The Crown Bottling Company has just installed a new bottling process that will f
ID: 3361559 • Letter: T
Question
The Crown Bottling Company has just installed a new bottling process that will fill 16-ounce bottles of the popular Crown Classic Cola soft drink. Both overfilling and underfilling bottles are undesirable: Underfilling leads to customer complaints and overfilling costs the company considerable money. In order to verify that the filler is set up correctly, the company wishes to see whether the mean bottle fill, , is close to the target fill of 16 ounces. To this end, a random sample of 32 filled bottles is selected from the output of a test filler run. If the sample results cast a substantial amount of doubt on the hypothesis that the mean bottle fill is the desired 16 ounces, then the filler’s initial setup will be readjusted.
(a) The bottling company wants to set up a hypothesis test so that the filler will be readjusted if the null hypothesis is rejected. Set up the null and alternative hypotheses for this hypothesis test.
H0 : (Click to select)= 16 versus Ha : (Click to select)= 16 PLEASE SEE THIS!! (THIS IS WHERE I THINK I AM WRONG)
CI [, ] (Click to select)ReadjustDo not readjust
CI [, ] (Click to select)Do not readjustReadjust
CI [, ] (Click to select)ReadjustDo not readjust
CI [, ] (Click to select)ReadjustDo not readjust
Explanation / Answer
Given that,
population mean(u)=16
standard deviation, =1
sample mean, x =16.05
number (n)=32
null, Ho: =16
alternate, H1: !=16
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.58
since our test is two-tailed
reject Ho, if zo < -2.58 OR if zo > 2.58
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 16.05-16/(1/sqrt(32)
zo = 0.2828
| zo | = 0.2828
critical value
the value of |z | at los 1% is 2.58
we got |zo| =0.2828 & | z | = 2.58
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 0.2828 ) = 0.7773
hence value of p0.01 < 0.7773, here we do not reject Ho
ANSWERS
---------------
a.
null, Ho: =16
alternate, H1: !=16
b.
I)at X= 16.05
test statistic: 0.2828
critical value: -2.58 , 2.58
decision: do not reject Ho
p-value: 0.7773
TRADITIONAL METHOD
given that,
standard deviation, =1
sample mean, x =16.05
population size (n)=32
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 1/ sqrt ( 32) )
= 0.177
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 0.177
= 0.455
III.
CI = x ± margin of error
confidence interval = [ 16.05 ± 0.455 ]
= [ 15.595,16.505 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =1
sample mean, x =16.05
population size (n)=32
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
value of z table is 2.576
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 = [ 16.05 ± Z a/2 ( 1/ Sqrt ( 32) ) ]
= [ 16.05 - 2.576 * (0.177) , 16.05 + 2.576 * (0.177) ]
= [ 15.595,16.505 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [15.595 , 16.505 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 16.05
standard error =0.177
z table value = 2.576
margin of error = 0.455
confidence interval = [ 15.595 , 16.505 ]
II)
X = 15.91
Given that,
population mean(u)=16
standard deviation, =1
sample mean, x =15.91
number (n)=32
null, Ho: =16
alternate, H1: !=16
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.58
since our test is two-tailed
reject Ho, if zo < -2.58 OR if zo > 2.58
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 15.91-16/(1/sqrt(32)
zo = -0.5091
| zo | = 0.5091
critical value
the value of |z | at los 1% is 2.58
we got |zo| =0.5091 & | z | = 2.58
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -0.5091 ) = 0.6107
hence value of p0.01 < 0.6107, here we do not reject Ho
ANSWERS
---------------
null, Ho: =16
alternate, H1: !=16
test statistic: -0.5091
critical value: -2.58 , 2.58
decision: do not reject Ho
p-value: 0.6107
TRADITIONAL METHOD
given that,
standard deviation, =1
sample mean, x =15.91
population size (n)=32
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 1/ sqrt ( 32) )
= 0.177
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 0.177
= 0.455
III.
CI = x ± margin of error
confidence interval = [ 15.91 ± 0.455 ]
= [ 15.455,16.365 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =1
sample mean, x =15.91
population size (n)=32
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
value of z table is 2.576
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 = [ 15.91 ± Z a/2 ( 1/ Sqrt ( 32) ) ]
= [ 15.91 - 2.576 * (0.177) , 15.91 + 2.576 * (0.177) ]
= [ 15.455,16.365 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [15.455 , 16.365 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 15.91
standard error =0.177
z table value = 2.576
margin of error = 0.455
confidence interval = [ 15.455 , 16.365 ]
III).
X = 16.04
Given that,
population mean(u)=16
standard deviation, =1
sample mean, x =16.04
number (n)=32
null, Ho: =16
alternate, H1: !=16
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.58
since our test is two-tailed
reject Ho, if zo < -2.58 OR if zo > 2.58
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 16.04-16/(1/sqrt(32)
zo = 0.2263
| zo | = 0.2263
critical value
the value of |z | at los 1% is 2.58
we got |zo| =0.2263 & | z | = 2.58
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 0.2263 ) = 0.821
hence value of p0.01 < 0.821, here we do not reject Ho
ANSWERS
---------------
null, Ho: =16
alternate, H1: !=16
test statistic: 0.2263
critical value: -2.58 , 2.58
decision: do not reject Ho
p-value: 0.821
TRADITIONAL METHOD
given that,
standard deviation, =1
sample mean, x =16.04
population size (n)=32
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 1/ sqrt ( 32) )
= 0.177
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 0.177
= 0.455
III.
CI = x ± margin of error
confidence interval = [ 16.04 ± 0.455 ]
= [ 15.585,16.495 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =1
sample mean, x =16.04
population size (n)=32
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
value of z table is 2.576
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 = [ 16.04 ± Z a/2 ( 1/ Sqrt ( 32) ) ]
= [ 16.04 - 2.576 * (0.177) , 16.04 + 2.576 * (0.177) ]
= [ 15.585,16.495 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [15.585 , 16.495 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 16.04
standard error =0.177
z table value = 2.576
margin of error = 0.455
confidence interval = [ 15.585 , 16.495 ]
IV)
X = 15.97
Given that,
population mean(u)=16
standard deviation, =1
sample mean, x =15.97
number (n)=32
null, Ho: =16
alternate, H1: !=16
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.58
since our test is two-tailed
reject Ho, if zo < -2.58 OR if zo > 2.58
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 15.97-16/(1/sqrt(32)
zo = -0.1697
| zo | = 0.1697
critical value
the value of |z | at los 1% is 2.58
we got |zo| =0.1697 & | z | = 2.58
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -0.1697 ) = 0.8652
hence value of p0.01 < 0.8652, here we do not reject Ho
ANSWERS
---------------
null, Ho: =16
alternate, H1: !=16
test statistic: -0.1697
critical value: -2.58 , 2.58
decision: do not reject Ho
p-value: 0.8652
TRADITIONAL METHOD
given that,
standard deviation, =1
sample mean, x =15.97
population size (n)=32
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 1/ sqrt ( 32) )
= 0.177
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 0.177
= 0.455
III.
CI = x ± margin of error
confidence interval = [ 15.97 ± 0.455 ]
= [ 15.515,16.425 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =1
sample mean, x =15.97
population size (n)=32
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
value of z table is 2.576
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 = [ 15.97 ± Z a/2 ( 1/ Sqrt ( 32) ) ]
= [ 15.97 - 2.576 * (0.177) , 15.97 + 2.576 * (0.177) ]
= [ 15.515,16.425 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [15.515 , 16.425 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 15.97
standard error =0.177
z table value = 2.576
margin of error = 0.455
confidence interval = [ 15.515 , 16.425 ]
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