CHAPTER 5
STATISTICAL INFERENCE:
ESTIMATION
1. Statistical Inference—an Introduction.
2. Statistical Estimation
2.1.
Point Estimation
2.2.
Interval Estimation
3. Confidence Interval For the Population Mean
3.1.
Interval Estimate of μ
3.2.
The Margin of Error and Precision of a Confidence Interval
3.2.1. The Impact of Changing the Error Probability α
3.2.2. The Impact of Changes in σ
3.2.3. The Impact of Changing the Sample Size n
3.3.
Confidence Interval for μ when Population Standard Deviation is Unknown
3.3.1. The t-Distribution
3.3.1.1. How to Find the t Score Corresponding to a Given Tail Area
3.3.1.1.1.
Using the t-Table
3.3.1.1.2.
Using Excel
3.3.2. The Option of Using z in Place of t for Large Sample Sizes
3.4.
Confidence Intervals for μ when Population is not Normal
3.5.
Minimum Sample Size Needed to Estimate μ to Within a Desired Margin of Error
4. Confidence Interval For The Population Proportion π
4.1.
Minimum Sample Size Needed to Estimate π to Within a Desired Margin of Error
1. Statistical Inference—an Introduction
In the previous chapters it was shown that features of a data set can be described by summary characteristics
that are measures of location (for example, the mean) or measures of dispersion (such as the standard
deviation). If the data set represents a population, then these summary characteristics, the mean μ and the
standard deviation σ, each define a population parameter. These parameters are constants whose values are
most often unknown. If the data set represents a sample, then the summary characteristics mean x̅ and
standard deviation s each define a sample statistic.
The summary characteristics of a population data set, the parameters, provide very useful information about
the data set. Thus in statistical analysis attention is focused on these parameters. For example, a population
may consist of data showing the number of hours light bulbs manufactured in a plant last. The manufacturer
may be interested in the average life, the population parameter μ. Theoretically, the value of the parameter μ
could be computed by measuring the lifetime of each unit in the population. Practically, however, this would
not be feasible. Instead, a sample of light bulbs could be selected and their lifetime used to estimate the value
of μ. Here, the sample mean x̅ , a sample statistic, is used as the estimator of the population mean μ, the
population parameter.
In many cases like the above it may be impossible or impractical to determine the value of population
parameter by analyzing the entire population values. The process of determining the value of the parameter
may destroy the population units or it may simply be too expensive in money or resources to analyze each
unit. In these cases, we use statistical inference to obtain information about the values of the population
parameters. The objective of statistical inference is to make inferences about a population based on the
sample drawn from the population.
Chapter 5—Interval Estimates
Page 1 of 21
In this course, we will consider statistical inference for two population parameters: the population mean μ
and the population proportion π. There are two basic, closely related, ways to draw inferences about the
value of these two parameters.
•
The first way is to estimate the value of the parameter, as mentioned above. Here the idea is to select
a sample statistic whose value will be used as an estimate of the population parameter. This process
involves point estimation and interval estimation (building a confidence interval).
•
The second way is called hypothesis testing. In this process, we hypothesize a value for the
population parameter and use the sample information to make a decision as to whether to accept or
reject the hypothesis.
In this chapter estimation will be discussed. Hypothesis testing is explained in the next chapter.
2. Statistical Estimation
Statistical estimation involves two main categories: point estimation and interval estimation.
2.1.
Point Estimation
In point estimation a single number computed from the sample data is used as an estimate of the population
parameter. It is called a "point" estimate because one point on the real number line is used to estimate the
population parameter. In interval estimation two points are used defining an interval on the real number line,
which hopefully will contain the value of the population parameter. For example, if the parameter is the
population mean lifetime of light bulbs, based on sample information we might arrive at a point estimate of μ,
say, 𝑥̅ = 923 hours, and an interval estimate of μ of from 913 hours to 933 hours.
The point estimate is obtained as the mean of a single sample of size 𝑛 randomly selected from the
population. In the previous chapter it was shown that, because infinite number of samples of a given size may
be obtained from the population, 𝑥̅ , the sample mean, is a random variable with infinite number of values.
Thus, a given point estimate is only one of these infinite number of values the random variable 𝑥̅ can take on.
The point estimate, therefore, may or may not be close to the population parameter. For example, given that
the distribution of 𝑥̅ is normal, 95% of all 𝑥̅ ‘s fall within ±1.96 standard errors from the population mean,
that is,
P(µ − 1.96se(𝑥̅ ) ≤ 𝑥̅ ≤ µ + 1.96se(𝑥̅ )) = 0.95
Therefore, one cannot tell where a single sample mean falls on the number line relative to the population
mean. If we only provide a point estimate of the parameter μ, we do not have any information about the
reliability of the estimation process.
2.2.
Interval Estimation
To overcome the drawback associated with point estimation, we can obtain an interval or range of values
which may contain the population parameter. Associated with this interval is a percentage value reflecting
the level of confidence that the interval may contain the actual population parameter. For example, we may
state that we are 95% confident that the population mean lifetime of light bulbs is between 913 to 933 hours.
This is why such interval estimates are called confidence intervals. Confidence intervals provide specific
information about the accuracy of the estimation.
Chapter 5—Interval Estimates
Page 2 of 21
Building confidence intervals for population parameters μ and π are routine and involve simple formulas.
The following is an explanation of how these formulas are obtained and what they imply.
3. Confidence Interval For the Population Mean
To explain confidence intervals, let us refer back to the sampling distribution of x̅ . In the last chapter it was
shown that x̅ is a normally distributed random variable1 with a mean of µ𝑥̅ = µ and standard error of se(𝑥̅ ) =
σ⁄√𝑛 . Now consider the following example involving the sampling distribution of x̅ .
Example 1
Assume we know the population mean lifetime of light bulbs manufactured in a plant to be µ = 920 hours. The
standard deviation of population is also known to be σ = 20 hours. The population is normally distributed.
We know that we can take infinite samples of size n from the population. If n = 25, then the sample mean
values are normally distributed with a mean of 𝜇𝑥̅ = 𝜇 = 920 and a standard error of se(𝑥̅ ) = σ⁄√𝑛 = 4.
