34. Suppose that a lightbulb manufacturing plant produces bulbs with a mean life
ID: 3364234 • Letter: 3
Question
34. Suppose that a lightbulb manufacturing plant produces bulbs with a mean life of 2000 hours and a standard deviation of 200 hours. An inventor claims to have developed an improved process that produces bulbs with a longer mean life and the same standard deviation. The plant manager randomly selects 100 bulbs produced by the process. She says that she will believe the inventor's claim if the sample mean life of the bulbs is greater than 2100 hours; otherwise, she will conclude that the new process is no better than the old process. Let denote the mean of the new process. Consider the null and alternative hypothesis H0 : 2000 vs H1 > 2000 (e) Wt is the probablity of making a Type I error if the new procedture is o better than the old one? (b) Suppose the new process is in fact better and has a mean bulb life of 2150 hours. What is the power of the plant manager's testing procedure? (c) What testing procedure should the plant manager use if she wants the size of her test to be 5%?Explanation / Answer
34.
a.
Given that,
Standard deviation, =200
Sample Mean, X =2100
Null, H0: <=2000
Alternate, H1: >2000
Level of significance, = 0.05
From Standard normal table, Z /2 =1.64
Since our test is right-tailed
Reject Ho, if Zo < -1.64 OR if Zo > 1.64
Reject Ho if (x-2000)/200/(n) < -1.64 OR if (x-2000)/200/(n) > 1.64
Reject Ho if x < 2000-328/(n) OR if x > 2000-328/(n)
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Suppose the size of the sample is n = 100 then the critical region
becomes,
Reject Ho if x < 2000-328/(100) OR if x > 2000+328/(100)
Reject Ho if x < 1967.2 OR if x > 2032.8
Suppose the true mean is 2000
Probability of Type I error,
P(Type I error) = P(Reject Ho | Ho is true )
= P(1967.2 < x OR x >2032.8 | 1 = 2000)
= P(1967.2-2000/200/(100) < x - / /n OR x - / /n >2032.8-2000/200/(100)
= P(-1.64 < Z OR Z >1.64 )
= P( Z <-1.64) + P( Z > 1.64)
= 0.0505 + 0.0505 [ Using Z Table ]
= 0.101
b.
Given that,
Standard deviation, =200
Sample Mean, X =2150
Null, H0: <2000
Alternate, H1: >2000
Level of significance, = 0.05
From Standard normal table, Z /2 =1.6449
Since our test is right-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-2000)/200/(n) < -1.6449 OR if (x-2000)/200/(n) > 1.6449
Reject Ho if x < 2000-328.98/(n) OR if x > 2000-328.98/(n)
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Suppose the size of the sample is n = 100 then the critical region
becomes,
Reject Ho if x < 2000-328.98/(100) OR if x > 2000+328.98/(100)
Reject Ho if x < 1967.102 OR if x > 2032.898
Implies, don't reject Ho if 1967.102 x 2032.898
Suppose the true mean is 2000
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(1967.102 x 2032.898 | 1 = 2000)
= P(1967.102-2000/200/(100) x - / /n 2032.898-2000/200/(100)
= P(-1.6449 Z 1.6449 )
= P( Z 1.6449) - P( Z -1.6449)
= 0.95 - 0.05 [ Using Z Table ]
= 0.9
For n =100 the probability of Type II error is 0.9
power of the test = 1-beta
= 1-0.9 = 0.1
c.
Given that,
population mean(u)=2000
standard deviation, =200
sample mean, x =2100
number (n)=100
null, Ho: <=2000
alternate, H1: >2000
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 2100-2000/(200/sqrt(100)
zo = 5
| zo | = 5
critical value
the value of |z | at los 5% is 1.645
we got |zo| =5 & | z | = 1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : right tail - ha : ( p > 5 ) = 0
hence value of p0.05 > 0, here we reject Ho
ANSWERS
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null, Ho: <=2000
alternate, H1: >2000
test statistic: 5
critical value: 1.645
decision: reject Ho
p-value: 0
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