Download S160 #18 - Sampling Distribution of the Mean

S160 #18
Sampling Distribution of the Mean
JC Wang
March 29, 2016
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
Outline
1
Sampling Distribution of the Mean
Sample Mean Versus Individual Values
Sampling From Normal Population
iClicker Questions
2
Estimating the Population Mean µ
Estimating the Population Mean µ
iClicker Questions
3
Confidence Interval for Population Mean
Confidence Interval for Population Mean
JC Wang (WMU)
S160 #18
S160, Lecture 18
2 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
Sample Mean
versus individual values
The variability in x’s (individual values) has to account for extremes in
the data but the variability in the possible sample means will be less
because any extreme value will be averaged with other values in the
sample. Therefore, as you increase the size of the sample, you have
more information; consequently, the sample mean is more accurate.
JC Wang (WMU)
S160 #18
S160, Lecture 18
3 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
Standard Error of the Mean
Standard Error of the sample mean is the variation in X :
SD
σX = SE(= SEX ) = √
n
where SD is the variability (the standard deviation) in the individual
values and n is the sample size.
Note: The sample mean X has expected value of µ ( the population
mean). That is, the average of the sample means of all size-n samples
is the population mean.
JC Wang (WMU)
S160 #18
S160, Lecture 18
4 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
Men’s Height Example
selecting one man
Suppose that male’s height is
approximately N(mean = 69, SD = 3).
Using empirical rule
0.68 ≈ P(66 ≤ X ≤ 72)
60
JC Wang (WMU)
S160 #18
65
70
height (in.)
n=1
75
S160, Lecture 18
5 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
Men’s Height Example
selecting nine men
Nine (n = 9) men
√ are randomly selected.
Now, SE = 3/ 9 = 1. Using empirical
rule,
0.997 ≈ P(66 ≤ X ≤ 72)
66
JC Wang (WMU)
S160 #18
67
68 69 70 71
average height (in.)
n=9
S160, Lecture 18
72
6 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
Men’s Height Example
discussion
Therefore, when considering sample sizes of 1 or 9, our probability
went from 68.27% to 99.73%.
JC Wang (WMU)
S160 #18
S160, Lecture 18
7 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
Distribution of the Sample Mean
when sampling from normal population
If a population has a normal shaped histogram with mean = µ, and
standard deviation, SD = σ, then n-member averages will have a
normal shaped
histogram with mean, µ and standard error of the
√
mean, σ/ n.
σ
If X1 , X2 , . . . , Xn are sampled from N(µ, σ) then X ∼ N(µ, √ )
n
JC Wang (WMU)
S160 #18
S160, Lecture 18
8 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
Men’s Height Example
(Example 1, page 88)
(a) n = 1; P(X > 71) =? z =
71−69
3
= 0.67
P(X > 71) = P(Z > 0.67) = 1−P(Z ≤ 0.67) = 1−.7486 = .2514.
(b) n = 1; P(X > 71) = .2514
(c) n = 9; P(X > 71) =? SE =
√3
9
= 1, z =
71−69
1
= 2.00
P(X > 71) = P(Z > 2.00) = 1−P(Z ≤ 2.00) = 1−.9772 = .0228.
(d) n = 9; P(X > 71) = .0228
JC Wang (WMU)
S160 #18
S160, Lecture 18
9 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
Men’s Height Example
(Example 1, page 88)
(a) n = 1; P(X > 71) =? z =
71−69
3
= 0.67
P(X > 71) = P(Z > 0.67) = 1−P(Z ≤ 0.67) = 1−.7486 = .2514.
(b) n = 1; P(X > 71) = .2514
(c) n = 9; P(X > 71) =? SE =
√3
9
= 1, z =
71−69
1
= 2.00
P(X > 71) = P(Z > 2.00) = 1−P(Z ≤ 2.00) = 1−.9772 = .0228.
(d) n = 9; P(X > 71) = .0228
JC Wang (WMU)
S160 #18
S160, Lecture 18
9 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
Men’s Height Example
(Example 1, page 88)
(a) n = 1; P(X > 71) =? z =
71−69
3
= 0.67
P(X > 71) = P(Z > 0.67) = 1−P(Z ≤ 0.67) = 1−.7486 = .2514.
(b) n = 1; P(X > 71) = .2514
(c) n = 9; P(X > 71) =? SE =
√3
9
= 1, z =
71−69
1
= 2.00
P(X > 71) = P(Z > 2.00) = 1−P(Z ≤ 2.00) = 1−.9772 = .0228.
