Question

An engineer studying the tensile strength of a composite material knows that tensile strength is approximately normally distributed with σ = 60 psi. A random sample of 20 specimens has a mean tensile strength of 3450 psi.

(a) Test the hypothesis that the mean tensile strength is 3500 psi, using α = 0.01

(b) What isthe smallest level of significance at which you would be willing to reject the null hypothesis?

(c) What is the β error for the test in part (a) if the true mean tensile strength is 3470 psi?

(d) Suppose that you wanted to reject the null hypothesis with probability of at least 0.8 if mean strength μ = 3470 psi. What sample size would be necessary?

(e) Explain how you would answer the question in part (a) with a two sided CI on mean tensile strength.

Answer 1

Given that,
population mean(u)=3500
standard deviation, sigma =60
sample mean, x =3450
number (n)=20
null, Ho: μ=3500
alternate, H1: μ!=3500
level of significance, alpha = 0.01
from standard normal table, two tailed z alpha/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 = 3450-3500/(60/sqrt(20)
zo = -3.727
| zo | = 3.727
critical value
the value of |z alpha| at los 1% is 2.576
we got |zo| =3.727 & | z alpha | = 2.576
make decision
hence value of | zo | > | z alpha| and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -3.727 ) = 0
hence value of p0.01 > 0, here we reject Ho
ANSWERS
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a.
null, Ho: μ=3500
alternate, H1: μ!=3500
test statistic: -3.727
critical value: -2.576 , 2.576
b.
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that mean tensile strength is 3500psi.
c.
Given that,
Standard deviation, σ =60
Sample Mean, X =3450
Null, H0: μ=3500
Alternate, H1: μ!=3500
Level of significance, α = 0.01
From Standard normal table, Z α/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)/60/√(n) < -2.5758 OR if (x-3500)/60/√(n) > 2.5758
Reject Ho if x < 3500-154.548/√(n) OR if x > 3500-154.548/√(n)
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Suppose the size of the sample is n = 20 then the critical region
becomes,
Reject Ho if x < 3500-154.548/√(20) OR if x > 3500+154.548/√(20)
Reject Ho if x < 3465.442 OR if x > 3534.558
Implies, don't reject Ho if 3465.442≤ x ≤ 3534.558
Suppose the true mean is 3470
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(3465.442 ≤ x ≤ 3534.558 | μ1 = 3470)
= P(3465.442-3470/60/√(20) ≤ x - μ / σ/√n ≤ 3534.558-3470/60/√(20)
= P(-0.3397 ≤ Z ≤4.8119 )
= P( Z ≤4.8119) - P( Z ≤-0.3397)
= 1 - 0.367 [ Using Z Table ]
= 0.633
For n =20 the probability of Type II error is 0.633
d.
type 2 error is 0.8
n= ((σ(Zalpha +Zbeta))/(U-Uo))^2
n =((60*(Z0.01+Z0.8)/(3500-3470))^2
n= ((60*(2.32+0.8416))/(3500-3470))^2
n =39.98
n= 40
e.
TRADITIONAL METHOD
given that,
standard deviation, σ =60
sample mean, x =3450
population size (n)=20
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 60/ sqrt ( 20) )
= 13.42
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 13.42
= 34.56
III.
CI = x ± margin of error
confidence interval = [ 3450 ± 34.56 ]
= [ 3415.44,3484.56 ]
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DIRECT METHOD
given that,
standard deviation, σ =60
sample mean, x =3450
population size (n)=20
level of significance, α = 0.01
from standard normal table, two tailed z α/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 = [ 3450 ± Z a/2 ( 60/ Sqrt ( 20) ) ]
= [ 3450 - 2.576 * (13.42) , 3450 + 2.576 * (13.42) ]
= [ 3415.44,3484.56 ]
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interpretations:
1. we are 99% sure that the interval [3415.44 , 3484.56 ] 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