Math/Stat2330 Elementary Statistics and Probability Lecture 14: Sample Variability Gejun Zhu

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Math/Stat2330 Elementary Statistics and Probability
Lecture 14: Sample Variability
Gejun Zhu
Department of Mathematics
University of Texas - Pan American
Edinburg, Texas 78539
[email protected]
June 24th, 2013
Last Class Review
What did we learn in last class?
Applications of Normal Distribution
Notations z(α)
Normal Approximation of Binomial Distribution
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Example 1: Given that x is normally distributed random variable with a mean of
60 and a standard deviation of 10, find the following probabilities:
a. P(x > 60)
d. P(65 < x < 82)
b. P(60 < x < 72)
e. P(38 < x < 78)
c. P(57 < x < 78)
f. P(x < 38)
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Example 1: Given that x is normally distributed random variable with a mean of
60 and a standard deviation of 10, find the following probabilities:
a. P(x > 60)
d. P(65 < x < 82)
b. P(60 < x < 72)
e. P(38 < x < 78)
c. P(57 < x < 78)
f. P(x < 38)
Solution:
a. 0.5000
d. 0.2946
b. 0.3849
e. 0.9502
c. 0.6072
f. 0.0139
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Notations
Example 2: z(0.05) (read “z of 0.05”) is the algebraic name for the z such that
the area to the right and under the standard normal curve is exactly 0.05, as
shown below:
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Notations
Example 2: z(0.05) (read “z of 0.05”) is the algebraic name for the z such that
the area to the right and under the standard normal curve is exactly 0.05, as
shown below:
Using z-table,
z(0.05) = 1.65
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Rule
The normal distribution provides a reasonable approximation to a binomial
probability distribution whenever the values of np and n(1 − p) both equal or
exceed 5.
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Rule
exceed 5.
Example 3: Find the normal approximation for the binomial probability P(x ≤ 8),
where n = 14 and p = 0.4.
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Rule
exceed 5.
Example 3: Find the normal approximation for the binomial probability P(x ≤ 8),
Using z-table,
np = 14 × 0.4 = 5.6, n(1 − p) = 14 × 0.6 = 8.4
µ=
√
√np = 14 × 0.4 = 5.6
√
σ = npq = 14 × 0.4 × 0.6 = 3.36 = 1.83
8.5 − 5.6
P(x ≤ 8) = P(z ≤
) = 1.58
1.83
P(x ≤ 8) = 0.9430
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Application of Normal Distribution
Example 4: Find the normal approximation for the binomial probability P(x ≥ 8),
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Application of Normal Distribution
Example 4: Find the normal approximation for the binomial probability P(x ≥ 8),
np = 14 × 0.4 = 5.6, n(1 − p) = 14 × 0.6 = 8.4
µ = np = 14 × 0.4 = 5.6
√
√
√
σ = npq = 14 × 0.4 × 0.6 = 3.36 = 1.83
7.5 − 5.6
7.5 − 5.6
P(x ≥ 8) = P(z ≥
) = 1 − P(z ≤
) = 1 − P(z < 1.04) = 0.1492
1.83
1.83
Using z-table,
P(x ≥ 8) = 0.1492
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Sample Variability
What will we cover in this chapter?
We are going to use the sample mean x¯ to estimate the population µ.
Actually, we don’t expect that the value of sample mean x¯ is exactly equal to the
value of the population mean µ, but we will satisfied with our sample results if the
sample mean is “close” to the value of the population mean.
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Sample Variability
What is “close”? How to determine the closeness?
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Sample Variability
What is “close”? How to determine the closeness?
Sampling distribution of a sample statistic
The distribution of values for a sample statistic obtained from repeated samples,
all of the same size and all drawn from the same population.
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Sampling Distribution of the Sample Mean
Example: Heights of Starting Players Suppose that the population of interest
consists of five starting players on a men’s basketball team, who we will call A, B,
C, D, and E. Further suppose that the variable of interest is height, in inches.
Here is the data of the players.
Player
Height
A
76
B
78
C
79
D
81
E
86
a. Obtain the sampling distribution of the sample mean from samples of size 2.
b. Make some observations about sampling error when the mean height of a
random sample of two players is used to estimate the population mean.
