1. A civil engineer is analyzing the compressive strength of concrete. Compressi

Question

1. A civil engineer is analyzing the compressive strength of concrete. Compressive strength is approximately normally distributed with variance of 1000 psi2. A random sample of 12 specimens has a mean compressive strength of 3255.42 psi. a) Test the hypothesis that mean compressive strength is not 3500 psi. Use =0.01. b) Draw conclusions using the P-value approach. c) Construct a 99% two-sided CI and draw conclusions. d) What is the -value for this test if the true mean compressive strength is 3300 psi?

Explanation / Answer

1.

Given that,
population mean(u)=3500
standard deviation, =31.6227
sample mean, x =3255.42
number (n)=12
null, Ho: =3500
alternate, H1: !=3500
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
reject Ho, if zo < -2.576 OR if zo > 2.576
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 3255.42-3500/(31.6227/sqrt(12)
zo = -26.79246
| zo | = 26.79246
critical value
the value of |z | at los 1% is 2.576
we got |zo| =26.79246 & | z | = 2.576
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -26.79246 ) = 0
hence value of p0.01 > 0, here we reject Ho
ANSWERS
---------------
a.
null, Ho: =3500
alternate, H1: !=3500
test statistic: -26.79246
critical value: -2.576 , 2.576
decision: reject Ho
b.
p-value: 0
c.
TRADITIONAL METHOD
given that,
standard deviation, =31.6227
sample mean, x =3255.42
population size (n)=12
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 31.6227/ sqrt ( 12) )
= 9.129
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 * 9.129
= 23.515
III.
CI = x ± margin of error
confidence interval = [ 3255.42 ± 23.515 ]
= [ 3231.905,3278.935 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =31.6227
sample mean, x =3255.42
population size (n)=12
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 = [ 3255.42 ± Z a/2 ( 31.6227/ Sqrt ( 12) ) ]
= [ 3255.42 - 2.576 * (9.129) , 3255.42 + 2.576 * (9.129) ]
= [ 3231.905,3278.935 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [3231.905 , 3278.935 ] 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 = 3255.42
standard error =9.129
z table value = 2.576
margin of error = 23.515
confidence interval = [ 3231.905 , 3278.935 ]
d.
Given that,
Standard deviation, =31.6227
Sample Mean, X =3255.42
Null, H0: =3500
Alternate, H1: !=3500
Level of significance, = 0.01
From Standard normal table, Z /2 =2.5758
Since our test is two-tailed
Reject Ho, if Zo < -2.5758 OR if Zo > 2.5758
Reject Ho if (x-3500)/31.6227/(n) < -2.5758 OR if (x-3500)/31.6227/(n) > 2.5758
Reject Ho if x < 3500-81.4538/(n) OR if x > 3500-81.4538/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 12 then the critical region
becomes,
Reject Ho if x < 3500-81.4538/(12) OR if x > 3500+81.4538/(12)
Reject Ho if x < 3476.4863 OR if x > 3523.5137
Implies, don't reject Ho if 3476.4863 x 3523.5137
Suppose the true mean is 3300
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(3476.4863 x 3523.5137 | 1 = 3300)
= P(3476.4863-3300/31.6227/(12) x - / /n 3523.5137-3300/31.6227/(12)
= P(19.3332 Z 24.4848 )
= P( Z 24.4848) - P( Z 19.3332)
= 1 - 1 [ Using Z Table ]
= 0
For n =12 the probability of Type II error is 0

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