Question

A manufacturer is interested in the output voltage of a power supply used in a PC. Output voltage is assumed to be normally distributed, with standard deviation 0.25 volt. The manufacturer wishes to test H0: μ=5 volts against H1: μ≠5 volts using α=0.10. Suppose a sample size of 8 power supply units is used and you are interested in computing the probability the test will fail to detect a true mean output voltage of 5.1 volts. What is the appropriate notation for the referred-to probability?

Selected Answer: power Answers: P-value power

Answer 1

a.
Given that,
population mean(u)=5
standard deviation, σ =0.25
sample mean, x =5.1
number (n)=8
null, Ho: μ=5
alternate, H1: μ!=5
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
reject Ho, if zo < -1.645 OR if zo > 1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 5.1-5/(0.25/sqrt(8)
zo = 1.131
| zo | = 1.131
critical value
the value of |z α| at los 10% is 1.645
we got |zo| =1.131 & | z α | = 1.645
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 1.131 ) = 0.258
hence value of p0.1 < 0.258, here we do not reject Ho
ANSWERS
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null, Ho: μ=5
alternate, H1: μ!=5
test statistic: 1.131
critical value: -1.645 , 1.645
decision: do not reject Ho
p-value: 0.258
we do not have enough evidence to support the claim that population mean is equal to 5.
b.
Given that,
Standard deviation, σ =0.25
Sample Mean, X =5.1
Null, H0: μ=5
Alternate, H1: μ!=5
Level of significance, α = 0.1
From Standard normal table, Z α/2 =1.6449
Since our test is two-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-5)/0.25/√(n) < -1.6449 OR if (x-5)/0.25/√(n) > 1.6449
Reject Ho if x < 5-0.4112/√(n) OR if x > 5-0.4112/√(n)
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Suppose the size of the sample is n = 8 then the critical region
becomes,
Reject Ho if x < 5-0.4112/√(8) OR if x > 5+0.4112/√(8)
Reject Ho if x < 4.8546 OR if x > 5.1454
Implies, don't reject Ho if 4.8546≤ x ≤ 5.1454
Suppose the true mean is 5.1
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(4.8546 ≤ x ≤ 5.1454 | μ1 = 5.1)
= P(4.8546-5.1/0.25/√(8) ≤ x - μ / σ/√n ≤ 5.1454-5.1/0.25/√(8)
= P(-2.7764 ≤ Z ≤0.5136 )
= P( Z ≤0.5136) - P( Z ≤-2.7764)
= 0.6962 - 0.0027 [ Using Z Table ]
= 0.6935
For n =8 the probability of Type II error is 0.6935
power = 1-type 2 error
power = 1-0.6935 =0.3065