Chapter 7: Inference for A Single Sample
Inferences About the Mean and Variance of a Normal Population
• Monitor the mean of a manufacturing process to determine if the
process is under control.
• Evaluate the precision of a laboratory instrument measured by
the variance of its readings.
• By using the central limit theorem (CLT), inference procedures
for the mean of a normal population can be extended to the mean
of a non-normal population when a large enough sample is
available.
• Inference procedures for the variance using the Chi-square
distribution depend upon the assumption of a normal population.
• Prediction intervals and tolerance intervals are methods for
estimating future observations from a population.
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Inferences on Mean (Large Samples)
Inferences on µ will be based on the sample mean X ,
which is an unbiased estimator of µ with variance σ2/n.
For a large enough sample size n, the CLT tells us that
is approximately N(µ, σ2/n ) distributed, even if the
population is not normal.
X
Also for large n, the sample variance S2 may be taken as
an accurate estimator of σ2 with negligible sampling error.
If n ≥ 30, substitute estimated standard deviation s for σ in
the formulas which follow.
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Confidence Intervals on the Mean
Assume that σ2 is known or estimated from a large sample.


X −µ
P  − za / 2 ≤ Z =
≤ za / 2  = 1 − α
σ n


x − za / 2
Similarly,
x − za
σ
n
≤ µ ≤ x + za / 2
σ
n
[See Tables A.3 or A.4 (T∞)]
100(1-α)% two-sided
confidence interval
(0 ≤ ) µ ≤ x + za σn
Upper one-sided 100(1-α)%
confidence interval (limit)
σ
Lower one-sided 100(1-α)%
confidence interval (limit)
n
≤ µ (≤ ∞)
Note: zα in one-sided confidence intervals, not zα/2
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Sample Size Determination for a C.I.
Suppose that we require a (1-α)-level two-sided C.I. for µ of
the form ( x − E , x + E ) , with specified width 2E.
Set E = zα / 2
σ
n
2
 zα / 2 σ 
and solve for n, obtaining n = 
.
 E 
n increases as confidence (1-α) and σ increases or E decreases.
Calculation is done at the design stage so a sample estimate
of σ is usually not available.
An estimate for σ can be obtained by anticipating the range of
the observations and dividing by 4.
Based on assuming normality, such that 95% of the
observations are expected to fall in ( µ − 2σ , µ + 2σ ) .
• See Example 7.1, p. 239, Airline Revenue
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Hypothesis Test on a Mean
Example 7.2, Test H0 : µ = 42,000 vs H1 : µ ≠ 42,000 at α=10%
Can perform test in three ways:
1. Calculate z-statistics (z-score):
x − µ 42196 − 42000
z=
=
= 2.217
500 / 32
σ/ n
Reject H0 if |z| > z.05 = 1.645
2. Calculate the P-value:
P-value=P{ |Z| ≥ 2.217 } = 2[1-Φ(2.217) ] = 2[1-0.987] = 0.026
Reject if P-value is less than 0.1.
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Hypothesis Test on a Mean (continued)
3. Calculate 90% C.I. for µ:
500
x ± zα / 2 σn = 42196 ± 1.645
= 42196 ± 341 = ( 42050, 42340)
32
Reject H0 if 42000 does not fall in the C.I.
All three methods are equivalent and indicate rejection of H0,
that is, it is not likely at the 10% level that the mean is 42000.
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Power Calculation for One-sided Test
π(µ) = P{ Test rejects H0 | µ }
Consider the problem of testing H0: µ ≤ µ0 vs. H1: µ > µ0 .
• Derivation in Eqn 7.7, p. 242
• Figures 7.1 and 7.2, p. 243
Properties of power curve for one-sided test:
1. π(µ) > π(µ0) = α for µ > µ0
2. π(µ) ≤ α for all µ ≤ µ0 (justification for composite vs. simple H0)
3. As µ increases, power increases (the greater ∆=µ-µ0 , the
greater probability of detecting the difference)
4. For fixed ∆=µ-µ0, σ, and α, power increases with increasing n
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Power Calculation for One-sided Test
Example 7.3, p. 244, SAT Coaching
- see calculations for power against H1 : µ = 30 (∆=30-15=15)
Also, can use JMP (Sample Size and Power Facility):(no data file can be open)
Index tab, scroll to Power, run script, one sample mean, and then
enter α (double for one-sided test), Std. Dev., δ, sample size
“Continue” for power calculation
Or use α =0.05, δ = -15,
Or leave out δ for power curve
“Animation Script” and
then “High Side”
Double of 5%
since is a
one-sided test
Approx.
50%, as
in book
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Power Calculation for Two-sided Test
π(µ) = P{ Test rejects H0 | µ }
Consider the problem of testing H0: µ = µ0 vs. H1: µ ≠ µ0 .
• Derivation in Eqn 7.9, pp. 244-245
• Figures 7.3 and 7.4, pp. 245-246
Properties of power curve for two-sided test:
1. π(µ) is symmetrical around µ0
2. π(µ) increases as |µ - µ0| increases, with the lowest value at
π(µ0) = α
3. For fixed ∆=µ-µ0, σ, and α, power increases with increasing n
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Power Calculation for Two-sided Test
Example 7.4, p. 246, cereal box weights
- see calculations for power
Also, can use JMP (Sample Size and Power Facility):(no data file can be open)
Index tab, scroll to Power, run script, one sample mean, and then
enter α (double for one-sided test), Std. Dev., δ, sample size
“Continue” for power calculation OR leave out δ for power curve
Not the same
as .223 in book
Difference probably due to using T-statistic vs. Z-statistic, that is, the
standard deviation is estimated vs. being known, Section 7.2.2, pp.253-254.
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Sample Sizing for One-sided Z-Test
Determine the sample size so that a study will have sufficient
power to detect an effect of practical importance magnitude.
If the goal of the study is to show that the mean response µ
under a treatment is higher than the mean response µ0 without
the treatment, then µ - µ0 is called the treatment effect.
Let δ > 0 denote a practically important treatment effect and let
1 - β denote the minimum power required to detect it.
The goal is to find the minimum sample size n which would
guarantee that an α-level test of H0 has at least 1 - β power to
reject H0 when the treatment effect is at least δ.
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Sample Sizing for One-sided Z-Test (cont)
Because Power is an increasing function of µ - µ0 , it is only
necessary to find n that makes the power 1 - β at µ = µ0 + δ.