Given this information, find the percent of all x̅ values that are between 916 and 924. That is, find:
P(916 < 𝑥̅ < 924)
For 𝑥̅ = 916, 𝑧 =
𝑥̅ −µ
se(𝑥̅ )
= −1.00, and for 𝑥̅ = 924, 𝑧 = 1.00. Using the 𝑧 table, we find that
P(−1.00 < 𝑧 < 1.00) = 0.6827. This means that approximately 68% of x values fall within one standard error
from the population mean. This deviation from the mean is ±𝑧 ∙ se(𝑥̅ ) = ±1.00 × 4 = ±4 hours. That is, 68%
of all 𝑥̅ values deviate from the population mean by no more than ±4 hours.
The deviation ±𝑧 ∙ 𝑠𝑒(𝑥̅ ) = ±4 is the margin of error (MOE) or, to be precise, the 68% MOE.
Next you will be asked to find the 95% MOE:
Example 2
Find the two boundaries of the interval, symmetric about the mean, that contains 95% of all possible 𝑥̅ ‘s.
Since the interval would contain 95% of all sample means, the 5% of 𝑥̅ values would fall outside this interval.
This 5% is the “error probability”: 𝛼 = .05. Here, you must first find the z scores that bound an area of 2.5%
at each tail of the z curve. These z scores are ±𝑧𝛼⁄2 = 𝑧0.025 = ±1.96. Thus, the 95% MOE is
𝑀𝑂𝐸 = ±𝑧0.025 se(𝑥̅) = 1.96 × 4 = 7.84
and the lower and upper boundaries are, respectively:
𝑥̅1 = 𝜇 − 𝑀𝑂𝐸 = 920 − 7.84 = 912.16
𝑥̅2 = 𝜇 + 𝑀𝑂𝐸 = 920 + 7.84 = 927.84
Therefore, P(912.16 < 𝑥̅ < 927.84) = 0.9500, which means that 95 percent of all sample mean values fall
within the interval (912.15, 927.84) formed by 𝑀𝑂𝐸 = 7.84.
1
The sampling distribution of x̅ is normal if the population is normal or, per Central Limit Theorem, is approximately
normal if the sample size is 30 or more
Chapter 5—Interval Estimates
Page 3 of 21
P(μ − z0.025 se(x̅) ≤ x̅ ≤ μ − z0.025 se(x̅ )) = 0.95
0.95
x̅
μ
95% of all x̄ values fall
within this interval
μ − z0.025 se(x̄)
3.1.
μ + z0.025 se(x̄)
Interval Estimate (Confidence Interval) for μ
Now suppose a single sample of size n = 25 is selected and the mean 𝑥̅ = 925 hours. Using the MOE
computed above, ±1.96 × 4 = ±7.84, we can build an interval around this sample mean with the following
Lower and Upper boundaries:
𝐿 = 𝑥̅ − 𝑀𝑂𝐸 = 925 − 7.84 = 917.16
𝐿 = 𝑥̅ + 𝑀𝑂𝐸 = 925 + 7.84 = 932.84
Note that this interval (917.16, 932.84) captures the population mean µ = 920.
An Interval Estimate Built Around x̄
x̄ ± MOE
912.16
927.84
917.6
x̅ = 925
932.84
μ = 920
We can repeat the sampling process infinite number of times, obtaining infinite number of 𝑥̅ values. As stated
above, 95% of these would fall within ±𝑀𝑂𝐸 from the population mean. Logically, then, if we built infinite
Chapter 5—Interval Estimates
Page 4 of 21
intervals 𝑥̅ ± 𝑀𝑂𝐸, then of every 100 such intervals 95 should capture the population mean. In the diagram
below, some of the possible intervals are shown, one of which, where 𝑥̅ = 929, does not capture μ. The
percentage of all intervals that do not capture the population mean is 5%. In the diagram below several of
these infinite number of intervals are shown. Note that as long as 𝑥̅ falls within the 𝑀𝑂𝐸 (within the
boundaries 𝜇 ± 𝑀𝑂𝐸), then the interval estimate would capture μ. One of the intervals in the diagram is
shown not to contain μ because 𝑥̅ = 929 is outside the boundary form by ±𝑀𝑂𝐸.
Multiple Interval Estimates Built Around x̅
x̅
x̄ = 925
x̄ = 915
x̄ = 922
x̄ = 918
x̄ = 915
x̄ = 919
x̄ = 929
x̄ = 913
912.16
μ = 920
927.84
In the estimation process, only one sample of size n is selected. Using the formula
𝐿, 𝑈 = 𝑥̅ ± 𝑀𝑂𝐸
we build only one interval. This interval may or may not capture the population mean. However, since we
know that 95 percent of such intervals would capture the population mean, we can say that we are 95%
confident that the interval we have built would capture µ. This interval, hence, is called a 95% confidence
interval.
Generally, the fraction (proportion, or percentage) of intervals that do not contain μ is denoted by α. Thus,
1 − 𝛼 proportion will contain the μ. The term 1 − 𝛼 is called the confidence coefficient (when expressed as a
percentage, it is called the confidence level) and 𝛼 is called the error probability.
Example 3
A normal population has a standard deviation of 𝜎 = 10. A sample of size 𝑛 = 25 is taken. The sample mean
𝑥̅ = 30. Build a 95% confidence interval for the population mean μ.
Chapter 5—Interval Estimates
Page 5 of 21
The 95% confidence interval means that the confidence coefficient 1 − 𝛼 = 0.95. Thus 𝛼 = 0.05, and we
know 𝑧𝛼⁄2 = 𝑧0.025 = 1.96, and se(𝑥̅ ) = σ⁄√𝑛 = 10⁄5 = 2. The lower and upper bounds of the confidence
interval are then:
𝐿, 𝑈 = 𝑥̅ ± 𝑀𝑂𝐸
𝑀𝑂𝐸 = 𝑧𝛼⁄2 se(𝑥̅ )
𝑀𝑂𝐸 = 1.96 × 2 = 3.92
𝐿 = 30 − 3.92 = 26.08
𝑈 = 30 + 3.92 = 33.92
We are 95% confident that:
3.2.
26.08 ≤ µ ≤ 33.92
The Margin of Error and Precision of a Confidence Interval
In the confidence interval for the population mean,
𝐿, 𝑈 = 𝑥̅ ± 𝑀𝑂𝐸
The two ends of the interval deviate from 𝑥̅ by the margin of error, where 𝑀𝑂𝐸 = ±𝑧𝛼⁄2 se(𝑥̅ ), and se(𝑥̅ ) =
σ⁄√𝑛. Therefore,
𝑀𝑂𝐸 = 𝑧α⁄2
σ
√𝑛
The width, hence the precision, of the interval is determined by the size of MOE. The size of the margin of
error, in turn, is determined by the three factors visible in the formula. These are,



α, the error probability (the proportion or percentage of all possible intervals not capturing the
population mean);
σ, the population standard deviation; and
n, the sample size.
