The life in hours of a 100 watt light bulb is known to be normally distributed w
ID: 3363859 • Letter: T
Question
The life in hours of a 100 watt light bulb is known to be normally distributed with 20 hours. A random sample 25 bulbs has a mean life xbar- 1025 hours. 2. (a) Construct a 90 % two-sided confidence interval on the mean life. Interpret the confidence interval. (b) Construct a 90% lower-confidence bound on the mean life. Interpret the result. (c) Reconstruct the two sided 90% CI ifo was not known but estimated from sample of size 25 (s-20). Compare the half-width with that of part (a) and comment on the result. What sample size should be used? from part (a) and comment on impact of level of confidence on the length of the confidence (d) Suppose that in part (a) you want the total width of the confidence interval to be 80 hours. (e) Set-up a 99% two sided confidence interval for mean. Compare the interval with answer interval.Explanation / Answer
2.
a.
TRADITIONAL METHOD
given that,
standard deviation, =20
sample mean, x =1025
population size (n)=25
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 20/ sqrt ( 25) )
= 4
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
since our test is two-tailed
value of z table is 1.645
margin of error = 1.645 * 4
= 6.58
III.
CI = x ± margin of error
confidence interval = [ 1025 ± 6.58 ]
= [ 1018.42,1031.58 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =20
sample mean, x =1025
population size (n)=25
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
since our test is two-tailed
value of z table is 1.645
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 = [ 1025 ± Z a/2 ( 20/ Sqrt ( 25) ) ]
= [ 1025 - 1.645 * (4) , 1025 + 1.645 * (4) ]
= [ 1018.42,1031.58 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [1018.42 , 1031.58 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 1025
standard error =4
z table value = 1.645
margin of error = 6.58
confidence interval = [ 1018.42 , 1031.58 ]
b.
TRADITIONAL METHOD
given that,
standard deviation, =20
sample mean, x =1025
population size (n)=25
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 20/ sqrt ( 25) )
= 4
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table,left tailed z /2 =1.282
since our test is left-tailed
value of z table is 1.282
margin of error = 1.282 * 4
= 5.128
III.
CI = x ± margin of error
confidence interval = [ 1025 ± 5.128 ]
= [ 1019.872,1030.128 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =20
sample mean, x =1025
population size (n)=25
level of significance, = 0.1
from standard normal table,left tailed z /2 =1.282
since our test is left-tailed
value of z table is 1.282
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 = [ 1025 ± Z a/2 ( 20/ Sqrt ( 25) ) ]
= [ 1025 - 1.282 * (4) , 1025 + 1.282 * (4) ]
= [ 1019.872,1030.128 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [1019.872 , 1030.128 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 1025
standard error =4
z table value = 1.282
margin of error = 5.128
confidence interval = [ 1019.872 , 1030.128 ]
c.
TRADITIONAL METHOD
given that,
sample mean, x =1025
standard deviation, s =20
sample size, n =25
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 20/ sqrt ( 25) )
= 4
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 24 d.f is 1.711
margin of error = 1.711 * 4
= 6.844
III.
CI = x ± margin of error
confidence interval = [ 1025 ± 6.844 ]
= [ 1018.156 , 1031.844 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =1025
standard deviation, s =20
sample size, n =25
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 24 d.f is 1.711
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 = [ 1025 ± t a/2 ( 20/ Sqrt ( 25) ]
= [ 1025-(1.711 * 4) , 1025+(1.711 * 4) ]
= [ 1018.156 , 1031.844 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 1018.156 , 1031.844 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
d.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.1% LOS is = 1.645 ( From Standard Normal Table )
Standard Deviation ( S.D) = 20
ME =80
n = ( 1.645*20/80) ^2
= (32.9/80 ) ^2
= 0.169 ~ 1
e.
TRADITIONAL METHOD
given that,
standard deviation, =20
sample mean, x =1025
population size (n)=25
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 20/ sqrt ( 25) )
= 4
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 * 4
= 10.304
III.
CI = x ± margin of error
confidence interval = [ 1025 ± 10.304 ]
= [ 1014.696,1035.304 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =20
sample mean, x =1025
population size (n)=25
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 = [ 1025 ± Z a/2 ( 20/ Sqrt ( 25) ) ]
= [ 1025 - 2.576 * (4) , 1025 + 2.576 * (4) ]
= [ 1014.696,1035.304 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [1014.696 , 1035.304 ] 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
[ANSWERS]
best point of estimate = mean = 1025
standard error =4
z table value = 2.576
margin of error = 10.304
confidence interval = [ 1014.696 , 1035.304 ]
from part a)confidence interval = [ 1025 ± 6.58 ]
= [ 1018.42,1031.58 ] compare with part(e) at level of significance is 0.01 then confidence interval changes.
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