Transcript Section 6 - Confidence Intervals

William Christensen, Ph.D.
Using Confidence Intervals to Estimate
Population Parameters


Do you understand what a “population parameter” is? We
use the word “parameter” as a general way to describe one
or all of the characteristics of a population such as;
average/mean, proportion, standard deviation and variance.
In the “real world” we usually do not know the true
population parameters (such as the mean) because it is too
expensive and time consuming to collect data on every
member of a population. Therefore, we most often use
sample data to estimate things about the population like
mean and standard deviation (population parameters)
Using Confidence Intervals to Estimate
Population Parameters

The most common method for using sample data to
estimate a population parameter is to create Confidence
Intervals


Basically, a confidence interval allows us to say something like this:
“We are 90% confident that the true population mean is between
24.5 and 27.8.”
Here you can see the two main parts of a confidence interval:
1.
2.
A level of confidence, such as 90% or 95% or 99%. By the way, there
is no 100% confidence level in statistics
A range of values. Using the Confidence Interval methods you are
about to learn, we will establish a range of values that we think the true
population parameter falls between
Using Confidence Intervals to Estimate
Population Parameters


Remember: the whole reason for calculating
confidence intervals is that we usually only have
sample data which is only a small subset of the
population we are interested in. Since there are
probably some differences between our sample
data (which we have) and the true population data
(which we don’t have), we need to be able to
estimate what the true population parameters are.
Confidence Intervals allow us to use our sample
data to estimate the true population parameters,
such as mean and standard deviation.
Using Confidence Intervals to Estimate
Population Parameters

In this Section you will learn to create confidence intervals to
estimate the following population parameters:

Confidence intervals for a Population Mean



Confidence intervals for a Population Mean


When our sample size is small (less than or equal to 30)
Confidence intervals for a Population Proportion


When our sample size is large (more than 30)
You will also learn how to calculate the sample size that would be
necessary to estimate a population mean with a given level of accuracy
A proportion is kind of like a mean, but expressed as a probability
(between 0 and 1) or percentage
Confidence intervals for a Population Variance and/or Standard
Deviation
Point Estimate of Population Parameter

Without a confidence interval, the best estimate
(Point Estimate) of a population parameter is simply
whatever we calculate from the sample data. For
example, if we have a sample of women’s weights
with a mean of 143 lbs., then this is the best “Point
Estimate” we have of the true population mean.

Confidence Intervals allow us to create a better
estimate
Two Parts of a Confidence Interval

Let’s re-visit the two main parts of a confidence
interval:
1. A level of confidence, such as 90% or 95% or 99%.
 The 3 most common confidence levels are 90%, 95%, and 99%
 Associated with any confidence level is a value called alpha ()
  is simply the difference between the confidence level and 1
 For a confidence level of 90%,  = 0.10
 For a confidence level of 95%,  = 0.05
 For a confidence level of 99%,  = 0.01
 You can view a confidence level as the chance we are right about
our confidence interval, and  (alpha) as the chance that we are
wrong
Two Parts of a Confidence Interval
2. A range of values. A Confidence Interval is defined as a range
(or an interval) of values used to estimate the true value of the
population parameter.

The correct form for expressing the range or interval of values is:
Lower #  population parameter  Upper #
Note: the population parameter must always be
expressed by using the appropriate symbol:
 (mu) for population mean
 (sigma) for population standard deviation
2 (sigma squared) for population variance
p for population proportion
Example: 24.3    27.8
Confidence Intervals
for Population Means
in large samples (n > 30)
Estimating Population Means
Calculating the Lower & Upper Limits
Lower #  µ  Upper #
x-E
x+E
We calculate the lower and upper limits of a confidence
interval for a population mean by taking the sample
mean (x-bar) - / + the margin of error (E). Where:
E=z
σ
α/2 • n
Before proceeding to use this formula, let’s learn a little more
about this zα/2, or what is called the CRITICAL VALUE
The Critical Value zα/2
α/2
α/2
-zα

/2
z=0
zα/2
The Critical Value zα/2 is a z-score and
number that separates an area α/2 in the
each (left and right) tail of the standard
normal distribution.
The Critical Value zα/2
95%
α = 5%
α/2 = 2.5% = .025
.95
.025
-zα/2

.025
zα/2
The critical value +/- zα/2 sets apart the area or
probability for our confidence interval. In this case,
we are looking for a 95% confidence interval, so α =
0.05 and α/2 = 0.025
The Critical Value zα/2
95%
.95
.025
-zα/2

