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7.5 Estimation of the Population Standard Deviation

Most populations are so large; the majority of studies have little to
no information about the population variance
deviation  .
 2 or standard
Why do we need to estimate standard deviation?
- GE Company manufactures cat-scan machines with an
automatic injection of iodine to blood stream. The consistency
of the amount of iodine injected each time is subject to strict
safety measurements.
- Bolts used in constructing air planes undergo a tremendous
amount of abuse (weather, pressure, friction etc.). How can a
manufacturer guarantee that each bolt they manufacture will
consistently function properly under such conditions?
When a sample is drawn from a normally distributed parent
population, the quantity (n  1)
s2
2
is known to follow a Chi-Square
distribution with n-1 degrees of freedom, denoted as  2n1 ( the proof
of this fact is beyond the scope of this course)
The Chi-Square Distribution
Is a right skewed distribution with mean n-1 and a variance of
2(n-1).
The  2n1 distribution is a family of distributions determined by the
sample size (Table 6 page 832).
Let’s Do It! 1
Determine the mean and standard deviation of  322 .
Confidence Interval for the Population Variance 
2
A (1  )% confidence interval for variance  2 of a normally
distributed population is given by:
(n  1)s 2
2
 right
 2 
(n  1)s 2
2
 left
Confidence Interval for the Population Standard Deviation 
A (1  )% confidence interval for the standard deviation  of a
normally distributed population is given by:
(n  1)s 2
2
 right
 
(n  1)s 2
2
 left
Let’s Do It! 2
Use textbook Table G, to find the following chi-square values
.299,12 
Example
.201,12 
Find the 95% confidence interval for the variance and standard
deviation of the nicotine content of cigarettes manufactured if a
sample of 20 cigarettes has a standard deviation of 1.6 milligrams.
Solution
Since   0.05, the two critical values, respectively, for the 0.025
and 0.975 levels for 19 degrees of freedom are 32.852 and 8.907.
The 95% confidence interval for the variance is found by
substituting in the formula.
Hence, one can be 95% confident that the true variance for the
nicotine content is between 1.5 and 5.5.
For the standard deviation, the confidence interval is
Hence, one can be 95% confident that the true standard deviation
for the nicotine content of all cigarettes manufactured is between
1.2 and 2.3 milligrams based on a sample of 20 cigarettes.
Let’s Do It! 3
A sample of 20 cigarettes of a certain brand has a standard
deviation of 1.6mg. Estimate the standard deviation of nicotine
content of this brand at a 90% confidence level. Assume normality
of the nicotine content in cigarettes.
Estimate the variance of nicotine content of this brand at a 90%
confidence level.
Let’s Do It! 4
Find the 90% confidence interval for the standard deviation for the
price in dollars of an adult single-day ski lift ticket. The data
represent a selected sample of nationwide ski resorts. Assume the
variable is normally distributed.
59 54 53 52 51
39 49 46 49 48
Interpret the interval
Homework 7.5 Page 382: 3-12 all