upper H 0H0.
The value
11.0
is
outside ofoutside of
the confidence interval.
d. Compare the results of (a) and (c). Are the results the same?
Yes
No
e. Compare the results of parts (a) through (d) to those when the standard deviation is
0.0200.020.
When the standard deviation is
0.0200.020,
Upper Z Subscript STATZSTATequals=negative 1.681.68,
the p-value is
0.0940.094,
and the
9595%
confidence interval is
0.98920.9892less than or equalsmuless than or equals1.00081.0008.
Using the critical value approach,
H0.
This result is
the result found in part (a), because
Upper Z Subscript STATZSTAT
has
decreased
increased
in absolute value.Using the p-value approach,
reject
do not reject
H0.
This result is
the same as
different from
the result found in part (b), because thep-value has
decreased.
increased.
Using the confidence interval approach,
do not reject
reject
H0.
This result is
the same as
different from
the result found in part (c), because the confidence interval has become
wider.
narrower.
Explanation / Answer
Given that,
population mean(u)=11
standard deviation, =0.008
sample mean, x =0.995
number (n)=45
null, Ho: >=11
alternate, H1: <11
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.64
since our test is left-tailed
reject Ho, if zo < -1.64
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 0.995-11/(0.008/sqrt(45)
zo = -8389.448
| zo | = 8389.448
critical value
the value of |z | at los 5% is 1.64
we got |zo| =8389.448 & | z | = 1.64
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : left tail - ha : ( p < -8389.448 ) = 0
hence value of p0.05 > 0, here we reject Ho
ANSWERS
---------------
null, Ho: >=11
alternate, H1: <11
test statistic: -8389.448
critical value: -1.64
decision: reject Ho
a.
Reject Upper H 0H0.There is sufficient evidence that the mean amount is different from 11.0 gallon.
b.
p-value: 0
c.
TRADITIONAL METHOD
given that,
standard deviation, =0.008
sample mean, x =0.995
population size (n)=45
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 0.008/ sqrt ( 45) )
= 0.001
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 0.001
= 0.002
III.
CI = x ± margin of error
confidence interval = [ 0.995 ± 0.002 ]
= [ 0.993,0.997 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =0.008
sample mean, x =0.995
population size (n)=45
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
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 = [ 0.995 ± Z a/2 ( 0.008/ Sqrt ( 45) ) ]
= [ 0.995 - 1.96 * (0.001) , 0.995 + 1.96 * (0.001) ]
= [ 0.993,0.997 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [0.993 , 0.997 ] 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 = 0.995
standard error =0.001
z table value = 1.96
margin of error = 0.002
confidence interval = [ 0.993 , 0.997 ]
d.
yes
e.
Given that,
population mean(u)=11
standard deviation, =0.02
sample mean, x =0.995
number (n)=45
null, Ho: >=11
alternate, H1: <11
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.64
since our test is left-tailed
reject Ho, if zo < -1.64
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 0.995-11/(0.02/sqrt(45)
zo = -3355.779
| zo | = 3355.779
critical value
the value of |z | at los 5% is 1.64
we got |zo| =3355.779 & | z | = 1.64
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : left tail - ha : ( p < -3355.779 ) = 0
hence value of p0.05 > 0, here we reject Ho
ANSWERS
---------------
null, Ho: >=11
alternate, H1: <11
test statistic: -3355.779
critical value: -1.64
decision: reject Ho
p-value: 0
TRADITIONAL METHOD
given that,
standard deviation, =0.02
sample mean, x =0.995
population size (n)=45
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 0.02/ sqrt ( 45) )
= 0.003
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 0.003
= 0.006
III.
CI = x ± margin of error
confidence interval = [ 0.995 ± 0.006 ]
= [ 0.989,1.001 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =0.02
sample mean, x =0.995
population size (n)=45
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
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 = [ 0.995 ± Z a/2 ( 0.02/ Sqrt ( 45) ) ]
= [ 0.995 - 1.96 * (0.003) , 0.995 + 1.96 * (0.003) ]
= [ 0.989,1.001 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [0.989 , 1.001 ] 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 = 0.995
standard error =0.003
z table value = 1.96
margin of error = 0.006
confidence interval = [ 0.989 , 1.001 ]
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