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ID: 3363775 • Letter: B
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B 1 u-x, x, l A··A-E-EEIjs. | . . 11N emal 1 Spac.. Heading 1 Heading 2 Title Subtitle Subtle En. Emphan iter Font Paragraph Styles 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. Construct a 90 % two-sided confidence interval on the mean life. Interpret the confidence interval (a) (b) Construct a 90% lower-confidence bound on the mean life. Interpret the result. (c) Reconstruct the two sided 90% CI if was not known but estimated from sample of size 2 (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 (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 intervalExplanation / Answer
PART A.
given that,
standard deviation, =20
sample mean, x =1025
population size (n)=100
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 ( 100) ) ]
= [ 1025 - 1.645 * (2) , 1025 + 1.645 * (2) ]
= [ 1021.71,1028.29 ]
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interpretations:
1. we are 90% sure that the interval [1021.71 , 1028.29 ] 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
PART B.
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
confidence interval = [ 1025 ± Z a/2 ( 20/ Sqrt ( 100) ) ]
= [ 1025 - 1.282 * (2) , 1025 + 1.282 * (2) ]
= [ 1022.436,1027.564 ]
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interpretations:
1. we are 90% sure that the interval [1022.436 , 1027.564 ] 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
PART C.
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 ]
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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
PART 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) = 80
ME =20
n = ( 1.645*80/20) ^2
= (131.6/20 ) ^2
= 43.296 ~ 44
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