(d) n = 9; P(X > 71) = .0228
JC Wang (WMU)
S160 #18
S160, Lecture 18
9 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
Men’s Height Example
(Example 1, page 88)
(a) n = 1; P(X > 71) =? z =
71−69
3
= 0.67
P(X > 71) = P(Z > 0.67) = 1−P(Z ≤ 0.67) = 1−.7486 = .2514.
(b) n = 1; P(X > 71) = .2514
(c) n = 9; P(X > 71) =? SE =
√3
9
= 1, z =
71−69
1
= 2.00
P(X > 71) = P(Z > 2.00) = 1−P(Z ≤ 2.00) = 1−.9772 = .0228.
(d) n = 9; P(X > 71) = .0228
JC Wang (WMU)
S160 #18
S160, Lecture 18
9 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
Example 1, page 88
continued
(e) n = 25; P(X > 71) =? SE =
√3
25
= 0.6, z =
71−69
0.6
= 3.33
P(X > 71) = P(Z > 3.33) = 1−P(Z ≤ 3.33) = 1−.9996 = .0004.
(f) n = 9; P(X > a) = 0.9, a =?
Note that P(Z ≥ −1.28)(= P[Z ≤ 1.28] = .8997) ≈ 0.9.
a = 69 + 1 × (−1.28) = 67.72.
(g) P(X > a) = 0.9 → a = 67.72.
JC Wang (WMU)
S160 #18
S160, Lecture 18
10 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
Example 1, page 88
continued
(e) n = 25; P(X > 71) =? SE =
√3
25
= 0.6, z =
71−69
0.6
= 3.33
P(X > 71) = P(Z > 3.33) = 1−P(Z ≤ 3.33) = 1−.9996 = .0004.
(f) n = 9; P(X > a) = 0.9, a =?
Note that P(Z ≥ −1.28)(= P[Z ≤ 1.28] = .8997) ≈ 0.9.
a = 69 + 1 × (−1.28) = 67.72.
(g) P(X > a) = 0.9 → a = 67.72.
JC Wang (WMU)
S160 #18
S160, Lecture 18
10 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
Example 1, page 88
continued
(e) n = 25; P(X > 71) =? SE =
√3
25
= 0.6, z =
71−69
0.6
= 3.33
P(X > 71) = P(Z > 3.33) = 1−P(Z ≤ 3.33) = 1−.9996 = .0004.
(f) n = 9; P(X > a) = 0.9, a =?
Note that P(Z ≥ −1.28)(= P[Z ≤ 1.28] = .8997) ≈ 0.9.
a = 69 + 1 × (−1.28) = 67.72.
(g) P(X > a) = 0.9 → a = 67.72.
JC Wang (WMU)
S160 #18
S160, Lecture 18
10 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
iClicker Question 18.1
It is suggested that the substrate concentration (mg/cm3 ) of influent to
a domestic-waste biofilm reactor is normally distributed with mean 0.30
and standard deviation of 0.06. A sample of 9 reactors is taken. What
is the chance that the mean substrate concentration of influent of
these reactors is in between 0.26 mg/cm3 and 0.34 mg/cm3 ?
A. 68%
B. 95%
C. 99.7%
D. 90%
E. none of the previous
JC Wang (WMU)
S160 #18
S160, Lecture 18
11 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
Outline
1
Sampling Distribution of the Mean
Sample Mean Versus Individual Values
Sampling From Normal Population
iClicker Questions
2
Estimating the Population Mean µ
Estimating the Population Mean µ
iClicker Questions
3
Confidence Interval for Population Mean
Confidence Interval for Population Mean
JC Wang (WMU)
S160 #18
S160, Lecture 18
12 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
Estimating the Population Mean µ
The population mean µ is estimated using the sample mean X . This
estimate tends to miss by an amount called
the standard error (SE) of
√
the mean. which is calculated as SD/ n. Note that SD is the sample
standard deviation
s
Sum of (Xi − X )2
SD =
n−1
JC Wang (WMU)
S160 #18
S160, Lecture 18
13 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
WMU Undergraduates’ Average GPA
Suppose a sample of n = 25 WMU students yielded an average GPA
of X = 3.05 and a standard deviation of 0.40. Then the WMU true
population
√ average GPA, µ, is estimated by 3.05 with a standard error
of 0.40/ 25 = 0.08.