π ( µ0 + δ ) = Φ  − za +

δ n
 = 1− β
σ 
Using the fact that Φ(zβ) = 1 - β gives the relation
− za +
δ n
= zβ
σ
Solving for n, we obtain
2
 ( z + zβ )σ 
n= α

δ


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Sample Size Calculation for One-sided Test
Example 7.5, p. 248, SAT Coaching
- see calculations
Also, can use JMP (Sample Size and Power Facility):(no data file can be open)
Index tab, scroll to Power, run script, one sample mean, and then
enter α (double for one-sided test), Std. Dev., δ, power
“Continue” for sample size calculation OR leave out δ for sample
size curve
Approx. 60
samples
Similar to
60.92 in book
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Sample Size Calculation for Two-sided Test
Example 7.6, p. 249, cereal box weights
- see calculations
2
 ( zα / 2 + z β )σ 
n=

δ


Also, can use JMP (Sample Size and Power Facility):(no data file can be open)
Index tab, scroll to Power, run script, one sample mean, and then
enter α , Std. Dev., δ, power
“Continue” for sample size calculation OR leave out δ for sample
size curve
Approx. 20
samples
Not the same as
13.51 in book
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Inferences on Mean (Small Samples)
The sampling variability of σ2 may be sizable if the sample is
small (less than 30). Inference methods must take this variability
into account when σ2 is unknown.
Assume that X1, ..., Xn is a random sample from a N(µ, σ2)
distribution. Then
T=
X −µ
S/ n
has a T-distribution with n -1 degrees of freedom (d.f.).
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Confidence Intervals on Mean (Small Samples)