.025
zα/2
There are two ways to determine the zα/2 critical
value(s). Note that we have +/- zα/2. These are the
same number, only one is positive and the other
negative. Therefore, if we find one, we know the
other by just changing the sign.
The Critical Value zα/2
There are two ways to determine zα/2

1.
The first and easiest way is to use the Excel function we already
learned, NORMSINV(probability)

Example: to find the zα/2 critical value for a 90% confidence interval:



2.
If the confidence level is 90% then we know alpha = 0.10 (the difference
between the confidence level and 1)
Therefore, α/2 = 0.10 / 2 = 0.05
Using Excel =NORMSINV(0.05) we get an answer of negative 1.64485 (this
is the left-side critical value). The right-side critical value is simply +1.64485
(just change the – to +), or you could calculate the right-side critical value by
=NORMSINV(0.95) = 1.64485.
The second method is the old-style or traditional method which
involves looking up zα/2 in a “normal distribution” table. Since tables
are not always available, I suggest you stick with the Excel method
Estimating Population Means
Calculating the Lower & Upper Limits
Lower #  µ  Upper #
x-E
x+E
E=z
σ
α/2 • n
Now that we understand zα/2 and how to use Excel to find its value, we should be able
to construct a confidence interval for a population mean. Important note: In this
formula, we only need the positive value of zα/2, NOT the negative value. We take
the negative value into account later when we subtract E from the sample mean to
calculate the Lower# for the confidence interval
Estimating Population Means
Calculating the Lower & Upper Limits
Confidence Interval for a population mean:
(sample mean – E)  µ  (sample mean + E), where E = zα/2*(σ/sqrt(n))
EXAMPLE: Given a sample of 50 women in which we find an average or
mean weight of 143 lbs., with a standard deviation of 29 lbs., construct a
95% confidence interval for the population mean
To solve this problem we must first calculate E (margin of error).
The formula for margin of error is: E = zα/2 * (σ / sqrt(n))
Our sample data already provided us the info that s = 29 lbs, and n=50 women, so the
only thing missing is to find zα/2. With a confidence level of 95%, we know α = 0.05, so
α/2 = 0.025. Using the Excel function NORMSINV
We calculate zα/2 as follows: =NORMSINV(0.025) = -1.96 (we ONLY use the positive
value in the formula for calculating E)
Estimating Population Means
Calculating the Lower & Upper Limits
EXAMPLE: Given a sample of 50 women in which we find an average or
mean weight of 143 lbs., with a standard deviation of 29 lbs., construct a
95% confidence interval for the population mean
Now the we have all the pieces, we can solve for E and then construct the confidence
interval. E = zα/2 * (σ / sqrt(n)) = 1.96 * (29 / sqrt(50)) = 1.96 * 4.1012 = 8.04 lbs.
Finally, knowing E we can construct our confidence interval as follows:
(sample mean – E)  µ  (sample mean + E)
(143 – 8.04)  µ  (143 + 8.04)
134.96  µ  151.04
We did it. This is the correct form for a confidence interval. We can read this as
follows: we are 95% confident that for womens’ weights the true population mean is
between 134.96 lbs. And 151.04 lbs.
Estimating Population Means
Calculating the Lower & Upper Limits
PRACTICE, PRACTICE, PRACTICE:
You must know how to do and interpret all kinds of confidence interval
problems. For confidence intervals that estimate population means of large
sample (sample size greater than 30), here are some sample problems.
Practice constructing 90%, 95%, and 99% confidence intervals for the
population means.
1.
2.
3.
A sample of 54 bears in Yellowstone National Park has a mean weight of 182.9
lbs., with a standard deviation of 121.8 lbs.
A study of hospital costs among 40 automobile accident victims who were
wearing seat belts showed an average hospital cost $9000 with a standard
deviation of $5600.
Use other data from the data sets provided for this course to create and solve
your own problems
Calculating E When σ Is Unknown

You may have noticed in the sample
problems that we did that the formula for E
includes σ, which is actually the population
standard deviation, NOT the sample
standard deviation.

