Eight breasts of 7- to 8-week-old broiler chickens weighing 0.9 to 1.2 kg were t

Question

Eight breasts of 7- to 8-week-old broiler chickens weighing 0.9 to 1.2 kg were taken processing plant. The concentration of sarcoplasmic protein (mg N/g tissue) in the was measured. The data are presented below. Specimen # 1 2 3 4 5 6 7 8 in (x) 12.8 13.5 13.6 13.3 13.9 13.9 13.3 13.5 ens weighing 0.9 to 1.2 kg were taken at random from a poultry- coplasmic protein (mg N/g tissue) in the breast muscles of these specimens Compute a 95% confidence interval for the mean sarcoplasmic protein. OW Would you do part (a) if you knew that the population standard deviation were o = 0.37 c) Again assum n assuming the population standard deviation is g = 0.3, if a different researcher were to report the margin of error as 0.1856, what is the confidence level this researcher is using? 6. A simple rando mple random sample of batteries is used to test the one sided hypothesis that the mean lifetime exceeds 160 against the null hypothesis that the mean lifetime is less than or equal to 160 hours. Assume that the true population standard deviation is o = 10 hours. The sample size is 400. Assume the sample of 400 batteries yields a sample mean lifetime of 160.8 hours. What is the p-value? How would the rejection region be defined so that the significance level of the test is 0.01? That is, for what subset of the number line would i have to fall so that you would reject the null hypothesis, keeping the sample size at 400? C) Using the rejection region of part (b), what is the probability of a type II error if the mean lifetime is actually 161.0 hours. d) How large a sample must be taken so that a significance level 0.05 test also has BAI61.0) 50.10?

Explanation / Answer

5.
a.
TRADITIONAL METHOD
given that,
sample mean, x =13.475
standard deviation, s =0.3576
sample size, n =8
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.3576/ sqrt ( 8) )
= 0.126
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 = 7 d.f is 2.365
margin of error = 2.365 * 0.126
= 0.299
III.
CI = x ± margin of error
confidence interval = [ 13.475 ± 0.299 ]
= [ 13.176 , 13.774 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =13.475
standard deviation, s =0.3576
sample size, n =8
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 7 d.f is 2.365
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 = [ 13.475 ± t a/2 ( 0.3576/ Sqrt ( 8) ]
= [ 13.475-(2.365 * 0.126) , 13.475+(2.365 * 0.126) ]
= [ 13.176 , 13.774 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 13.176 , 13.774 ] 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

b.
Assume population stanadard deviation = 0.3
TRADITIONAL METHOD
given that,
standard deviation, =0.3
sample mean, x =13.475
population size (n)=8
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 0.3/ sqrt ( 8) )
= 0.106
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 * 0.106
= 0.208
III.
CI = x ± margin of error
confidence interval = [ 13.475 ± 0.208 ]
= [ 13.267,13.683 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =0.3
sample mean, x =13.475
population size (n)=8
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 = [ 13.475 ± Z a/2 ( 0.3/ Sqrt ( 8) ) ]
= [ 13.475 - 1.96 * (0.106) , 13.475 + 1.96 * (0.106) ]
= [ 13.267,13.683 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [13.267 , 13.683 ] 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 = 13.475
standard error =0.106
z table value = 1.96
margin of error = 0.208
confidence interval = [ 13.267 , 13.683 ]

c.
Assume level of significance = 0.08
TRADITIONAL METHOD
given that,
standard deviation, =0.3
sample mean, x =13.475
population size (n)=8
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 0.3/ sqrt ( 8) )
= 0.1061
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.08
from standard normal table, two tailed z /2 =1.751
since our test is two-tailed
value of z table is 1.751
margin of error = 1.751 * 0.1061
= 0.1857
III.
CI = x ± margin of error
confidence interval = [ 13.475 ± 0.1857 ]
= [ 13.2893,13.6607 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =0.3
sample mean, x =13.475
population size (n)=8
level of significance, = 0.08
from standard normal table, two tailed z /2 =1.751
since our test is two-tailed
value of z table is 1.751
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 = [ 13.475 ± Z a/2 ( 0.3/ Sqrt ( 8) ) ]
= [ 13.475 - 1.751 * (0.1061) , 13.475 + 1.751 * (0.1061) ]
= [ 13.2893,13.6607 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 92% sure that the interval [13.2893 , 13.6607 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 92% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 13.475
standard error =0.1061
z table value = 1.751
margin of error = 0.1857
confidence interval = [ 13.2893 , 13.6607 ]

6.
a.
Given that,
population mean(u)=160
standard deviation, =10
sample mean, x =160.8
number (n)=400
null, Ho: >=160
alternate, H1: <=160
level of significance, = 0.01
from standard normal table,left tailed z /2 =2.326
since our test is left-tailed
reject Ho, if zo < -2.326
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 160.8-160/(10/sqrt(400)
zo = 1.6
| zo | = 1.6
critical value
the value of |z | at los 1% is 2.326
we got |zo| =1.6 & | z | = 2.326
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : left tail - ha : ( p < 1.6 ) = 0.945
hence value of p0.01 < 0.945, here we do not reject Ho
ANSWERS
---------------
null, Ho: >=160
alternate, H1: <=160
test statistic: 1.6
critical value: -2.326
decision: do not reject Ho
p-value: 0.945

b.
sample mean falls critical reject region is do not falls. it is do not reject the null hypothesis

c.
Given that,
Standard deviation, =10
Sample Mean, X =160.8
Null, H0: >=160
Alternate, H1: <=160
Level of significance, = 0.01
From Standard normal table, Z /2 =2.3263
Since our test is left-tailed
Reject Ho, if Zo < -2.3263 OR if Zo > 2.3263
Reject Ho if (x-160)/10/(n) < -2.3263 OR if (x-160)/10/(n) > 2.3263
Reject Ho if x < 160-23.263/(n) OR if x > 160-23.263/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 400 then the critical region
becomes,
Reject Ho if x < 160-23.263/(400) OR if x > 160+23.263/(400)
Reject Ho if x < 158.837 OR if x > 161.163
Implies, don't reject Ho if 158.837 x 161.163
Suppose the true mean is 161
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(158.837 x 161.163 | 1 = 161)
= P(158.837-161/10/(400) x - / /n 161.163-161/10/(400)
= P(-4.326 Z 0.326 )
= P( Z 0.326) - P( Z -4.326)
= 0.6278 - 0 [ Using Z Table ]
= 0.628
For n =400 the probability of Type II error is 0.628

d.
beta(161) <= 0.10
n= (((Zalpha +Zbeta))/(U-Uo))^2
Z alpha at 0.05 = 1.96
Z beta at 0.10 = 1.28
n= ((10(1.96+1.28))/(161-160)^2
n =32.4 =33

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