JC Wang (WMU)
S160 #18
S160, Lecture 18
14 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
WMU Undergraduates’ Average Stay
Example 2 on Page 89
A sample of n = 25 graduating students were randomly selected and
asked about their length of stay. Suppose that the sample averaged
5.3 years, with an SD of 1.5 years. Then the true WMU average stay,
µ, is estimated
√ 5.3 years give or take
√ as X =
SE = SD/ n = 1.5/ 25 = 0.3 year.
JC Wang (WMU)
S160 #18
S160, Lecture 18
15 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
WMU Undergraduates’ Average Stay
continued, page 90
A second sample of 100 students were interviewed. The mean and SD
for the second sample were also 5.3 years and 1.5 years, respectively.
Then the true average
stayõ is estimated as X = 5.3 years give or
√
take SE = SD/ n = 1.5/ 100 = 0.15 year. Hence
Effect of Sample Size on Standard Error
The standard error of the mean decreases like the square root of the
sample size.
JC Wang (WMU)
S160 #18
S160, Lecture 18
16 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
WMU Undergraduates’ Average Stay
continued, page 90
A second sample of 100 students were interviewed. The mean and SD
for the second sample were also 5.3 years and 1.5 years, respectively.
Then the true average
stayõ is estimated as X = 5.3 years give or
√
take SE = SD/ n = 1.5/ 100 = 0.15 year. Hence
Effect of Sample Size on Standard Error
The standard error of the mean decreases like the square root of the
sample size.
JC Wang (WMU)
S160 #18
S160, Lecture 18
16 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
iClicker Question 18.2
The HDL (high-density lipoprotein, a.k.a., good cholesterol) among
adults is normally distributed. A sample of 25 adults was selected
which yielded a sample standard deviation of 5.3 mg/dL. A second
sample of 100 adults was selected which also yielded a sample
standard deviation of 5.3 mg/dL. Which of the following is true about
the SEs (standard errors of the mean) of the two samples? (SE1 = SE
of size-25 sample, SE2 = SE of size-100 sample)
A. SE1 = 4SE2
B. SE2 = 2SE1
C. SE1 = 2SE2
D. SE2 = 4SE1
E. none of the previous
JC Wang (WMU)
S160 #18
S160, Lecture 18
17 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
Outline
1
Sampling Distribution of the Mean
Sample Mean Versus Individual Values
Sampling From Normal Population
iClicker Questions
2
Estimating the Population Mean µ
Estimating the Population Mean µ
iClicker Questions
3
Confidence Interval for Population Mean
Confidence Interval for Population Mean
JC Wang (WMU)
S160 #18
S160, Lecture 18
18 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
Confidence Interval for µ
A 95% confidence interval for µ:
SD
X ± 1.96 √
n
√
That is, calculate the margin of error ME = 1.96SE = 1.96 × SD/ n
first. Then a 95% confidence interval is
(X − ME, X + ME)
JC Wang (WMU)
S160 #18
S160, Lecture 18
19 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
WMU Undergraduates’ Average GPA
continued
Recall that the sample of n = 25 students yielded a sample mean of
X = 3.05 with a standard error of SE = 0.08. The margin of error is
then ME = 1.96 × 0.08 = 0.157. Hence a 95% c.i. for the true average
GPA, µ, is:
(3.05 − 0.157, 3.05 + 0.157) = (2.893, 3.207).
Interpretation: based on the sample, we are 95% confident that the
true mean GPA is in between 2.9 and 3.2, approximately.
JC Wang (WMU)
S160 #18
S160, Lecture 18
20 / 21
Sampling Distribution of the Mean
Estimating the Population Mean µ
Confidence Interval for Population Mean
A Note About Confidence Interval
A 95% confidence
as the one for µ:
√ interval such √
(X − 1.96SD/ n, X + 1.96SD/ n) is like an extremely loaded coin in
which a head occurs 95% of the time. Flipping a head means that the
confidence interval contains the true mean µ. Once the coin is flipped,
a head or a tail is observed: the interval contains µ or not. If the same
coin is flipped a large number of times (i.e., size-n samples is repeated
a large number of times), then approximately 95% of times are heads
(i.e., approximately 95% of confidence intervals cover µ).
JC Wang (WMU)
S160 #18
S160, Lecture 18
21 / 21