The following explains the impact of each of these factors on the precision of the confidence interval. The
wider the interval, the less precise, and hence, the less meaningful it is. The information about the population
parameter may be so wide and imprecise to make the confidence interval meaningless.
3.2.1.
The Impact of Changing the Error Probability α
The term 𝑧𝛼⁄2 in the margin of error formula is the 𝑧 score that bounds a tail area of 𝛼⁄2 under the 𝑧 curve.
The location of the 𝑧 score thus depends on the choice for 𝛼, the error probability. If we want to lower the
risk of missing the population parameter, then a smaller 𝛼 must be chosen. A smaller 𝛼 will increase 𝑧α⁄2 ,
thus widening the confidence interval. The following table shows this relationship between 𝛼 and the
interval width for a given 𝜎 = 10 and sample size of 𝑛 = 25.
Chapter 5—Interval Estimates
Page 6 of 21
Holding 𝜎 = 10 and 𝑛 = 25, the interval width increases as α
decreases.
𝑧α⁄2
𝑀𝑂𝐸 = 𝑧α⁄2 σ⁄√𝑛 𝑤 = 2 × 𝑀𝑂𝐸
Α
0.20
0.10
0.05
0.01
1.28
1.64
1.96
2.58
2.6
3.3
3.9
5.2
5.1
6.6
7.8
10.3
Thus, the width of the confidence interval varies inversely with the error probability. The interval narrows
(becomes more precise) as α increases. But this precision is obtained by increasing the risk of having the
population parameter fall outside the interval. On the other hand, reducing the probability of error widens
the interval and makes it less precise.
3.2.2. The Impact of Differences in σ
The second factor affecting the margin of error is σ, the population standard deviation, the measure of
dispersion of population data. Note that there is a direct relationship between σ and 𝑀𝑂𝐸. When holding n
and α constant, as σ rises, 𝑀𝑂𝐸 gets bigger and, consequently, the interval becomes wider and less precise
Holding α = 0.05 and n = 25, the interval width increases directly and
proportionately with σ
se(𝑥̅ ) = σ⁄√𝑛 𝑀𝑂𝐸 = 𝑧α⁄2 se(𝑥̅ ) 𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑊𝑖𝑑𝑡ℎ = 2 × 𝑀𝑂𝐸
σ
5
1
1.96
3.92
10
2
3.92
7.84
20
4
7.84
15.68
40
8
15.68
31.36
Thus, if the population standard deviation is doubled, the margin of error will also double. However, Unlike
α, which can be determined by the statistician, σ is not a controlled factor. It is a characteristic of the
population. The more dispersed the population data, the wider the confidence interval.
3.2.3. The Impact of Changing the Sample Size n
The sample size is the most important factor affecting 𝑀𝑂𝐸 because it is controlled by the statistician and can
be used to improve the precision of the interval without increasing the error probability. As the formula for
𝑀𝑂𝐸 shows, there is an inverse relationship between 𝑀𝑂𝐸 and n: 𝑀𝑂𝐸 = 𝑧α⁄2 σ⁄√𝑛. As n increases, 𝑀𝑂𝐸
gets smaller, thus narrowing the interval. By increasing n the confidence interval becomes more precise
without increasing the error probability. For a given population σ, and holding α constant, the larger the
sample size, the more precise the interval becomes.
Chapter 5—Interval Estimates
Page 7 of 21
Holding α = 0.05 and σ = 10, the interval width decreases as n
increases.
𝑀𝑂𝐸 = 𝑧α⁄2 se(𝑥̅ ) 𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑊𝑖𝑑𝑡ℎ = 2 × 𝑀𝑂𝐸
n
5
8.77
17.53
10
6.20
12.40
25
3.92
7.84
50
2.77
5.54
100
1.96
3.92
500
0.88
1.75
1000
0.62
1.24
3.3.
Confidence Interval for μ when Population Standard Deviation is Unknown
So far, to explain the theory of confidence intervals we have used examples in which the population standard
deviation was assumed to be known. This, obviously, is an unrealistic assumption because the value of σ is
obtained using the population mean µ, the unknown parameter we are attempting to estimate:
σ=√
∑(𝑥 − µ)2
𝑁
In practical inferential statistical analyses, population standard deviation is also an unknown parameter
which must be estimated using the sample data. The estimator of the population parameter σ is the sample
statistic 𝑠, the sample standard deviation, which is computed from the sample data according to the formula,
∑(𝑥 − 𝑥̅)2
𝑠=√
𝑛−1
Since 𝑠 is used in place of σ, in the 𝑀𝑂𝐸 formula the standard error of 𝑥̅ changed from se(𝑥̅) = σ ⁄√𝑛 to:
se(𝑥̅) =
s
√𝑛
3.3.1. The t-Distribution
When 𝑠 is used in place of σ a peculiar thing happens to the shape of the sampling distribution of 𝑥̅. The
sampling distribution is still bell shaped, but the area under the curve for a given interval of 𝑥̅ values is not
the same as when the known σ is used. To illustrate, consider the following example:
First, suppose the population mean is µ = 100 and the standard deviation is σ = 20. The fraction (proportion,
or percentage) of 𝑥̅ values for samples of size n = 16 taken from this population that fall between, say, 90.2
and 109.8 are determined as follows:
P(90.2 < 𝑥̅ < 109.8)
se(𝑥̅) = 𝜎⁄√𝑛 = 20⁄√16 = 5
𝑧=
𝑥̅ − µ 90.2 − 100
=
= −1.96
se(𝑥̅ )
5
Chapter 5—Interval Estimates
Page 8 of 21
P(−1.96 < z < 1.96) = 0.95
Now, instead of using σ, let the standard deviation 20 be as if determined from a sample. That is, let 𝑠 = 20.
Hence, se(𝑥̅ ) = s⁄√𝑛 = 5. However, when we attempt to transform 𝑥̅ to z using the formula
𝑧 = (𝑥̅ − µ)⁄se(𝑥̅ ) a problem arises. The new random variable obtained through this transformation no
longer has a z distribution (with mean 0 and standard deviation 1). This problem was observed by William S.
