A simple random sample of size n is drawn from a population that is known to be normally distributed. The sample variance,
s squareds2,
is determined to be
12.512.5.
Complete parts (a) through (c).
(a) Construct a 90% confidence interval for
sigma squaredσ2
if the sample size, n, is 20.The lower bound is
nothing.
(Round to two decimal places as needed.)The upper bound is
nothing.
(Round to two decimal places as needed.)(b) Construct a 90% confidence interval for
sigma squaredσ2
if the sample size, n, is 30.The lower bound is
nothing.
(Round to two decimal places as needed.)The upper bound is
nothing.
(Round to two decimal places as needed.)
How does increasing the sample size affect the width of the interval?
The width increases
The width does not change
The width decreases
(c) Construct a 98% confidence interval for
sigma squaredσ2
if the sample size, n, is 20.The lower bound is
nothing.
(Round to two decimal places as needed.)The upper bound is
nothing.
(Round to two decimal places as needed.)
Compare the results with those obtained in part (a). How does increasing the level of confidence affect the confidence interval?
The width decreases
The width increases
The width does not change
Solution:
Confidence interval for population variance formula is given as below:
(n – 1)*S2 / χ2α/2, n – 1 < σ2 < (n – 1)*S2 / χ21 -α/2, n– 1
Part a
We are given
Confidence level = 90%, α = 0.10
n = 20, df = n – 1 = 19
Sample variance = S2 = 12.5
χ2α/2, n – 1 = 30.1435
χ21 -α/2, n– 1 = 10.1170
(20 – 1)* 12.5/ 30.1435< σ2 < (20 – 1)* 12.5/ 10.1170
7.8790 < σ2 < 23.4753
Lower bound = 7.88
Upper bound = 23.48
Part b
We are given
Confidence level = 90%, α = 0.10
n = 30, df = n – 1 = 29
Sample variance = S2 = 12.5
χ2α/2, n – 1 = 42.5570
χ21 -α/2, n– 1 = 17.7084
(30 – 1)* 12.5/ 42.5570< σ2 < (30 – 1)* 12.5/ 17.7084
8.5180< σ2 < 20.4706
Lower bound = 8.52
Upper bound = 20.47
Part c
We are given
Confidence level = 98%, α = 0.02
n = 20, df = n – 1 = 19
Sample variance = S2 = 12.5
χ2α/2, n – 1 = 36.1909
χ21 -α/2, n– 1 = 7.6327
(20 – 1)* 12.5/ 36.1909 < σ2 < (20 – 1)* 12.5/ 7.6327
6.5624< σ2 < 31.1160
Lower bound = 6.56
Upper bound = 31.12
How does increasing the level of confidence affect the confidence interval?
The width increases
(By comparing part a and c)
(All critical values are taken from Chi square table)
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