However, it is OK to use the sample standard
deviation (s) in place of σ, when the
population standard deviation is not known
(and it usually is not known)
SUMMARY - Procedure for Constructing
a Confidence Interval for µ
( Based on a Large Sample: n > 30 )
1.
Find the critical value zα/2 using Excel NORMSINV
2.
Calculate E, the margin of error: E = zα/2 * σ/sqrt(n)

If σ (population standard deviation) is unknown, use
s (sample standard deviation)
3.
Find the lower and upper limits of the confidence interval
(x - E and x + E). Use the correct form:
Lower#  µ  Upper#
4.
Round the final answer (do not round intermediate calculations)
to one more decimal place than is used in the original sample data
Confidence Intervals
for Population Means
in small samples (n  30)
Estimating Population Means
for small samples




Confidence intervals for population means when we have
small samples (n  30) is very similar to what we just
learned about large samples (n  30)
We still calculate the confidence interval as:
(sample mean – E)  µ  (sample mean + E)
The main difference is that we now have a slightly different
formula for E (margin of error)
For small samples (defined as less than or equal to 30):
E=t
s
α/2
n
(where tα/2 has n - 1 degrees of freedom)
degrees of freedom is discussed a little later
Use σ (population std. dev.) if
available. Otherwise, use s
(sample std. dev.)
The Critical Value tα/2



tα/2 is similar to a zα/2, but rather than coming from
the standard normal distribution, it comes from a
distribution called the “student t distribution”.
It should make sense to you that when we have a
smaller sample from which to estimate the
population mean, our estimate cannot be as
accurate as when we have a large sample.
The tα/2 value adjusts our E (margin of error) to
account for our smaller sample size
The Critical Value tα/2

There are two ways to determine tα/2
1. The first and easiest way is to use a new Excel function
2.
=TINV(probability,deg_freedom) which we will discuss in more detail
in just a minute.
The second method is the old-style or traditional method which
involves looking up tα/2 in a “student t distribution” table. Since tables
are not always available, I suggest you stick with the Excel method
Using TINV to find the Critical Value tα/2

To use Excel’s TINV function to
find the critical value tα/2 we must
input 2 things
1.
2.
The area or probability, which is
represented by α (alpha) – (NOT
α / 2). This means we simply put
in the value of alpha
Degrees of freedom. This is how
we adjust for our small sample.
Degrees of Freedom is the
sample size (n) minus 1.
df = n -1.
 Example: for a small sample of

25, df = 25 – 1 = 24
Example: for a sample of 11,
degrees of freedom (df) = 10
Using TINV to find the Critical Value tα/2

Here is an example where we have a 95% confidence
interval (α = 0.05) and a sample size of 20. Probability is
represented by α (NOT α / 2) when using TINV (assumes
two-tail test)
df = n – 1 = 20 – 1 = 19
Estimating Population Means
for small samples
Confidence Interval for a population mean:
(sample mean – E)  µ  (sample mean + E), where E = tα/2*(s/sqrt(n))
EXAMPLE: Given a sample of 15 women (a small sample) in which we find
an average or mean weight of 143 lbs., with a standard deviation of 29 lbs.,
construct a 95% confidence interval for the population mean
To solve this problem we must first calculate E (margin of error).
Our formula for margin of error is: E = tα/2 * (s / sqrt(n))
Our sample data already provided us the info that s = 29 lbs, and n=15 women, so the
only thing missing is to find tα/2. With a confidence level of 95%, we know α = 0.05.
UNLIKE when calculating zα/2, we DO NOT need to divide α/2 when using TINV
because the TINV function automatically assumes there are two tails, or that the alpha
is split evenly between the left and right sides.
Thus, using the Excel function TINV we calculate tα/2 as follows: =TINV(0.05,14) =
2.1448 (Note that df = n-1 = 15-1 = 14. Also note that
TINV always returns a positive value which we can input
directly into our margin of error formula)
Estimating Population Means
for small samples
EXAMPLE: Given a sample of 15 women in which we find an average or
mean weight of 143 lbs., with a standard deviation of 29 lbs., construct a
95% confidence interval for the population mean
Now the we have all the pieces, we can solve for E and then construct the confidence
interval. E = tα/2 * (s / sqrt(n)) = 2.1448 * (29 / sqrt(15)) = 2.1448 * 7.4877 = 16.06 lbs.
Finally, knowing E we can construct our confidence interval as follows:
(sample mean – E)  µ  (sample mean + E)
(143 – 16.06)  µ  (143 + 16.06)
126.94  µ  159.06
We did it. This is the correct form for a confidence interval. We can read this as follows: given
our sample of 15 women, we are 95% confident that the true population mean of womens’ weights
is between 126.94 lbs. And 159.06 lbs. You might notice that this estimate is not as accurate as
our estimate when our sample size was 50. That’s indeed how it works – smaller samples
yield less-precise estimates of population means
Estimating Population Means
for small samples
PRACTICE, PRACTICE, PRACTICE:
You must know how to do and interpret all kinds of confidence interval
problems. For confidence intervals that estimate population means of small
samples (n  30), here are some sample problems. Practice constructing
90%, 95%, and 99% confidence intervals for the population means.
1.
2.
3.
A sample of 24 bears in Yellowstone National Park has a mean weight of 182.9
lbs., with a standard deviation of 121.8 lbs.
A study of hospital costs among 20 automobile accident victims who were
wearing seat belts showed an average hospital cost $9000 with a standard
deviation of $5600.
Use other data from the data sets provided for this course to create and solve
your own problems
Determining the Sample
Size required for a given
margin or error (E)
Determining Sample Size given E