Gosset (1867-1937), a British chemist/statistician, in a paper published in 1908. Gosset showed that, when
the sample size is small, the standard normal table z does not provide the accurate area under the curve for
𝑥̅ −µ
𝑥̅ −µ
the scores obtained from the conversion formula
. In the above example, if
= 1.96, the area under
se(𝑥̅ )
se(𝑥̅ )
the curve is bounded by the two scores ±1.96 is no longer 0.95. In fact, that area is less than 0.95. To get 0.95,
the boundaries must be extended further from zero in either direction. He developed an alternative table to
obtain the more accurate areas or probability values for the scores thus calculated. The new table of
probabilities he provided is now called the t table. And the random variable obtained from this
transformation is said to have a t distribution, where,
𝑡=
𝑥̅ − µ
se(𝑥̅ )
The difference between the z and t distributions is shown in the following diagram. Like the z distribution,
the t distribution is symmetric about the mean of 0. However, unlike z, which has a unique, unchanging shape
due to its fixed standard deviation 1, the t distribution acquires different shapes depending on a parameter
called degrees of freedom. In estimations involving μ, the degrees of freedom is df = n − 1, the denominator
used in computing the sample variance. The smaller the degrees of freedom, the larger the tail areas. As the
df increases, the tail area under the t curve approaches the tail area under the z.2
As the degrees of freedom increases, the distinction between z and t practically disappears. For any df > 2, the standard
deviation of the t distribution is σ𝑡 = √𝑑𝑓 ⁄(𝑑𝑓 − 2). For example, if df = 9, then σ𝑡 = 1.134. As df rises, the standard
deviation approaches 1, which is the standard deviation of z. For example, let df = 1000, then the standard deviation is
practically 1 (√1000⁄998 = 1.001). The fact that t has a larger standard deviation than z makes the tail area under the t
curve relatively larger. Thus, using a computer, it can be shown that, while the tail area for the z score 1.96 is 0.025, the
tail area associated with a t score of 1.96 (with df of, say, 9) is 0.0408. Having a larger standard deviation and tail area
than z is a reflection of the fact that the t distribution applies to situations with a greater inherent uncertainty. The
̅−µ
𝑥
uncertainty arises from the fact that σ is unknown and is estimated by the random variable s. The t distribution, 𝑡 = ⁄ ,
2
thus reflects the uncertainty in two random variables, 𝑥̅ and s, while 𝑧 =
̅
𝑥−µ
σ⁄√𝑛
𝑠 √𝑛
reflects only an uncertainty due to x̄ . The
greater uncertainty in t (which makes confidence intervals based on t wider than those based on z) is the price we pay for
not knowing σ and having to estimate it from sample data.
Chapter 5—Interval Estimates
Page 9 of 21
0
In the above diagram, the tail area corresponding to the score 1.96 is shown under the two curves. The tail
area corresponding to 1.96 under the curve 𝑡𝑑𝑓=9 (0.0408) is larger than that under the z curve (0.025). The t
score which bounds a tail area of 0.025 under 𝑡𝑑𝑓=9 is 2.262, which is further away from the center 0 than
1.96.
The t distribution is a family of curves with a mean of 0 and a standard deviation which is determined by the
degrees of freedom. The tail area for each 𝑡 score thus depends on the degrees of freedom. Conversely, the t
score corresponding to a given tail area is determined by the degrees of freedom of the t distribution.
Going back to the example, after the conversions, using df = n − 1 = 9, we find:
P(−1.96 < t < 1.96) = 0.9184.3
Conversely,
P(−2.262 < t < 2.262) = 0.95
As we can see, in the above example if the population standard deviation is known, the proportion of x̅ values
that fall within 1.96 standard errors from μ is 0.95, but when s is used the proportion falling within ±1.96se(x̅ )
is 0.9184. Or, conversely, 95% of 𝑥̅ values fall within ±2.262se(𝑥̅) from the population mean.
Back to the Confidence Intervals for μ
To build a 95% confidence interval when the population standard deviation is estimated from the sample, the
margin of error is,
𝑀𝑂𝐸 = 𝑡𝛼⁄2,(𝑛−1) se(𝑥̅ )
For example, suppose you are asked to build a 95% confidence interval for the population mean where you
have to estimate the standard deviation using a sample of size 𝑛 = 10. In this case you need to find the t score
with 𝑛 − 1 = 9 degrees of freedom corresponding to a tail area of 𝛼⁄2 = 0.025, that is, tα/2,(n − 1) = t0.025,(9).
The t-score that bounds a tail area of 0.025 for a t-distribution with 9 degrees of freedom is 𝑡0.025,9 = 2.262.
Next section explains how to obtain this t-score.
This probability is found by using the T.DIST function in Excel. The T.DIST syntax is:
=T. DIST(𝐱, deg_freedom, cumulative). Using = T. DIST(­𝟏. 𝟗𝟔, 9,1), we find the tail area 0.0408 under the t-curve. Then,
the area P(−1.96 < t < 1.96) = 1 − 2 × 0.0408 = 0.9184.
3
Chapter 5—Interval Estimates
Page 10 of 21
3.3.1.1.
How to Find the t Score Corresponding to a Given Tail Area
3.3.1.1.1.
Using the t-Table
In the absence of a computer, you have to use a t table. The following shows a portion of a typical t table. The
full table is at the end of the chapter.
df
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
t scores corresponding to different tail areas
0.1
0.05
0.025
0.01
0.005
3.078
6.314
12.706
31.821
63.656
1.886
2.920
4.303
6.965
9.925
1.638
2.353
3.182
4.541
5.841
1.533
2.132
2.776
3.747
4.604
1.476
2.015
2.571
3.365
4.032
1.440
1.943
2.447
3.143
3.707
1.415
1.895
2.365
2.998
3.499
1.397
1.860
2.306
2.896
3.355
1.383
1.833
2.262
2.821
3.250
1.372
1.812
2.228
2.764
3.169
1.363
1.796
2.201
2.718
3.106
1.356
1.782
2.179
2.681
3.055
1.350
1.771
2.160
2.650
3.012
1.345
1.761
2.145
2.624
2.977
1.341
1.753
2.131
2.602
2.947
To find the t score associated with a tail area of 0.025 for a t-distribution with 9 degrees of freedom, you look
up the t score under the column associated with a right tail area of 0.025 with df = 9. In this case, 𝑡0.025,9 =
2.262. Clearly, the t-score for the left tail area would be -2.626.
3.3.1.1.2.
