6.7 Problems For problems 1-4 we consider a random sample of 10 recent graduates
ID: 3356811 • Letter: 6
Question
6.7 Problems For problems 1-4 we consider a random sample of 10 recent graduates from the given as: of engineering in a particular year. The data consist of their annual salaries (44600, 30800, 55200, 34500, 4200, 52300, 22000, 12500, 18300, 62200) For problems 1-3 below, build a 90%, 95% and a 99% confidence interval estimate for the mean salary. Explain briefly how we should interpret the results probabilis- tically. Thepopulation standard deviator forth esalaryofallgraduatesinthatyear is -15200. 2. The population standard deviation is unknown. 3. For 1 and 2, comment on whether the width of the interval increases or de creases as we increase the confidence level. Use the formula for the defini- tion of the confidence interval to explain this result. 4. Assume that the population variance is unknown and the data are normally lisòribuisxl. HBulil(ll an 95% C) the popula tion variance.Explanation / Answer
Given sample data:
(44600,30800,55200,34500,4200,52300,22000,12500,18300,62200)
Q1.
a.
DIRECT METHOD
given that,
sample mean, x =33660
standard deviation, s =19556.3119
sample size, n =10
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 9 d.f is 1.833
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 = [ 33660 ± t a/2 ( 19556.3119/ Sqrt ( 10) ]
= [ 33660-(1.833 * 6184.249) , 33660+(1.833 * 6184.249) ]
= [ 22324.272 , 44995.728 ]
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interpretations:
1) we are 90% sure that the interval [ 22324.272 , 44995.728 ] 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
b.
confidence interval = [ 33660 ± t a/2 ( 19556.3119/ Sqrt ( 10) ]
= [ 33660-(2.262 * 6184.249) , 33660+(2.262 * 6184.249) ]
= [ 19671.229 , 47648.771 ]
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interpretations:
1) we are 95% sure that the interval [ 19671.229 , 47648.771 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
c.
confidence interval = [ 33660 ± t a/2 ( 19556.3119/ Sqrt ( 10) ]
= [ 33660-(3.25 * 6184.249) , 33660+(3.25 * 6184.249) ]
= [ 13561.191 , 53758.809 ]
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interpretations:
1) we are 99% sure that the interval [ 13561.191 , 53758.809 ] 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
Q2.
a.
DIRECT METHOD
given that,
standard deviation, =15200
sample mean, x =33660
population size (n)=10
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 = [ 33660 ± Z a/2 ( 15200/ Sqrt ( 10) ) ]
= [ 33660 - 1.645 * (4806.662) , 33660 + 1.645 * (4806.662) ]
= [ 25753.041,41566.959 ]
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interpretations:
1. we are 90% sure that the interval [25753.041 , 41566.959 ] 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 = 33660
standard error =4806.662
z table value = 1.645
margin of error = 7906.959
confidence interval = [ 25753.041 , 41566.959 ]
b.
confidence interval = [ 33660 ± Z a/2 ( 15200/ Sqrt ( 10) ) ]
= [ 33660 - 1.96 * (4806.662) , 33660 + 1.96 * (4806.662) ]
= [ 24238.942,43081.058 ]
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interpretations:
1. we are 95% sure that the interval [24238.942 , 43081.058 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
c.
confidence interval = [ 33660 ± Z a/2 ( 15200/ Sqrt ( 10) ) ]
= [ 33660 - 2.576 * (4806.662) , 33660 + 2.576 * (4806.662) ]
= [ 21278.039,46041.961 ]
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interpretations:
1. we are 99% sure that the interval [21278.039 , 46041.961 ] 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
Q3.
width increases CI is wider
Q4.
CONFIDENCE INTERVAL FOR STANDARD DEVIATION
ci = (n-1) s^2 / ^2 right < ^2 < (n-1) s^2 / ^2 left
where,
s = standard deviation
^2 right = (1 - confidence level)/2
^2 left = 1 - ^2 right
n = sample size
since alpha =0.05
^2 right = (1 - confidence level)/2 = (1 - 0.95)/2 = 0.05/2 = 0.025
^2 left = 1 - ^2 right = 1 - 0.025 = 0.975
the two critical values ^2 left, ^2 right at 9 df are 19.0228 , 2.7
s.d( s )=19556.3119
sample size(n)=10
confidence interval for ^2= [ 9 * 382449335.1301/19.0228 < ^2 < 9 * 382449335.1301/2.7 ]
= [ 3442044016.1707/19.0228 < ^2 < 3442044016.1707/2.7004 ]
[ 180943079.6818 < ^2 < 1274642281.2068 ]
and confidence interval for = sqrt(lower) < < sqrt(upper)
= [ sqrt (180943079.6818) < < sqrt(1274642281.2068), ]
= [ 13451.5085 < < 35702.1327 ]
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