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Solution:
a. Obtain the sampling distribution of the sample mean from samples of size 2.
Sample
A, B
A, C
A, D
A, E
B, C
B, D
B, E
C, D
C, E
D, E
Height
76, 78
76, 79
76, 81
76, 86
78, 79
78, 81
78, 86
79, 81
79, 86
81, 86
x¯
77.0
77.5
78.5
81.0
78.5
79.5
82.0
80.0
82.5
83.5
The population is so small that we can list all the possible samples of size 2. The
first column of the above table gives the 10 possible samples, the second column
the corresponding heights and the third column the sample means.
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P
76 + 78 + 79 + 81 + 86
xi
Population Mean: µ =
=
= 80 inches.
n
5
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P
76 + 78 + 79 + 81 + 86
xi
=
= 80 inches.
n
5
From the table before, we can see that mean height of the two players selected
isn’t likely to the population mean of 80 inches. In fact, only 1 of 10 samples has
1
, or 10%, that x¯ will
a mean of 80 inches. The chances are, therefore, only
10
equal µ.
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P
76 + 78 + 79 + 81 + 86
xi
=
= 80 inches.
n
5
From the table before, we can see that mean height of the two players selected
isn’t likely to the population mean of 80 inches. In fact, only 1 of 10 samples has
1
, or 10%, that x¯ will
a mean of 80 inches. The chances are, therefore, only
10
equal µ.
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c. Find the probability that, for a random sample of size 4, the sampling error
made in estimating the population mean by the same mean will be 1 inch or
less; that is determine the probability that x¯ will be within 1 inch of µ.
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Solution:
Sample
Height
x¯
A, B, C, D 76, 78, 79, 81 78.50
A, B, C, E 76, 78, 79, 86 79.75
A, B, D, E 76, 78, 81, 86 80.25
A, C, D, E 76, 79, 81, 86 80.50
B, C, D, E 78, 79, 81, 86 81.00
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Solution:
Sample
Height
x¯
A, B, C, D 76, 78, 79, 81 78.50
A, B, C, E 76, 78, 79, 86 79.75
A, B, D, E 76, 78, 81, 86 80.25
A, C, D, E 76, 79, 81, 86 80.50
B, C, D, E 78, 79, 81, 86 81.00
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Solution:
Sample
Height
x¯
A, B, C, D 76, 78, 79, 81 78.50
A, B, C, E 76, 78, 79, 86 79.75
A, B, D, E 76, 78, 81, 86 80.25
A, C, D, E 76, 79, 81, 86 80.50
B, C, D, E 78, 79, 81, 86 81.00
4
The probability is , or
5
0.8, that the sampling
error made in estimating
µ by x¯ will be 1 inch or
less.
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The Sampling Distribution of Sample Means
Sampling Distribution of sample means (SDSM)
If all possible random samples, each of size n, are taken from any population with
mean µ and standard deviation σ, then the sampling distribution of sample means
will have the following:
1. A mean µx¯ equal to µ
σ
2. A standard deviation σx¯ equal to √
n
Furthermore, if the sampled population has a normal distribution, then the
sampling distribution of x¯ will also be normal for samples of all sizes.
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Sampling Distribution of sample means (SDSM)
If all possible random samples, each of size n, are taken from any population with
mean µ and standard deviation σ, then the sampling distribution of sample means
will have the following:
1. A mean µx¯ equal to µ
σ
2. A standard deviation σx¯ equal to √
n
Furthermore, if the sampled population has a normal distribution, then the
sampling distribution of x¯ will also be normal for samples of all sizes.
Central Limit Theorem (CLT)
The sampling distribution of sample means will more closely resemble the normal
distribution as the sample size increases.
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If the sampled distribution is normal, then the sampling distribution of sample
means is normal, as stated previously, and the central limit theorem is not needed.
But if the sampled population is not normal, the CLT tells us that the sampling
distribution will still be approximately normally distributed under the right
conditions.