Using Excel
In Excel, there are two functions to find the t scores associated with a given tail area and degrees of freedom.
You may use either one.


=T. INV(probability, deg _freedom)
=T. INV. 2T(probability, deg _freedom)
For example, to find 𝑡0.025,9 = 2.262, you can use the two Excel functions as follows:

=T.INV(0.025,9)
This would return the t-score for the left tail area, −2.262. You then ignore
the negative sign.
=T.INV(α/2,df)

=T.INV.2T(0.05,9)
=T.INV.2T(α,df)
The term “2T” in the function implies “two tails”. Therefore you must
double the tail area to get the t-score for the right tail.
Examples 4
Using Excel, find tα/2, df for each of the following error probabilities (α) and degrees of freedom df = n − 1.
Chapter 5—Interval Estimates
Page 11 of 21
(a)
𝛼 = 0.05, 𝑛 = 5
𝑡𝛼⁄2,𝑑𝑓 = 𝑡0.025,4
(b)
α = 0.10, n = 15
𝑡𝛼⁄2,𝑑𝑓 = 𝑡0.05,14
(c)
α = 0.05, n = 100
𝑡𝛼⁄2,𝑑𝑓 = 𝑡0.025,99
=T.INV(0.025,4) = -2.776
=T.INV.2T(0.05,4) = 2.776
=T.INV(0.05,14) = -1.761
=T.INV.2T(0.1,14) = 1.761
=T.INV(0.025,99) = -1.984
=T.INV.2T(0.05,99) = 1.984
Example 5
The Food and Drug Administration needs to estimate the average content of an additive in a given food
product. A random sample of 20 portions of the product yields a sample average of 𝑥̅ = 8.9 and sample
standard deviation of 𝑠 = 0.5 units. Construct a 95% confidence interval for the average number of units of
additive in any portion of this food product.
First, find the standard error, se(𝑥̅ ) = 𝑠⁄√𝑛 = 0.5⁄√20 = 0.112. The appropriate 𝑡 score is,
t0.025,19 = 2.093. The margin of error is then,
𝑀𝑂𝐸 = 𝑡𝛼⁄2,𝑑𝑓 se(𝑥̅ ) = 2.093 × 0.112 = 0.234
MOE = tα/2,(n − 1) se(x̅ ) = 2.093 × 0.112 = 0.234
The interval is then,
𝐿, 𝑈 = 8.9 ± 0.234 = (8.67,9.13)
Example 6
A sample of 16 five-pound bags of potatoes have the following weights (in pounds).
4.8
4.9
4.7
5.0
5.0
5.0
5.2
4.6
5.0
4.7
5.1
4.7
4.7
4.9
4.5
4.9
Construct a 90% confidence interval for the population mean μ.
In this example, you must first compute 𝑥̅ and 𝑠 from the sample data.
𝑥̅ = ∑𝑥 ⁄𝑛 = 77.7⁄16 = 4.856
𝑠 2 = ∑(𝑥 − 𝑥̅ )2 ⁄(𝑛 − 1) = 0.559⁄15 = 0.035
𝑠 = √0.035 = 0.193
Since 1 − 𝛼 = 0.90, 𝛼⁄2 = 0.05, then 𝑡0.05,15 = 1.753
se(𝑥̅ ) = 𝑠⁄√𝑛 = 0.193⁄√16 = 0.048
𝑀𝑂𝐸 = 𝑡𝛼⁄2,𝑑𝑓 se(𝑥̅ ) = 1.753 × 0.048 = 0.085
𝐿, 𝑈 = 𝑥̅ ± 𝑀𝑂𝐸 = 4.856 ± 0.085 = (4.77,4.94)
3.3.2. The Option of Using 𝒛 in Place of 𝒕 for Large Sample Sizes
As the sample size increases, the t score converges to the z score for a given tail area. The following table
shows this convergence. The t scores for larger degrees of freedom are obtained using Excel.
Chapter 5—Interval Estimates
Page 12 of 21
The t and z scores associated with a
right tail area of 0.025 and different
sample sizes (𝑑𝑓 = 𝑛 − 1)
n
t score
z score
10
2.262
1.96
25
2.064
1.96
50
2.010
1.96
100
1.984
1.96
500
1.965
1.96
1000
1.962
1.96
The practical impact of this convergence is that for large sample sizes you can use 𝑧𝛼⁄2 in place of 𝑡𝛼⁄2,𝑑𝑓
with a negligible impact on the 𝑀𝑂𝐸. The following table compares the margin of errors for 95% confidence
intervals for different sample sizes for z and t distributions. A sample standard deviation of s = 8 is used.
Margin of error for 95% confidence intervals associated with different n’s: A comparison of
t and z scores. (𝑠 = 8)
𝑡0.025,(𝑛−1)
𝑧0.025
𝑀𝑂𝐸 = 𝑡0.025,(𝑛−1) 8⁄√𝑛
𝑛
𝑀𝑂𝐸 = 𝑧0.025 8⁄√𝑛
10
25
30
50
100
500
1000
2.262
2.064
2.045
2.010
1.984
1.965
1.962
1.96
1.96
1.96
1.96
1.96
1.96
1.96
5.72
3.30
2.99
2.27
1.59
0.70
0.50
4.96
3.14
2.86
2.22
1.57
0.70
0.50
Note that after 𝑛 = 100, the difference between the two margin of error estimates practically disappears.
Therefore, for large sample sizes (𝑛 > 100) you may use the following formula for the margin of error:
𝑀𝑂𝐸 = 𝑧𝛼⁄2 se(𝑥̅ )
By convention, any sample of size greater than 100 is considered “large”. Therefore, use z whenever the
̅), if the sample size
sample size is greater than 101. You must use t in the MOE formula, 𝑴𝑶𝑬 = 𝒕𝜶⁄𝟐,𝒅𝒇 𝐬𝐞(𝒙
is 𝒏 ≤ 𝟏𝟎𝟎, which yields 𝒅𝒇 ≤ 𝟗𝟗.
3.4.
Confidence Intervals for μ when Population is not Normal
If the population distribution is not known or it is known not to be normally distributed, then to assure that
the sampling distribution of x̅ is approximately normal, the sample size, per Central Limit Theorem, must be
30 or more. The advantage afforded by the Central Limit Theorem makes it possible to construct confidence
intervals for the population mean regardless of the distribution of the population.