X −µ
≤ tn −1,α / 2 
1 − α = P − tn −1,α / 2 ≤ T =
S/ n


S


= P  X − tn −1,α / 2
≤ µ ≤ X + tn −1,α / 2 
n


x − tn −1,α / 2
s
n
≤ µ ≤ x + tn −1,α / 2
s
n
Two - sided 100(1- α )% C.I.
tn-1,α/2 > zα/2 ⇒ t-interval is wider than z-interval
• See Example 7.7, p. 250, density of earth
µ ≤ x + tn −1,α
s
µ ≥ x − tn −1,α
s
n
n
Upper one - sided 100(1- α )% C.L.
Lower one - sided 100(1- α )% C.L.
• See Example 7.8, pp. 251-252, tear strength
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T-test for the Mean (Small Samples)
t=
x − µ0
~ T with n-1 d.f.
s n
where x is the observed sample mean,
s is the observed sample standard deviation,
and µ0 is the postulated value of µ under H0 .
Tests and confidence intervals summarized in Table 7.5, p.253
• Example 7.9, pp. 252-253, tear strength
Power and Sample size calculations for T-tests
– these are probably what is calculated in JMP
– if the standard deviation is very well known (from a previous
study and can reasonably be assumed not to have changed),
then use the Z-based formulas
– T-based formulas will be more conservative (less power,
higher sample sizes)
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JMP T-test on Mean and P-values
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JMP T-test on Mean and P-values (continued)
Can also do a
Z-test by
specifying σ
Can also do a
non-parametric
test
Reject H0 at
any α level
bigger than
0.0027
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JMP T-test on Mean and Power Animation
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Inferences on Variance
Assume that X1, …, Xn is a random sample from a N(µ, σ2) distribution.
χ2 =
( n − 1) S 2
σ2
~ Chi-squared with n-1 d.f.
Necessary
assumption
• See Figure 7.8 for the distribution.
There are confidence intervals (two-sided and one-sided) on the
variance and the standard deviation.
- Eqns. 7.17, 7.18, 7.19, 7.20
- Example 7.10, p.256, thermometer
There are hypothesis tests on the variance (and standard deviation).
- Eqn. 7.21, p. 256
- Example 7.11, p. 256, thermometer
Tests and confidence intervals summarized in Table 7.5, p. 253
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JMP Chi-Square test on Std.Dev. and P-values
Probably would not reject either two-sided
or one-sided tests that σ = 1.
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Prediction Intervals
Many practical applications call for an interval estimate of an individual
(future) observation sampled from a population rather than of the mean
of the population.
An interval estimate for an individual observation is called a prediction
interval.
Prediction Interval Formula for a Single future observation:
x − tn −1,α / 2 s 1 +
1
n
≤ µ ≤ x + tn −1,α / 2 s 1 + n1
As n→∞, the interval converges to [ µ - zα/2 σ , µ + zα/2 σ ] .
Confidence Interval Formula for the Mean:
x − tn −1,α / 2 s
1
n
≤ µ ≤ x + tn −1,α / 2 s
1
n
As n→∞, the interval converges to a single point µ
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Prediction Intervals
Example 7.12, pp. 259-260, tear strength
First, look at a run chart to see if the process is under control
- should probably use a control chart
95% Prediction Interval for a Single future batch:
x − t14−1,.025 s 1 + 141 ≤ µ ≤ x + t14−1,.025 s 1 + 141
33.712 ± 2.160 × 0.798 1 + 141 = (31.93, 35.48)
95% Confidence Interval for the Mean:
x − t14 −1,.025 s
1
14
≤ µ ≤ x + t14−1,.025 s
33.712 ± 2.160 × 0.798
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14
1
14
= (33.25, 34.17)
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Smaller
interval
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Prediction Intervals using JMP
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Prediction Intervals using JMP (continued)
Called “Individual” Confidence Interval instead of “Prediction”
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Tolerance Intervals
Suppose we want an interval which will contain at least .90
= 1-γ of the strengths of the future batches (observations)
with 95% = 1 - α confidence.
[ x − Ks, x + Ks ]
Get the K factor from Table A.12, p. 685
33.712 ± 2.529 × 0.798 = (31.69, 35.73)
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Homework #5
Read Chapter 7
Exercises: (answers to odd problems in the back of the book)
(show work – no credit if calculations or printout not shown)
7.2 (a) 68 (b) (49.5, 50.4)
7.5, 7.7, 7.11, 7.13, 7.15, 7.17
7.20 (a) (1233.4, 1266.6)
(b) (1014.2, 1485.8)
(c) (1013.2, 1468.3) (note: use K for n=50)
Read or skim Chapter 8
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