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If the sampled distribution is normal, then the sampling distribution of sample
means is normal, as stated previously, and the central limit theorem is not needed.
But if the sampled population is not normal, the CLT tells us that the sampling
distribution will still be approximately normally distributed under the right
conditions.
By combining the preceding information, we can describe the sampling
distribution of x¯ completely:
(1) the location of the center (mean),
(2) a measure of spread indicating how widely the distribution is dispersed
(standard error of the mean), and
(3) an indication of how it is distributed.
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Example: Let’s consider all possible samples of size 2 that could be drawn from a
population that contains the three numbers 2, 4, 6. First let’s take a look at the
population itself. Construct a histogram to picture its distribution, calculate the
mean µ and standard deviation, σ.
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x
2
4
6
P
P(x)
1/3
1/3
1/3
1
xP(x)
2/3
4/3
6/3
12/3 = 4
x 2 P(x)
4/3
16/3
36/3
56/3 = 18.66
µ = 4.0
σ=
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p
√
18.66 − (4.0)2 = 2.66 = 1.63
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Here lists all the possible samples and the means of size 2 that can be drawn from
this population. (One number is drawn, observed, and then returned to the
population before the second number is drawn.)
Sample
2, 2
2, 4
2, 6
4, 2
4, 4
4, 6
6, 2
6, 4
6, 6
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x¯
2
3
4
3
4
5
4
5
6
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Here lists all the possible samples and the means of size 2 that can be drawn from
this population. (One number is drawn, observed, and then returned to the
population before the second number is drawn.)
Sample
2, 2
2, 4
2, 6
4, 2
4, 4
4, 6
6, 2
6, 4
6, 6
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x¯
2
3
4
3
4
5
4
5
6
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x
2
3
4
5
6
P
P(x)
1/9
2/9
3/9
2/9
1/9
1.0
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xP(x)
2/9
6/9
12/9
10/9
6/9
4.0
x 2 P(x)
4/9
18/9
48/9
50/9
36/9
17.33
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x
2
3
4
5
6
P
P(x)
1/9
2/9
3/9
2/9
1/9
1.0
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xP(x)
2/9
6/9
12/9
10/9
6/9
4.0
x 2 P(x)
4/9
18/9
48/9
50/9
36/9
17.33
σx¯ =
p µx¯ = 4.0
17.33 − (4.0)2 = 1.15
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P(x) xP(x) x 2 P(x)
1/9
2/9
4/9
p µx¯ = 4.0
2/9
6/9
18/9
σx¯ = 17.33 − (4.0)2 = 1.15
3/9
12/9
48/9
2/9
10/9
50/9
1/9
6/9
36/9
1.0
4.0
17.33
Let’s now check the truth of the three facts about the sampling distribution of
sampling means:
1. The mean µx¯ of the sampling distribution will equal to the mean µ of the
population: both µ and µx¯ have the value 4.0.
2. The standard error of the mean σx¯ for the sampling distribution will equal the
standard deviation σ of the population divided by the square root of the
σ
1.63
sample size, n: σx¯ = 1.15 and σ = 1.63, n = 2, √ = √ = 1.15; they are
n
2
equal.
3. The distribution will become approximately normally distributed: the
histogram very strongly suggests normality.
x
2
3
4
5
6
P
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Application of the Sampling Distribution of Sample Means
Example: Consider a normal distribution with µ = 100 and σ = 20. If random
sample size of 16 is selected, what is the probability that this sample will have a
mean value between 90 and 110? That is, what is P(90 < x¯ < 100)?
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Solution:
Since the population is normally distributed, the sampling distribution of x¯0 s is
normally distributed.
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Solution:
σ
20
µx¯ = µ = 100, σx¯ = √ = √ = 5
n
16
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Solution:
σ
20
µx¯ = µ = 100, σx¯ = √ = √ = 5
n
16
x¯ − µx¯
Using the standard transformation z =
:
σx¯
90 − 100
= −2.00
5
110 − 100
z-score for x¯ = 110: z =
= 2.00
5
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Solution:
σ
20
µx¯ = µ = 100, σx¯ = √ = √ = 5
n
16
x¯ − µx¯
Using the standard transformation z =
:
σx¯
90 − 100
= −2.00
5
110 − 100
= 2.00
5
Therefore, P(90 < x¯ < 110) = P(−2.00 < z < 2.00) = 0.9773 − 0.0228 = 0.9545
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Example: Kindergarten children have heights that are approximately normally
distributed about a mean of 39 inches and a standard deviation of 2 inches. A
random sample of size 25 is taken, and the mean x¯ is calculated. What is the
probability that this mean value will between 38.5 and 40.0 inches?