Example 7
To determine the profitability of used car sales in the Indianapolis metropolitan area a sample of 120 used car
sales provided a sample mean profit of $320 and a sample standard deviation of $164. Build a 95%
confidence interval for the mean profit for all used car sales in the Indianapolis metropolitan area.
First, compute the standard error.
Chapter 5—Interval Estimates
Page 13 of 21
se(𝑥) = 164⁄√120 = 14.971
To compare confidence intervals using z and t, first use the t distribution to determine the 𝑀𝑂𝐸. Using Excel,
the t score is:
𝑡0.025,119 = 1.98
The margin of error is:
𝑀𝑂𝐸 = 1.980 × 14.971 = 29.64
and
𝐿, 𝑈 = 320 ± 29.64 = (290.36,349.64)
Alternatively, since n is large, you may use the z distribution.
𝑀𝑂𝐸 = 1.96 × 14.971 = 29.34
and
𝐿, 𝑈 = 320 ± 29.34 = (290.66,349.34)
Note that using z alters the boundaries of the interval by only negligible amounts.
3.5.
Determining the Minimum Sample Size to Obtain a Desired Margin of
Error
Before embarking on building a confidence interval for the population mean, a researcher must decide the
appropriate sample size. What is the appropriate sample size? The answer to this question depends upon the
desired precision of the confidence interval. A wide confidence interval provides an imprecise and, therefore,
inadequate information. You may narrow the interval by reducing the confidence coefficient 1 − α. But this
would increase the error probability, raising the chances that the interval built would not capture the
population mean. The best alternative is then to increase the sample size. However, sampling requires
expenditure of resources. Therefore, you must choose the sample size subject to the resource constraint. If
the benefits of a more precise estimation outweigh the additional cost of a larger sample, then sample size
should be increased.
The width of a confidence interval is determined by the margin of error
𝑀𝑂𝐸 = 𝑡𝛼⁄2,(𝑛−1)
𝑠
√𝑛
As the formula shows, the margin of error is inversely related to the sample size. Solving for n, we have:
𝑡α⁄2,(𝑛−1) 𝑠 2
𝑛=(
)
𝑀𝑂𝐸
To determine the appropriate sample size for a desired margin of error, therefore, we have to solve for 𝑛 in
the 𝑀𝑂𝐸 formula. The problem with using the above formula is that it requires the use of the t distribution
for which a sample size must be known in order to determine the degrees of freedom. Also, the formula
involves 𝑠, the sample standard deviation, which implies to find the sample size, we must know the sample
size!
Chapter 5—Interval Estimates
Page 14 of 21
Therefore, in the 𝑛 formula above we need to replace 𝑡α⁄2,(𝑛−1) and 𝑠 with alternate quantities. The
alternative to 𝑡α⁄2,(𝑛−1) is clearly 𝑧α⁄2 . And in place of 𝑠, we use a planning value for the standard deviation.
Denoting the planning value by 𝜎̂ (𝜎-hat), the formula for solving for 𝑛 becomes:
𝑧𝛼⁄2 𝜎 2
𝑛=(
)
𝑀𝑂𝐸
We can determine a value for 𝜎̂ by using any of the following methods:
 Use a sample standard deviation from a previous sample.
 Conduct a pilot study to determine a preliminary estimate.
 In some cases, if the minimum and maximum value of the population data set is known or can be
reasonably estimated, an acceptable value for 𝜎̂ can be obtained by dividing the range (maximum –
minimum) by 4.
Example 8
In the example above regarding profitability of used car sales, the margin of error for a 95% confidence
interval for the population average profit per used car sales, using a sample size of 𝑛 = 120, was:
𝑀𝑂𝐸 = 1.96(164⁄√120 = 29.34
Suppose we are interested in building a more precise 95% confidence interval with a margin of error of
𝑀𝑂𝐸 = ±$10. What is the minimum sample size needed to achieve this margin of error? Use the sample
standard deviation $164 as a planning value (𝜎̂ = 164).
𝑧𝛼⁄2 𝜎 2
1.96 × 164 2
𝑛=(
) =(
) = 1,033.24
𝑀𝑂𝐸
10
Since you cannot have fractional samples, and because you are interested in the minimum sample size, you
have to round up the resulting 𝑛 value to a full integer. Thus, 𝒏 = 𝟏, 𝟎𝟑𝟒.
4. Confidence Interval For the Population Proportion π
To estimate the population proportion π, the sample statistic 𝑝̅ is used. The population proportion measures
the relative frequency of occurrence of some characteristic in the population. For example, the percentage of
voters voting for a candidate is the frequency of the number of voters voting for the candidate relative to the
total number of voters (𝜋 = 𝑥 ⁄𝑁). If you take a sample of 𝑛 voters, then the sample proportion of voters is
the number of voters favoring the candidate in the sample divided by the sample size: 𝑝̅ = 𝑥 ⁄𝑛. The sample
statistic 𝑝̅ is the estimator of the population parameter π. Since the value of 𝑝̅ is obtained from a random
sample, then the sample statistic 𝑝̅ is a random variable. And, since there are infinite number of possible
samples, there are infinite number of values of 𝑝̅. If these sample proportions are normally distributed, then
we can develop confidence intervals for the population proportion π in the same manner we build intervals
for the population mean μ.
For large sample sizes, the sampling distribution of 𝑝̅ is (approximately) normal. In Chapter 4 it was
explained that the binomially distributed random variable 𝑥 is approximately normal if 𝑛𝜋 ≥ 5 and 𝑛𝜋 (1 −
𝜋) ≥ 5 . The following shows the distribution of 𝑥 for 𝑛 = 100 and 𝜋 = 0.50.
Chapter 5—Interval Estimates
Page 15 of 21
Binomial Probability Distribution of x
n = 100, π = 0.50
x
Note that 𝑥 in a given sample represents the number of “successes” (say, the number of voters in a sample
favoring a candidate). Since the sample size is large, 𝑥 is approximately normal. Therefore, the random
variable 𝑝̅ = 𝑥 ⁄𝑛 is also approximately normally distributed. (𝑝̅ is a linear transformation of 𝑥, since 𝑥 is
multiplied by a constant 1⁄𝑛.) The mean (the expected value) and standard error of 𝑝̅, as shown in Chapter 4,
are:
E(𝑝̅) = 𝜋
π(1 − π)
se(𝑝̅ ) = √
𝑛
Example 9
Suppose in the population of households 20% of households spend $200 or more per week on groceries. A
sample of 𝑛 = 500 households is taken. What is the probability that between 15%-25% of households in this
sample spend $200 or more per week on groceries?
