Variance and Standard
Deviation Confidence
Intervals
Warm Up
The weights of a sample of 20
students ended in a confidence
interval of (120, 200).
What was the point estimate of the
mean?
Objective
Construct a Variance
and St. Deviation
Confidence Interval
Relevance
Can Estimate Population
Parameters from Sample
Statistics
Objective
Construct a
Variance and
Standard Deviation
Confidence Interval
Relevance
Be able to estimate
a population
parameter from
sample statistics.
Why do we use them?......
Variance and St. Deviation can be as
important as the mean.
For example, in industry variations on
pipe size diameters must be kept as
small as possible; otherwise the pipes
won’t fit together properly and they
would have to be scrapped.
To find the confidence intervals for
variances and st. deviations a new
statistical distribution is needed.
This distribution is called the
CHI-SQUARE Distribution
Chi-Square (  )
2
Similar to a t-distribution in that it is
based on degrees of freedom
It cannot be negative
It is NOT normal. It is positively
skewed. (Tail goes to the right)
2
At approximately 100 df, the 
distribution becomes somewhat
symmetrical.

2
values can be found in table 8 of
your textbook on p. 719
Because the distribution is not
symmetrical until you get to 100
degrees of freedom, the values for
2
 will be different on the right and left
side of the distribution.
Find the values in each tail for a 90%
confidence interval and n=25……
df=24
  1  .90  0.10
 2  0.10 2  0.05
Look in the table
for the chi-square
values on the right
and left.
Answer……

right

left
2
2
 36.4
 13.8
Example – You Try……
Find the chi-square
values on the right
and left for a
sample size of 6
and a 95%
confidence level.
Answer:
df  5
  1  .95  0.05
 2  0.05 2  0.025

right

left
2
2
 12.8
 0.831
Confidence Interval Formula…..
Variance :
(n  1) s 2

2
St. Deviation :
 2 
right
(n  1) s

2
right
2
(n  1) s 2

2
 
left
(n  1) s
2 left
2
Example……
Find the 95% confidence interval
for the variance and standard
deviation of the nicotene content of
cigarettes manufactured if a
sample of 20 cigarettes has a st.
deviation of 1.6 milligrams.
Answer……
Values:
n  20, df  19
s  1.6
  1  .95  0.05
 2  0.05 2  0.025

2

2
right
left
 32.9
 8.91
Plug Values Into Equations……
Variance :
(n  1) s 2
 2 
2 right
(n  1) s 2
2 left
2
(20  1)(1.6) 2
(
20

1
)(
1
.
6
)
Variance :
 2 
32.9
8.91
Variance : 1.5   2  5.5
St. Deviation :
(n  1) s 2

2
 
right
St. Deviation : 1.5    5.5
St.Deviation : 1.2    2.3
(n  1) s 2
2 left
Example…
A manufacturer measures 19
randomly selected dowels and finds
the standard deviation of the sample
to be s = 0.16. Find the 95%
confidence interval for the
A. Population Variance
B. Population Standard Deviation
A manufacturer
measures 19
randomly selected
dowels and finds the
standard deviation of
the sample to be s =
0.16. Find the 95%
confidence interval for
the
A. Population
Variance
Values:
n  19
s  0.16
df  18
  0.025
2
2right  31.5
2
left
 8.23
Variance Confidence Interval

n  1s 2
Variance :

2
right

19  10.16 2
31.5
2

n  1s 2

2
left
2


19

1
0
.
16
2 
0.0146   2  0.0560
8.23
Standard Deviation Confidence
Interval
Take the square root of the variance
figures.
0.0146    0.0560
0.121    0.237
Example….
The football coach
randomly selected
10 players and
timed how long
each player took to
perform a certain
drill. The drill times
in minutes were:
9, 5, 7,12,14,14,5,
11,6, 9
s 2  11.96
n  10
df  9

2
 0.025
 2right  19.0
2
left
 2.70
Answer….

n  1s 2
Var :
2right
2


n

1
s
 2 
2
left


10  1(11.96)
10  111.96 
2
Var :
 
19.0
Var : 5.665   2  39.867
St. Dev :
5.665    39.867
St. Dev : 2.38    6.31
2.70
Assignments……
Worksheet