Example 5
:
A researcher wants to estimate the variance for the amount of comp time accumulated per week in the
aerospace industry. He randomly samples 18 managers and obtains a sample mean of 13.5556 hours with a sample
standard deviation of 7.8006 hours. Use this data to construct a 99% confidence interval for the population variance of
comp time worked per week by managers in the aerospace industry.
From the text, we have s = 7.8006, n = 18, n-1 = 17.
With a 99% confidence level: α = 1 - .99 = .01
α/2 = .01/2 = .005
,
1 - α/2 = 1 - .005 = .995
Using Excel, we get the following:
2
=
2
=
28.9609
2
181.5697
P (28.9609
2
181.5697) = 99%
Interpretation: We’re 99% confident that the population variance of comp time hours per week is between 28.9609 and 181.5697 hours.
Note: In this problem we were given the sample standard deviation (s) which had to be squared in the numerator of the equation on both
sides to get the sample variance.
•
Left (Lower Limit)
Right (Upper Limit)
X
2
α/2, n-1
Excel Syntax
X
2
1- α/2, n-1
Excel Syntax
X
2
.005,17
35.7185
=CHIINV(0.005,17)
X
2
.995, 17
5.6972
=CHIINV(0.995,17)

Module 12: Confidence Intervals Part 2
B.
Confidence intervals to estimate the population variance.
Example 6
:
The business council of Greece is interested in estimating the variance of the hourly wages of production
workers in Greece. They take a random sample of 25 production workers and obtain a sample mean wage of $9.12 with a
sample standard deviation of $1.12. Use this information to construct a 95% confidence interval for the variance of the hourly
wages of production workers in Greece.
From the text, we have s = 1.12, n = 25, n-1 = 24.
With a 95% confidence level: α = 1 - .95 = .05
α/2 = .05/2 = .025
,
1 - α/2 = 1 - .025 = .975
Using Excel, we get the following:
2
=
2
=
.7648
2
2.4276
P (.7648
2
2.4276) = 95%
Interpretation: We’re 95% confident that the population variance of hourly wages for production workers in Greece is between $.7648 and
$2.4276.
Note: In example 6, as in example 5 we were given the sample standard deviation (s) which had to be squared in the numerator of the equation
on both sides to get the sample variance.
•
Left (Lower Limit)
Right (Upper Limit)
X
2
α/2, n-1
Excel Syntax
X
2
1- α/2, n-1
Excel Syntax
X
2
.025,24
39.3641
=CHIINV(0.025,24)
X
2
.975, 24
12.4012
=CHIINV(0.975,24)

Module 12: Confidence Intervals Part 2
Module 12 Exercises
Exercise 1
: A random sample of 85 fast food fans is obtained. 40 of those sampled preferred McDonald's over Burger King. Use
this information to construct a 99% confidence for the population proportion of fast-food fans who prefer McDonalds over
Burger King.
Exercise 2
:
A random sample of 24 pieces of standard chalk from a factory which produces chalk resulted in a sample mean
diameter of .375” with a sample variance of .001 inches. Use this data to construct a 99% confidence interval for the
population variance of the chalk’s diameter.
Exercise 3
:
A random sample of 15 wine refrigerators of a certain model resulted in a sample mean temperature of 42
degrees with a sample standard deviation of 1.4142 degrees. Use this data to construct a 98% confidence interval for the
population variance of this wine refrigerator model’s temperature variance.

Module 12: Confidence Intervals Part 2
Solutions to Module 12 Exercises
Exercise 1
: A random sample of 850 fast food fans is obtained. 400 of those sampled preferred McDonald's over Burger King.

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- Spring '14
- DebraACasto
- Normal Distribution, BURGER KING, α