Question

1. A composites manufacturer is having serious problems with porosity in their parts. A Quality Engineer samples 300 parts and finds 58 defective.

1. Test the hypothesis that defective rate (proportion defective) exceeds 15%. Test at a = 0.05.

What is the parameter of interest?   What assumptions are made? Show mathematical evidence to support assumption.

i. Write the null and alternative hypotheses.

ii. Calculate the test statistic.

iii. Determine the reject region. Find the p-value. Show normal graph including reject region and test statistic.

iv. Make a decision and write a thorough interpretation in context of the problem.

v. If in reality the true proportion defective is 12%, what type of error, if any, occurred?

2. Compute the power of the test if the true defective rate is 0.18.
3. Suppose that you wanted to reject the null hypothesis with probability at least 0.9 if true defective rate p = .18. What sample size should be used?
4. Construct a 95% Confidence Interval for p, the true proportion defective.

Answer 1

a.
Given that,
possible chances (x)=58
sample size(n)=300
success rate ( p )= x/n = 0.1933
success probability,( po )=0.15
failure probability,( qo) = 0.85
null, Ho:p=0.15
alternate, H1: p>0.15
level of significance, α = 0.05
from standard normal table,right tailed z α/2 =1.64
since our test is right-tailed
reject Ho, if zo > 1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.19333-0.15/(sqrt(0.1275)/300)
zo =2.102
| zo | =2.102
critical value
the value of |z α| at los 0.05% is 1.64
we got |zo| =2.102 & | z α | =1.64
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: right tail - Ha : ( p > 2.10198 ) = 0.01778
hence value of p0.05 > 0.01778,here we reject Ho
ANSWERS
---------------
i.
null, Ho:p=0.15
alternate, H1: p>0.15
ii.
test statistic: 2.102
iii.
critical value: 1.64
decision: reject Ho
p-value: 0.01778
iv.
we have enough evidence to support the claim defective rate (proportion defective) exceeds 15%.
v.
Given that,
possible chances (x)=58
sample size(n)=300
success rate ( p )= x/n = 0.1933
success probability,( po )=0.12
failure probability,( qo) = 0.88
null, Ho:p=0.12
alternate, H1: p>0.12
level of significance, α = 0.05
from standard normal table,right tailed z α/2 =1.64
since our test is right-tailed
reject Ho, if zo > 1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.19333-0.12/(sqrt(0.1056)/300)
zo =3.9087
| zo | =3.9087
critical value
the value of |z α| at los 0.05% is 1.64
we got |zo| =3.909 & | z α | =1.64
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: right tail - Ha : ( p > 3.90868 ) = 0.00005
hence value of p0.05 > 0.00005,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.12
alternate, H1: p>0.12
test statistic: 3.9087
critical value: 1.64
decision: reject Ho
p-value: 0.00005
in this context,
type 1 error is possible because it is reject the null hypothesis.
b.
power of the test
beta= p(Z>=((po-p)-Zalpha*(sqrt(po*(1-po)/n))/(p*(1-p)/n)
beta = p(Z>=((0.15-0.18)-Z0.05*(sqrt(0.15*(1-0.15)/300)/(0.18*(1-0.18)/300))
beta= p(Z>=((0.15-0.18)-1.645*((sqrt((0.15*(1-0.15)/300)/(0.18*(1-0.18)/300))))
beta =p(Z>= -1.55889)
beta = 0.9405
power of the test = 1- type 2 error
power of the test = 1- 0.9405
power of the test =0.0595
c.
probability =0.9
type 2 error =1-0.9
type 2 error = 0.1
n=p(1−p)*((z1−α/2+z1−β/(p−p0))^2
n = 0.18*(1-0.18)*(1.645+1.28)/(0.18-0.15))^2
n =207.1
n =207
d.
TRADITIONAL METHOD
given that,
possible chances (x)=58
sample size(n)=300
success rate ( p )= x/n = 0.1933
I.
sample proportion = 0.1933
standard error = Sqrt ( (0.1933*0.8067) /300) )
= 0.0228
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.0228
= 0.0447
III.
CI = [ p ± margin of error ]
confidence interval = [0.1933 ± 0.0447]
= [ 0.1486 , 0.238]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=58
sample size(n)=300
success rate ( p )= x/n = 0.1933
CI = confidence interval
confidence interval = [ 0.1933 ± 1.96 * Sqrt ( (0.1933*0.8067) /300) ) ]
= [0.1933 - 1.96 * Sqrt ( (0.1933*0.8067) /300) , 0.1933 + 1.96 * Sqrt ( (0.1933*0.8067) /300) ]
= [0.1486 , 0.238]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 95% sure that the interval [ 0.1486 , 0.238] contains the true population proportion
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 proportion