Sometimes we want to determine in advance how much
error (i.e., E or the margin or error) we are willing to have in
our estimate of the population mean
In fact, we can obtain whatever margin of error we would
like, IF we are willing and able to adjust our sample size
The relationship between sample size and margin of error
is illustrated on the next slide
Sample Size for Estimating Mean µ

Using our basic formula for calculating E (margin
of error) we can also find n, given E
E=
σ
zα/2 
Where:
n
(solve for n by algebra)
n=
zα/2  σ
E
2
zα/2 is based on the desired level
of confidence
E = desired margin of error
Use σ if available, otherwise us s
(sample std. dev.)
Round-Off Rule for Sample Size n
When finding the sample size n, if the
calculated n does not result in a whole
number, always increase the value of n to
the next larger whole number.
n = 116.009 = 117 (round up)
Determining Sample Size given E
Example: If we want to estimate the true population mean IQ for statistics students, how
many statistics students would we need to test (i.e., what sample size is needed) so that
our estimate is within 2 IQ points of the true population mean with a confidence level of
95%? From previous studies, we believe a conservative estimate of the population
standard deviation (σ) is 15.
α = 0.05
zα/2 = NORMSINV(.025) = 1.96
E (desired margin of error) = 2
σ = 15
n = zα/2 σ
E
2
2
= (1.96)(15)
2
= 216.09 = 217 students
We would need to randomly select 217 statistics students and obtain
their IQ scores. We would then be 95% confident that our sample
mean would be within 2 IQ points of the true mean IQ score for the
entire population of statistics students.
Determining Sample Size given E
In determining sample size (given a desired margin of error) we have
assumed that some value or estimate of the population standard deviation
(σ) is available. However, many times we have no estimate of σ. In such
cases, we have three alternatives:
1. Use the range rule of thumb to estimate a standard deviation as follows:
est. standard deviation ≈ range / 4
2. Conduct a pilot study by starting the sampling process. Based on the first
collection of at least 31 randomly selected sample values, calculate the
sample standard deviation s and use it in place of σ. That value can be
refined as more sample data are obtained.
3. Estimate the value of σ by using the results of some other study that was
done earlier
Confidence Intervals for
Population Proportion
Estimating Population Proportion
•
Often we are interested in being able to
estimate a population “proportion”
•
Proportion is kind of like an average, but is
expressed as a probability (p) or percentage
•
For example, we might want to estimate what
proportion (%) of households in the U.S. who
are watching the Olympics on television
Estimating Population Proportion
We use the following notation to express the
confidence interval or estimate for a population
proportion:
Lower#  p  Upper#
“p” represents the true population proportion
ˆ
We use the symbol p (p-hat) to represent the sample
proportion. Another symbol, q is defined as p -1
ˆ
ˆ
Confidence Interval for
Population Proportion
pˆ - E  p  pˆ + E
where
E = zα/2



pˆ qˆ
n
Zα/2 is again found using Excel function
NORMSINV(probability) where “probability” is α / 2
and the absolute value of the result is used in the
above formula
Round the confidence interval limits to three
significant digits.
Example: The CBS television show 60 Minutes has a share of 20, meaning that
among the TV sets in use, 20% are typically tuned to 60 Minutes (based on Nielsen
Media Research data). Assume a sample size of 4,000 (typical for Nielsen surveys).
Construct a 97% confidence interval estimate of the population proportion (the
proportion of all TV sets in the U.S. tuned to 60 Minutes).
ˆ
ˆ
p - E  p  p + E where
E = zα/2
ˆˆ
pq

