Question

In order to determine the life expectancy of a particular type of lightbulb manufactures by a lighting company, the corporate quality control officer randomly selected 10 bulbs from the pre-packaging section of the company's production line. The bulbs were subsequently tested in the quality control lab and the following data was recorded.

Bulb	Life(Hours)
1	3900
2	4200
3	4100
4	3800
5	4000
6	4300
7	3600

a. Calculate the mean life of bulbs in this sample. Mean=

b. Calculate the satndard deviation for the life of bulbs in this sample. Standard Deviation=

c. Assuming that the life of this type of bulb produced by this company is normally distributed, find the 98% confidence interval estimate for the mean life of bulbs manufactured by this company. Between:

d. What is the probability that a six-pack of bulbs (considered to be a random sample size of 6) manufactured by this company will have a mean life expectancy of at least 5000 hours? Probability =

Answer 1

Mean = Sum of observations/ Count of observations
Mean = (3900 + 4200 + 4100 + 3800 + 4000 + 4300 + 3600) / 7 = 3985.7143
Variance
Step 1: Add them up
3900 + 4200 + 4100 + 3800 + 4000 + 4300 + 3600 = 27900
Step 2: Square your answer
27900*27900 =778410000
…and divide by the number of items. We have 7 items , 778410000/7 = 111201428.5714
Set this number aside for a moment.
Step 3: Take your set of original numbers from Step 1, and square them individually this time
3900^2 + 4200^2 + 4100^2 + 3800^2 + 4000^2 + 4300^2 + 3600^2 = 111550000
Step 4: Subtract the amount in Step 2 from the amount in Step 3
111550000 - 111201428.5714 = 348571.4286
Step 5: Subtract 1 from the number of items in your data set, 7 - 1 = 6
Step 6: Divide the number in Step 4 by the number in Step 5. This gives you the variance
348571.4286 / 6 = 58095.2381
Step 7: Take the square root of your answer from Step 6. This gives you the standard deviation
241.0295

a.
sample mean of life of bulbs =3985.7143

b.
sample standard deviation of life of bulbs = 241.0.295

c.
TRADITIONAL METHOD
given that,
sample mean, x =3985.7143
standard deviation, s =241.0295
sample size, n =7
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 241.0295/ sqrt ( 7) )
= 91.101
II.
margin of error = t ?/2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, ? = 0.02
from standard normal table, two tailed value of |t ?/2| with n-1 = 6 d.f is 3.143
margin of error = 3.143 * 91.101
= 286.329
III.
CI = x ± margin of error
confidence interval = [ 3985.7143 ± 286.329 ]
= [ 3699.385 , 4272.043 ]
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DIRECT METHOD
given that,
sample mean, x =3985.7143
standard deviation, s =241.0295
sample size, n =7
level of significance, ? = 0.02
from standard normal table, two tailed value of |t ?/2| with n-1 = 6 d.f is 3.143
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 3985.7143 ± t a/2 ( 241.0295/ Sqrt ( 7) ]
= [ 3985.7143-(3.143 * 91.101) , 3985.7143+(3.143 * 91.101) ]
= [ 3699.385 , 4272.043 ]
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interpretations:
1) we are 98% sure that the interval [ 3699.385 , 4272.043 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population mean

d.
the PDF of normal distribution is = 1/? * ?2? * e ^ -(x-u)^2/ 2?^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 3985.7143
standard Deviation ( sd )= 241.0295/ Sqrt ( 6 ) =98.3999
sample size (n) = 6
the probability that a six-pack of bulbs ( random sample size of 6) manufactured by this company will have a mean life expectancy of at least 5000 hours
P(X < 5000) = (5000-3985.7143)/241.0295/ Sqrt ( 6 )
= 1014.2857/98.3999= 10.3078
= P ( Z <10.3078) From Standard NOrmal Table
= 1
P(X > = 5000) = 1 - P(X < 5000)
= 1 - 1 = 0