P(0.15 < 𝑝̅ < 0.25) = ______
Here the sample proportion is normally distributed with a mean of 𝐸(𝑝̅) = 𝜋 = 0.20, and standard error of
0.20(1 − 0.20)
se(𝑝̅ ) = √
= 0.0179
500
Using
𝑧=
𝑝̅ − π
se(𝑝̅ )
and computing 𝑧 = ±2.79, then
P(−2.79 < z < 2.79) = 0.9947.
Chapter 5—Interval Estimates
Page 16 of 21
Example 10
In the previous example, find the interval of 𝑝̅ values which contains the middle 95% of sample proportions.
To determine the interval, as the diagram below shows, you need to determine the margin of error MOE =
z0.025se(p̅ )
= 0.95
0.95
π
p̅
95% of all p̅ values fall
within this interval
Since 𝑧0.025 = 1.96 and se(𝑝̅) = 0.0179, then
𝑀𝑂𝐸 = 1.96(0.0179) = 0.035
Thus, 95% of 𝑝̅ values fall within the interval 0.20 ± 0.035, or, 95% of 𝑝̅’s fall in the interval (0.165,0.235)
P(0.20 − 0.035 < 𝑝̅ < 0.20 + 0.035) = 0.95
P(0.165 < 𝑝̅ < 0.235) = 0.95
Recall from the discussion of the confidence interval for μ that we can use the margin of error to build
intervals around each value of sample statistic, 𝑥̅ ± 𝑀𝑂𝐸, and state, with a given level of confidence, that this
interval would capture the population parameter. We can do the same here. Since there are infinite number
of 𝑝̅ values available, we can build infinite number of intervals 𝑝̅ ± 𝑀𝑂𝐸. Of these, 95% would capture the
population parameter π and 5% would not. The fraction of intervals that would not capture the population
parameter π is denoted by α.
To build confidence intervals for the population proportion π, the margin of error used in the confidence
interval formula
𝐿, 𝑈 = 𝑝̅ ± 𝑀𝑂𝐸
is
𝑀𝑂𝐸 = 𝑧𝛼⁄2 se(𝑝̅)
where,
Chapter 5—Interval Estimates
Page 17 of 21
𝑝̅(1 − 𝑝̅)
se(𝑝̅ ) = √
𝑛
Note that the formula used for the se(𝑝̅ ) differs from the one use in the above example where
π(1−π)
se(𝑝̅ ) = √
𝑛
. The reason we cannot use this formula in the margin of error for the confidence interval is
that this formula requires the knowledge of π, the unknown parameter for which we are developing an
interval estimate.
Example 11
In a poll of 1,000 registered voters, 380 favored candidate A. Construct a 95 percent confidence interval for
the proportion of all registered voters favoring candidate A.
Here you must first compute the proportion of voters in the sample who favor candidate A. Assigning “1” to
those who favor the candidate, then the sample proportion is:
𝑝̅ =
∑𝑥
380
=
= 0.38
𝑛
1000
𝑝̅(1 − 𝑝̅)
0.38(1 − 0.38)
se(𝑝̅ ) = √
=√
= 0.0153
𝑛
1000
Since you are building a 95% confidence interval, the confidence coefficient is 1 − α = 0.95. Then α = 0.05 and
zα∕2 = z0.025 = 1.96 . The margin of error is
MOE = 1.96(0.0153) = 0.03
The boundaries of the 95% confidence interval are:
L = 0.38 − 0.03 = 0.35
U = 0.38 + 0.03 = 0.41
Note that this is a typical example of the poll results reported in the media. The news media report the result
as: The poll results show that 38% of potential voters prefer candidate A with a margin of error ±3
percentage points. The media, however, do not generally report that this is a 95% confidence interval. The
New York Times, as an exception, usually provides an explanation in a box, stating, for example, that: "In
theory, in 19 cases out of 20 the results based on such samples will differ by no more than three percentage
points in either direction from what would have been obtained by seeking out all American adults."
The term "in 19 cases out of 20" is what we understand as a 95% confidence level (19 ∕ 20 = 0.95). That is, 19
of every 20 (95% of ) intervals built around all possible 𝑝̅ values would contain the actual population
proportion π.
Example 12
A medical researcher working for one of the pharmaceutical firms developing anti-ulcer drugs needs to
estimate the percentage of ulcers that are caused by bacteria. In a random sample of 860 ulcer patients 758
had ulcers caused by bacteria. Build a 95% confidence interval for the percentage of ulcers caused by
bacteria.
𝑝̅ =
758
= 0.881
860
Chapter 5—Interval Estimates
Page 18 of 21
0.881(1 − 0.881)
se(𝑝̅ ) = √
= 0.0110
860
MOE = z0.025 se(p̅ ) = 1.96 × 0.0110 = 0.022
L = 0.881 − 0.022 = 0.859
U = 0.881 + 0.022 = 0.903
4.1.
Minimum Sample Size Needed to Estimate π to Within a Desired Margin of
Error
Like the confidence interval for μ, the precision of the confidence interval for π is determined by the margin of
error. For a given confidence level, the confidence interval for π is then
L, U = p̅ ± MOE
where,
𝑝̅(1 − 𝑝̅)
MOE = 𝑧0.025 √
𝑛
Solving for n, we have
𝑛=(
𝑧α⁄2 2
) 𝑝̅(1 − 𝑝̅ )
MOE
This formula would allow us to find the required sample size for a desired margin of error. However, there is
one problem. To find n, you need a value for the sample proportion, p̅ = ∑x/n. This requires a prior
knowledge of n. Thus, we are dealing with a circular formula. To avoid this problem the formula for n is
changed to the following:
𝑛=(
𝑧α⁄2 2
) 𝜋̂(1 − 𝜋̂)
MOE
Here 𝑝̅ is replaced by π̂ , which is the planning value for 𝑝̅. The planning value may be obtained from the past
surveys. If no previous surveys are available, use 𝝅
̂ = 𝟎. 𝟓𝟎 for the planning value. For a given confidence
level, this will result in the largest sample for the desired MOE.
Example 13
What is the minimum sample size to estimate the population of registered voters who favor candidate A to
within ±3 percentage points with a 95 percent level of confidence. Assume a planning value of 0.50 for
population proportion.
