SECTION A The life span of 100 W light bulbs manufactured by a particular compan
ID: 3054465 • Letter: S
Question
SECTION A
The life span of 100 W light bulbs manufactured by a particular company follows a normal distribution with a standard deviation of 120 hours and its half-life is guaranteed under warranty for a minimum of 800 hours. At random, a sample of 50 bulbs from a lot is selected and the sample revealed that the half-life is 750 hours.
Create a 99% C.I. for the population mean
should the lot be rejected by not honoring the warranty?
A manufacturer of electric lamps is testing a new production method that will be considered acceptable if the lamps produced by this method result in a normal population with an average life of 2,400 hours and a standard deviation equal to 300. A sample of 100 lamps produced by this method has an average life of 2,320 hours.
Compute a 95% C.I. for ?.
Using the C.I. can the hypothesis of validity for the new manufacturing process be accepted with a risk n 5%?
The quality control division of a factory that manufactures batteries suspects defects in the production of a model of mobile phone battery which results in a lower life for the product. Until now, the time duration in phone conversation for the battery followed a normal distribution with a mean of 300 minutes and a standard deviation of 30. However, in an inspection of the last batch produced before sending it to market, it was found that the average time spent in conversation was 290 minutes in a sample of 60 batteries. Assuming that the time is still normal with the same standard deviation:
Can it be concluded with 99% Confidence Level that the quality control suspicions are true?
It is believed that the average level of prothrombin in a normal population is 20 mg/100 ml of blood plasma with a standard deviation of 4 milligrams/100 ml. To verify this, a sample is taken from 40 individuals in whom the average is 18.5 mg/100 ml. Can you use a 95% C.I. for the mean to see whether the belief is a valid one?
SECTION B
Question B.1
Questions 1 and 2 use almost the same information.
A random sample of 30 households was selected as part of a study on electricity usage, and the number of kilowatt-hours (kWh) was recorded for each household in the sample, the average usage was found to be 375kWh. In a very large study it was found that the population standard deviation of the usage was given to be 81kWh.
Assuming that the usage is normally distributed,
Calculate a 99% confidence interval for the mean usage.
Can you tell using result in (a) : is there evidence that the average consumption for all households is 400 kWh ?
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Question B.2
Questions 1 and 2 use almost the same information (almost !!!)
A random sample of 30 households was selected as part of a study on electricity usage, and the number of kilowatt-hours (kWh) was recorded for each household in the sample for the March quarter of 2006. The sample average usage was found to be 375kWh. If it was found that the sample standard deviation of the usage was 81kWh. Assuming that the usage is normal
Calculate a 99% confidence interval for the mean usage.
Use results in (b) to test the hypothesis in (a)
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In your own words what is the major differences between this problem and the previous one
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Note: Some of the HW questions may appear with different numbers in your next Test
Explanation / Answer
Q1.
create a 99% C.I. for the population mean
TRADITIONAL METHOD
given that,
standard deviation, ? =120
sample mean, x =750
population size (n)=50
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 120/ sqrt ( 50) )
= 16.97
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
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
margin of error = 2.576 * 16.97
= 43.72
III.
CI = x ± margin of error
confidence interval = [ 750 ± 43.72 ]
= [ 706.28,793.72 ]
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DIRECT METHOD
given that,
standard deviation, ? =120
sample mean, x =750
population size (n)=50
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 = [ 750 ± Z a/2 ( 120/ Sqrt ( 50) ) ]
= [ 750 - 2.576 * (16.97) , 750 + 2.576 * (16.97) ]
= [ 706.28,793.72 ]
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interpretations:
1. we are 99% sure that the interval [706.28 , 793.72 ] 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
should the lot be rejected by not honoring the warranty?
the lot can be rejected since the interval achived is below 800 hours mark[ 706.28,793.72 ]
Q2.
compute a 95% C.I. for ???
H0 : ? = 2,400
H1 : ? ?2,400
given that,
standard deviation, ? =300
sample mean, x =2320
population size (n)=100
level of significance, ? = 0.05
from standard normal table, two tailed z ?/2 =1.96
since our test is two-tailed
value of z table is 1.96
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 = [ 2320 ± Z a/2 ( 300/ Sqrt ( 100) ) ]
= [ 2320 - 1.96 * (30) , 2320 + 1.96 * (30) ]
= [ 2261.2,2378.8 ]
The nule hypothesis, H0, cannot be accepted with a significance level of 5%.
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