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Solution: What we are trying to find is P(38.5 < x¯ < 40.0). The values x¯, 38.5,
x¯ − µ
√ :
40.0 must be converted to z-scores using z =
σ/ n
x¯ = 38.5: z =
38.5 − 39.0
−0.5
x¯ − µ
√ =
√
=
= −1.25
0.4
σ/ n
2/ 25
x¯ = 40.0: z =
G. Zhu (UTPA)
x¯ − µ
40.0 − 39.0
1
√ =
√
=
= 2.50
0.4
σ/ n
2/ 25
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Solution: What we are trying to find is P(38.5 < x¯ < 40.0). The values x¯, 38.5,
x¯ − µ
√ :
40.0 must be converted to z-scores using z =
σ/ n
x¯ = 38.5: z =
38.5 − 39.0
−0.5
x¯ − µ
√ =
√
=
= −1.25
0.4
σ/ n
2/ 25
x¯ = 40.0: z =
x¯ − µ
40.0 − 39.0
1
√ =
√
=
= 2.50
0.4
σ/ n
2/ 25
Therefore,
P(38.5 < x¯ < 40.0) = P(−1.25 < z < 2.50) = 0.9938 − 0.1057 = 0.8881
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Example: Use the heights of kindergarten children given previously. Within what
limits the middle 90% of the sampling distribution of sample means for samples of
size 100 fall?
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size 100 fall?
Solution:
Using z-table, we find that the middle 90% is bounded by z = ±1.65.
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size 100 fall?
Solution:
x¯ − µ
√ :
use the formula z =
σ/ n
Then, we
x¯ − 39.0
√
2/ 100
x¯ − 39 = (−1.65)(0.2)
x¯ = 39 − 0.33 = 38.67
z = −1.65: −1.65 =
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size 100 fall?
Solution:
x¯ − µ
√ :
use the formula z =
σ/ n
x¯ − 39.0
√
2/ 100
x¯ − 39 = (−1.65)(0.2)
x¯ = 39 − 0.33 = 38.67
z = −1.65: −1.65 =
G. Zhu (UTPA)
Then, we
x¯ − 39.0
√
2/ 100
x¯ − 39 = (1.65)(0.2)
x¯ = 39 + 0.33 = 39.33
z = 1.65: 1.65 =
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size 100 fall?
Solution:
x¯ − µ
√ :
use the formula z =
σ/ n
x¯ − 39.0
√
2/ 100
x¯ − 39 = (−1.65)(0.2)
x¯ = 39 − 0.33 = 38.67
z = −1.65: −1.65 =
Then, we
x¯ − 39.0
√
2/ 100
x¯ − 39 = (1.65)(0.2)
x¯ = 39 + 0.33 = 39.33
z = 1.65: 1.65 =
Thus, P(38.67 < x¯ < 39.33) = 0.90.
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size 100 fall?
Solution:
x¯ − µ
√ :
use the formula z =
σ/ n
x¯ − 39.0
√
2/ 100
x¯ − 39 = (−1.65)(0.2)
x¯ = 39 − 0.33 = 38.67
z = −1.65: −1.65 =
Then, we
x¯ − 39.0
√
2/ 100
x¯ − 39 = (1.65)(0.2)
x¯ = 39 + 0.33 = 39.33
z = 1.65: 1.65 =
Thus, P(38.67 < x¯ < 39.33) = 0.90.
Therefore, 38.67 inches and 39.33 inches are the limits that captures the middle
90% of the sample means.
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Math/Stat2330 Elementary Statistics and Probability Lecture 14: Sample Variability Gejun Zhu

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