9-44. An engineer who is studying the tensile strength of a steel alloy intended

Question

9-44. An engineer who is studying the tensile strength of a steel alloy intended for use in golf club shafts knows that tensile strength is approximately normally dis- tributed with 60 psi. A random sample of 12 specimens has a sample mean tensile strength of X 3450 psi (a) Test the hypothesis that mean strength is 3500 psi against two sided alternative. Use -0.01. (b) What is the smallest level of significance at which you would be willing to reject the null hypothesis? (c) What is the -error for the test in part (a) if the true mean is 3470? (d) Suppose that we wanted to reject the null hypothesis with probability at least 0.2 if mean strength = 3470 and 0.01. What sample size should be used? (e) Explain how you could answer the question in part (a) with a two-sided confidence interval on mean tensile strength S1.

Explanation / Answer

A.
Given that,
population mean(u)=3500
standard deviation, =60
sample mean, x =3450
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 = 3450-3500/(60/sqrt(12)
zo = -2.88675
| zo | = 2.88675
critical value
the value of |z | at los 1% is 2.576
we got |zo| =2.88675 & | 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 != -2.88675 ) = 0.00389
hence value of p0.01 > 0.00389, here we reject Ho
ANSWERS
---------------
null, Ho: =3500
alternate, H1: !=3500
test statistic: -2.88675
critical value: -2.576 , 2.576
decision: reject Ho
p-value: 0.00389

C.
Given that,
Standard deviation, =60
Sample Mean, X =3450
Null, H0: =3500
Alternate, H1: !=3500
Level of significance, = 0.01
From Standard normal table, Z /2 =2.58
Since our test is two-tailed
Reject Ho, if Zo < -2.58 OR if Zo > 2.58
Reject Ho if (x-3500)/60/(n) < -2.58 OR if (x-3500)/60/(n) > 2.58
Reject Ho if x < 3500-154.8/(n) OR if x > 3500-154.8/(n)
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Suppose the size of the sample is n = 12 then the critical region
becomes,
Reject Ho if x < 3500-154.8/(12) OR if x > 3500+154.8/(12)
Reject Ho if x < 3455.3131 OR if x > 3544.6869
Implies, don't reject Ho if 3455.3131 x 3544.6869
Suppose the true mean is 3470
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(3455.3131 x 3544.6869 | 1 = 3470)
= P(3455.3131-3470/60/(12) x - / /n 3544.6869-3470/60/(12)
= P(-0.8479 Z 4.3121 )
= P( Z 4.3121) - P( Z -0.8479)
= 1 - 0.1982   [ Using Z Table ]
= 0.8018
For n =12 the probability of Type II error( Beta) value is 0.8018

D.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.01% LOS is = 2.576 ( From Standard Normal Table )
Standard Deviation ( S.D) = 60
ME =0.2
n = ( 2.576*60/0.2) ^2
= (154.56/0.2 ) ^2
= 597219.84 ~ 597220   

E.
given that,
standard deviation, =60
sample mean, x =3450
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 = [ 3450 ± Z a/2 ( 60/ Sqrt ( 12) ) ]
= [ 3450 - 2.576 * (17.321) , 3450 + 2.576 * (17.321) ]
= [ 3405.382,3494.618 ]
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interpretations:
1. we are 99% sure that the interval [3405.382 , 3494.618 ] 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 = 3450
standard error =17.321
z table value = 2.576
margin of error = 44.618
confidence interval = [ 3405.382 , 3494.618 ]

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