n
1. α = 0.03 (97% confidence level) and α/2 = 0.03/2 = 0.015
Thus, zα/2 is found using Excel =NORMSINV(0.015) = 2.17 (we take the absolute value which is 2.17)
2. p-hat is given as 20% or 0.20. q-hat is simply 1 - p-hat = 1
– 0.20 = 0.80 (remember: p-hat + q-hat = 1 always)
Thus, E = 2.17 * sqrt ((0.20*0.80) / 4000) = 2.17 * .0063245
= E = 0.0137241
note: often we are dealing with very small numbers – do not round any intermediate
calculations – wait until we have the confidence interval limits to round to 3 significant
digits)
Example: The CBS television show 60 Minutes has a share of 20, meaning that
among the TV sets in use, 20% are typically tuned to 60 Minutes (based on Nielsen
Media Research data). Assume a sample size of 4,000 (typical for Nielsen surveys).
Construct a 97% confidence interval estimate of the population proportion (the
proportion of all TV sets in the U.S. tuned to 60 Minutes).
ˆ
ˆ
p - E  p  p + E where
E = zα/2
ˆˆ
pq

n
1. Finally, our confidence interval is:
p-hat – E  p  p-hat + E
0.20 – 0.0137241  p  0.20 + 0.0137241
0.186  p  0.214 (rounded to 3 significant digits)
•
With this large sample size of 4,000, the margin or error is
quite small and we can be 97% confident that the
population proportion of all TV viewers in the U.S. tuned to
60 Minutes varies only between 0.186 (18.6%) and 0.214
(21.4%).
Determining Sample Size when
estimating p, given desired E
E=
pˆ qˆ
n
zα/2
(solving for n by algebra)
n=
(zα/2)2 pˆ qˆ
E2

Note: If p-hat (the
sample proportion)
is unknown, use a
p-hat = 0.50
Example: We want to determine, with a margin of error of four percentage
points (4%), the current percentage of U.S. households using e-mail.
Assuming that we want 90% confidence in our results, how many households
must we survey? A recent study indicates 28.9% of U.S. households used email.
ˆˆ
n = [zα/2 ]2 p q
Use absolute value of the Excel function
NORMSINV(probability), with probability = α / 2 =
0.10/2 = 0.05
E2
= [1.645]2 (0.289)(0.711)
0.042
= 347.5195
= 348 households
To be 90% confident that our
sample percentage is within
four percentage points of the
true percentage for all
households, we should
randomly select and survey
348 households.
Example: We want to determine, with a margin of error of four percentage
points (4%), the current percentage of U.S. households using e-mail.
Assuming that we want 95% confidence in our results, how many households
must we survey? We have no idea what percent of U.S. households may be
using email (i.e., p is unknown, therefore we should us p = 0.50)
ˆˆ
n = [zα/2 ]2 p q
E2
= [1.96]2 (0.5)(0.5)
0.042
= 600.25
= 601 households
Use absolute value of the Excel function
NORMSINV(probability), with probability = α / 2 =
0.05/2 = 0.025
To be 95% confident that our
sample percentage is within
four percentage points of the
true percentage for all
households, we should
randomly select and survey
601 households.
Note: There is an important relationship between margin of error and sample size.
That is, to reduce margin of error by half, sample size must be increased four
times. In other words, a little less error requires a lot bigger sample. You should
remember this.
Confidence Intervals for
Population Variance and
Standard Deviation
Estimating Population Variance and
Standard Deviation
• The last population parameters that we will learn how
•
to estimate are Variance and Standard Deviation
• Variance is simply Standard Deviation squared
• Thus, Standard deviation is simply the square root of Variance
The bad news is that we need a new type of
distribution and critical value in order to estimate
population variance and standard deviation
•
This new distribution is called the Chi-squared (2)
distribution (see next slide for a density curve graph of the
Chi-squared distribution)
Properties of the Distribution of
the Chi-Square Statistic

The chi-square distribution is not symmetric, unlike the normal and Student t
distributions. As the number of degrees of freedom increases, the distribution
becomes more symmetric

