Question

Page 16 Example: An electronic parts factory produces resistors. Assume the resistance follows a distribution with standard deviation 0.156 ohms. A random sample of 60 resistors has an average resistance of 0.45 ohms. The factory wishes to test that the population mean resistance is less than 0.5 ohms? a. Find the p-value b. Express the rejection region of the above test in terms of the sample mean. c. Find trpel error d. Find the probability of type II error if the actual resistance is 0.48.

Answer 1

Given that,
population mean(u)=0.5
standard deviation, σ =0.156
sample mean, x =0.45
number (n)=60
null, Ho: μ=0.5
alternate, H1: μ<0.5
level of significance, α = 0.05
from standard normal table,left tailed z α/2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 0.45-0.5/(0.156/sqrt(60)
zo = -2.483
| zo | = 2.483
critical value
the value of |z α| at los 5% is 1.645
we got |zo| =2.483 & | z α | = 1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value : left tail - ha : ( p < -2.483 ) = 0.007
hence value of p0.05 > 0.007, here we reject Ho
ANSWERS
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a.
null, Ho: μ=0.5
alternate, H1: μ<0.5
test statistic: -2.483
critical value: -1.645
a.
p-value: 0.007
b.
decision: reject Ho
we have enough evidence to support the claim that population mean resistance is less than 0.5 ohms.
c.
Given that,
Standard deviation, σ =0.156
Sample Mean, X =0.45
Null, H0: μ=0.5
Alternate, H1: μ<0.5
Level of significance, α = 0.05
From Standard normal table, Z α/2 =1.64
Since our test is left-tailed
Reject Ho, if Zo < -1.64 OR if Zo > 1.64
Reject Ho if (x-0.5)/0.156/√(n) < -1.64 OR if (x-0.5)/0.156/√(n) > 1.64
Reject Ho if x < 0.5-0.256/√(n) OR if x > 0.5-0.256/√(n)
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Suppose the size of the sample is n = 60 then the critical region
becomes,
Reject Ho if x < 0.5-0.256/√(60) OR if x > 0.5+0.256/√(60)
Reject Ho if x < 0.467 OR if x > 0.533
Suppose the true mean is 0.45
Probability of Type I error,
P(Type I error) = P(Reject Ho | Ho is true )
= P(0.467 < x OR x >0.533 | μ1 = 0.45)
= P(0.467-0.45/0.156/√(60) < x - μ / σ/√n OR x - μ / σ/√n >0.533-0.45/0.156/√(60)
= P(0.844 < Z OR Z >4.121 )
= P( Z <0.844) + P( Z > 4.121)
= 0.8007 + 0 [ Using Z Table ]
= 0.801
type 1 error =0.801
d.
Given that,
Standard deviation, σ =0.156
Sample Mean, X =0.45
Null, H0: μ=0.5
Alternate, H1: μ<0.5
Level of significance, α = 0.05
From Standard normal table, Z α/2 =1.6449
Since our test is left-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-0.5)/0.156/√(n) < -1.6449 OR if (x-0.5)/0.156/√(n) > 1.6449
Reject Ho if x < 0.5-0.2566044/√(n) OR if x > 0.5-0.2566044/√(n)
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Suppose the size of the sample is n = 60 then the critical region
becomes,
Reject Ho if x < 0.5-0.2566044/√(60) OR if x > 0.5+0.2566044/√(60)
Reject Ho if x < 0.46687251 OR if x > 0.53312749
Implies, don't reject Ho if 0.46687251≤ x ≤ 0.53312749
Suppose the true mean is 0.48
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(0.46687251 ≤ x ≤ 0.53312749 | μ1 = 0.48)
= P(0.46687251-0.48/0.156/√(60) ≤ x - μ / σ/√n ≤ 0.53312749-0.48/0.156/√(60)
= P(-0.65182757 ≤ Z ≤2.63797287 )
= P( Z ≤2.63797287) - P( Z ≤-0.65182757)
= 0.9958 - 0.2573 [ Using Z Table ]
= 0.7385
For n =60 the probability of Type II error is 0.7385