MOE = 0.03
zα ∕ 2 = z0.025 = 1.96
π̂ = 0.50
1.96 2
𝑛=(
) (0.5)(1 − 0.5) = 1067.11
0.03
Rounded up, n = 1,068.
Chapter 5—Interval Estimates
Page 19 of 21
Summary of Formulas
Confidence Interval for μ
Confidence Interval for π
L, U = x̅ ± MOE
L, U = p̅ ± MOE
a) When n ≥ 100
b) When n < 100
MOE = zα ∕2 se(x̅ )
MOE = tα ∕2,(n − 1) se(x̅ )
se(𝑥̅ ) =
𝑠=√
𝑠
√𝑛
𝑝̅(1 − 𝑝̅)
se(𝑝̅ ) = √
𝑛
∑(𝑥 − 𝑥̅ )2
𝑛−1
Minimum Sample Size for a Given MOE
𝑧α⁄2 σ 2
𝑛=(
)
MOE
Chapter 5—Interval Estimates
p̅ = ∑x ∕ n
MOE = zα ∕2 se(p̅ )
Minimum Sample Size for a Given MOE
𝑛=(
𝑧α⁄2 2
) π(1 − π)
MOE
Page 20 of 21
df
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
Chapter 5—Interval Estimates
0.10
3.078
1.886
1.638
1.533
1.476
1.440
1.415
1.397
1.383
1.372
1.363
1.356
1.350
1.345
1.341
1.337
1.333
1.330
1.328
1.325
1.323
1.321
1.319
1.318
1.316
1.315
1.314
1.313
1.311
1.310
1.309
1.309
1.308
1.307
1.306
1.306
1.305
1.304
1.304
1.303
1.303
1.302
1.302
1.301
1.301
1.300
1.300
1.299
1.299
1.299
0.05
6.314
2.920
2.353
2.132
2.015
1.943
1.895
1.860
1.833
1.812
1.796
1.782
1.771
1.761
1.753
1.746
1.740
1.734
1.729
1.725
1.721
1.717
1.714
1.711
1.708
1.706
1.703
1.701
1.699
1.697
1.696
1.694
1.692
1.691
1.690
1.688
1.687
1.686
1.685
1.684
1.683
1.682
1.681
1.680
1.679
1.679
1.678
1.677
1.677
1.676
Right Tail Area
0.025
0.01
12.706 31.821
4.303
6.965
3.182
4.541
2.776
3.747
2.571
3.365
2.447
3.143
2.365
2.998
2.306
2.896
2.262
2.821
2.228
2.764
2.201
2.718
2.179
2.681
2.160
2.650
2.145
2.624
2.131
2.602
2.120
2.583
2.110
2.567
2.101
2.552
2.093
2.539
2.086
2.528
2.080
2.518
2.074
2.508
2.069
2.500
2.064
2.492
2.060
2.485
2.056
2.479
2.052
2.473
2.048
2.467
2.045
2.462
2.042
2.457
2.040
2.453
2.037
2.449
2.035
2.445
2.032
2.441
2.030
2.438
2.028
2.434
2.026
2.431
2.024
2.429
2.023
2.426
2.021
2.423
2.020
2.421
2.018
2.418
2.017
2.416
2.015
2.414
2.014
2.412
2.013
2.410
2.012
2.408
2.011
2.407
2.010
2.405
2.009
2.403
0.005
63.657
9.925
5.841
4.604
4.032
3.707
3.499
3.355
3.250
3.169
3.106
3.055
3.012
2.977
2.947
2.921
2.898
2.878
2.861
2.845
2.831
2.819
2.807
2.797
2.787
2.779
2.771
2.763
2.756
2.750
2.744
2.738
2.733
2.728
2.724
2.719
2.715
2.712
2.708
2.704
2.701
2.698
2.695
2.692
2.690
2.687
2.685
2.682
2.680
2.678
df
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
0.10
1.298
1.298
1.298
1.297
1.297
1.297
1.297
1.296
1.296
1.296
1.296
1.295
1.295
1.295
1.295
1.295
1.294
1.294
1.294
1.294
1.294
1.293
1.293
1.293
1.293
1.293
1.293
1.292
1.292
1.292
1.292
1.292
1.292
1.292
1.292
1.291
1.291
1.291
1.291
1.291
1.291
1.291
1.291
1.291
1.291
1.290
1.290
1.290
1.290
1.290
Right Tail Area
0.05
0.025
0.01
1.675 2.008 2.402
1.675 2.007 2.400
1.674 2.006 2.399
1.674 2.005 2.397
1.673 2.004 2.396
1.673 2.003 2.395
1.672 2.002 2.394
1.672 2.002 2.392
1.671 2.001 2.391
1.671 2.000 2.390
1.670 2.000 2.389
1.670 1.999 2.388
1.669 1.998 2.387
1.669 1.998 2.386
1.669 1.997 2.385
1.668 1.997 2.384
1.668 1.996 2.383
1.668 1.995 2.382
1.667 1.995 2.382
1.667 1.994 2.381
1.667 1.994 2.380
1.666 1.993 2.379
1.666 1.993 2.379
1.666 1.993 2.378
1.665 1.992 2.377
1.665 1.992 2.376
1.665 1.991 2.376
1.665 1.991 2.375
1.664 1.990 2.374
1.664 1.990 2.374
1.664 1.990 2.373
1.664 1.989 2.373
1.663 1.989 2.372
1.663 1.989 2.372
1.663 1.988 2.371
1.663 1.988 2.370
1.663 1.988 2.370
1.662 1.987 2.369
1.662 1.987 2.369
1.662 1.987 2.368
1.662 1.986 2.368
1.662 1.986 2.368
1.661 1.986 2.367
1.661 1.986 2.367
1.661 1.985 2.366
1.661 1.985 2.366
1.661 1.985 2.365
1.661 1.984 2.365
1.660 1.984 2.365
1.660 1.984 2.364
0.005
2.676
2.674
2.672
2.670
2.668
2.667
2.665
2.663
2.662
2.660
2.659
2.657
2.656
2.655
2.654
2.652
2.651
2.650
2.649
2.648
2.647
2.646
2.645
2.644
2.643
2.642
2.641
2.640
2.640
2.639
2.638
2.637
2.636
2.636
2.635
2.634
2.634
2.633
2.632
2.632
2.631
2.630
2.630
2.629
2.629
2.628
2.627
2.627
2.626
2.626
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