The values of chi-squared can be zero or positive, but they cannot ever be
negative
df = 10
Not symmetric
df = 20
0
All values are nonnegative
General Chi-Square Distribution
x2
0
5 10 15 20
25 30 35 40 45
Chi-Square Distribution for df = 10 and df = 20
Confidence Interval for the Population
Std. Deviation  and Variance 2
(n - 1)s2
Right-tail CV
XR2
σ2
(n - 1)s2
X L2
Left-tail CV
Note: our Chi-square distribution has one critical value for the
left tail and a completely separate critical value for the right-tail
(n - 1)s2
XR2
σ
(n - 1)s2
X L2
Chi-Square Critical Values
These are the two probabilities
for which we must find chisquare values
0.975
0.025
0
For α = 0.05, there is
α/2 = 0.025 in each
tail. With 0.025 in
the left-tail, there is
0.975 to-the-right of
that area. With 0.025
in the right-tail, there
is only exactly that
much (0.025) to-theright remaining.
XL2 = 2.700
0.025
0.025
2
X2
(df = 9)
XR = 19.023
Important: The
area or probability
associated with a
chi-square value is
always the area (all
the area) to the
right of that chisquare value
Chi-Square Critical Values
These are the two probabilities
for which we must find chisquare values.
0.975
0.025
0.025
0.025
0
XL2 = 2.700
2
X2
(df = 9)
XR = 19.023
We can use the
Excel function
CHIINV to find
these chi-square
values
Using Excel CHIINV function to find
chi-square values
In this example with alpha = 0.05 and df = 9 (n = 10), the two values that we
plug-in as probabilities in CHIINV are 0.975 and 0.025. We solve these
one-at-a-time since we cannot input both of them at once.
Degrees of freedom for this example
was given as 9, which means n=10
since df = n -1. We use the same df
each time
Using Excel CHIINV function to find
chi-square values
Try actually using CHIINV to find the chi-square left
and chi-square right values.
Chi-square left
value (always the
smaller value)
Chi-square right
value (always the
larger value)
Roundoff Rule for Confidence Interval
Estimates of σ or σ2
1. When using the original set of data to construct
a confidence interval, round the confidence
interval limits to one more decimal place than is
used for the original set of data.
2. When the original set of data is unknown and
only the summary statistics (n, s) are used,
round the confidence interval limits to the same
number of decimals places used for the sample
standard deviation or variance.
Example: SAT Match scores were collected from 15 randomly
selected women. The mean score is 496, with a standard
deviation of 108. Construct a 99% confidence interval for the
population standard deviation SAT Match score for all women.
Confidence Interval for the Population Standard Deviation 
(n - 1)s 2
X
•
•
•
2
R
 σ 
(n - 1)s 2
2
XL
n = 15
df = n – 1 = 15 – 1 = 14 = df
α = 0.01, so α / 2 = 0.005 with area-to-right of left tail =
0.995 and are-to-right of right tail = 0.005. These are the
two probabilities for which we need to find chi-square values
• s = 108 (given in above problem description)
Example: SAT Match scores were collected from 15 randomly
selected women. The mean score is 496, with a standard
deviation of 108. Construct a 99% confidence interval for the
population standard deviation SAT Match score for all women.
Confidence Interval for the Population Standard Deviation 
(n - 1)s 2
X
2
R
 σ 
(n - 1)s 2
2
XL
1. Use Excel’s CHIINV to find the chi-square right and
chi-square left values
Example: SAT Match scores were collected from 15 randomly
selected women. The mean score is 496, with a standard
deviation of 108. Construct a 99% confidence interval for the
population standard deviation SAT Match score for all women.
Confidence Interval for the Population Standard Deviation 
(n - 1)s 2
X
2
R
 σ 
(n - 1)s 2
2
XL
2. Plug-in the values and construct the confidence interval
(15-1)108 2
(15-1)108 2
31.3194
4.07466
 σ 
Note: DO NOT square the “chi-square” values
Example: SAT Match scores were collected from 15 randomly
selected women. The mean score is 496, with a standard
deviation of 108. Construct a 99% confidence interval for the
population standard deviation SAT Match score for all women.
2. Plug-in the values and construct the confidence interval
(15-1)108 2
31.3194
163296
 σ 
31.3194
 σ 
5213.8929
 σ 
72.2
 σ 
(15-1)108 2
4.07466
163296
4.07466
40075.981
200.2
We are 99%
confident that
the population
standard
deviation of
women’s SAT
Math scores is
between 72.2
and 2090.2
William Christensen, Ph.D.