(1) Confidence intervals are used to estimate population parameters based on sample data. They provide a range of plausible values for the parameter with a specified level of confidence, such as 95%. (2) The document discusses confidence intervals for several common statistics including the normal population mean when the variance is known or unknown, the proportion, differences and ratios of population means, and the normal population variance. Formulas and examples are provided for calculating each type of confidence interval. (3) Large samples allow approximating confidence intervals for any population mean using the normal distribution. Confidence intervals for proportions require minimum values for sample size and successes/failures.

1. INTRODUCTION TO CHAPTER 6
CONFIDENCE INTERVALS
 NOTION: (1 - α) CONFIDENCE INTERVAL
 CI FOR NORMAL POPULATION MEAN µ
 CI FOR ANY POPULATION MEAN :
LARGE SAMPLE
 CI FOR PROPORTION (OR PROBABILITY) p
 CI FOR NORMAL TWO POPULATION MEANS
 CI FOR NORMAL POPULATION VARIANCE σ 2
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2. NOTION - DEFINITION
CONFIDENCE INTERVAL

Definition
 Consider
 Definition:
The Interval
is called to be the (1 - α ) CI for a
parameter θ
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3. CI FOR NORMAL MEAN
WHEN VARIANCE IS KNOWN
The Key:
The (1- α) CI for the normal mean µ is:
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4. Example 1:
Find 95% CI for µ of X: content (liters) of certain containers.
Suppose that X is normally distributed with standard deviation
σ = 0.3. A sample of 16 containers gives a mean of 15.124 liters.
Solution:
Problem: CI for the mean µ of X. X: Normally Distributed
Variance is known: Standard Deviation, σ = 0.3
• Apply [ X − ∆, X + ∆] , where ∆ = z (σ / n )
• Given: size n = 16, Standard Deviation, σ = 0.3, x = 15.124 liter.
• 1 -  = 0.95,  = 0.05,  / 2 = 0.025, z = 1.96
• Calculate: ∆ = z σ / n = 1.96(0.3) / 16 = 0.147
α
2
0.025
α /2
x − ∆ = 15.124 − 0.147 = 14.977
x + ∆ = 15.124 + 0.147 = 15.271
• Answer: The 95% CI for µ is
[14.977, 15.271] liters.
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5. CI FOR NORMAL MEAN
WHEN VARIANCE UNKNOWN
 The key:
X−μ
T=
~ t(n − 1) , n < 30
S/ n
 The (1- α) CI for the normal mean µ is:
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6. Example 2:
Find the 90%-CI for the mean of gas mileage, in (miles/gallon).
Suppose the mileage is normally distributed.A sample of size 25
gives a sample mean 36.524 and a standard deviation 2.41.
Solution:
Problem: CI for the mean µ of gas mileage, X.
X: Normally Distributed,Variance is unknown:
• Apply [ X −∆, X +∆] , where ∆=t ( S / n )
• Given: size n = 25, Standard Deviation, S = 2.41, x = 36.524
• 1 -  = 0.90,  = 0.10,  / 2 = 0.05, t = 1.711
• Calculate: ∆ = t (S / n ) = (1.711)(2.41 / 5) = 0.825 , X = 36.524
α , n−
1
2
0.05,24
α , n −1
2
x − ∆ = 36.524 − 0.825 = 35.699 ;
x + ∆ = 37.349
• Answer: The 90% CI for the mean of gas mileage is
[35.699, 37.349] miles/gallon
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7. Example 3:
1. Suppose that the following data are observations from
N (µ, σ 2), σ 2 is unknown:
15 81 41 27 23 16 29 30 32 22 25 18
Find a 95% confidence interval for µ.
2. How large a sample is required if we want the length L
of a 95% confidence interval for the mean µ of a
normal population with standard deviation σ = 0.6 is
at most 0.2?
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8. Solution:
1. Problem: CI for the mean µ of X.
X: Normally Distributed,Variance is unknown:
• Apply [ X −∆, X +∆] , where ∆=t ( S / n )
• Given: size n = 12, Standard Deviation, S = 17.68, x = 29.92
• 1 -  = 0.95,  = 0.05,  / 2 = 0.025, t = 2.201
• Calculate: ∆ = t (S / n ) = (2.201)(17.68 / 12 ) = 11.23 , x = 29.92
α , n−
1
2
0.025,11
α , n −1
2
x − ∆ = 29.92 − 11.23 = 18.69 ;
x + ∆ = 41.15
• Answer: The 95% CI for the mean µ is [18.69, 41.15]
2. The length of the (1- ) confidence interval for µ is L = 2.∆ =
2.z α .σ /
n
2
L ≤ 0.2,
2.z α .σ / n ≤ 0.2
2
or
n ≥ [2(1.96)(0.6) /(0.2)]2 = 138.3 ≈ 139
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9. CI FOR THE MEAN
OF ANY DISTRIBUTION:
LARGE SAMPLE
 Key:
 Then


X −µ
P − z α ≤
 2 σ / n ≤ z α  ≅ 1 −α
2 


 Therefore the Approximate (1 - α) CI for µ is
 σ will be replaced by S if σ is unknown
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10. Example 4:
Find the 99% CI for the mean µ of certain chemical substance (in
grams/liter) in a river, based on measurements in 36 randomly
selected locations: sample mean = 2.45, S = 0.3
Solution:
Problem: CI for the mean µ of X. X: Not normally distributed,
[Variance is unknown,wheren ∆ = zα / 2 ( S / n )
X − ∆ , X + ∆ ] , Large (n = 36),
• Apply
x
• Given: size n = 36, Standard Deviation, S = 0.3, = 2.45
• 1 -  = 0.99,∆ == z0.01,/  /n2 = (0.005,)(z .3 / =)2.576 , x = 2.45
(S
) = 2.576 0 6 = 0.129
• Calculate: x − ∆ = 2.45 − 0.129 = 2.321 ; x + ∆ = 2.579
• Answer: The 99% CI for the mean µ of the chemical
α
2
0.005
substance in a river is [2.321, 2.579] mg/l
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11. CI FOR PROPORTION p
 Key: X: number of S-items in an n- sample
 Condition:
ˆ
ˆ
np = X ≥ 5, n(1 − p ) = n − X ≥ 5


ˆ
p −p
≤ zα  ≅1−α
 Then P − z α ≤

2
2 
ˆ
ˆ
p(1 − p ) / n


 Therefore the Approximate (1 - α) CI for p is:
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12. Example 5:
Find the 90% CI for the proportion p of defective items produced
by a manufacturer if in a random sample of 250 items there are 12
defectives.
Solution:
Problem: CI for the proportion p of defective items.
Large sample.
ˆ
ˆ
• Apply [ p −∆, p +∆] , where ∆ = z
• Given Sample Data: n = 250, X = 12
• 1 -  = 0.90,  = 0.10,  / 2 = 0.05, z = 1.645
• Calculations:
α
2
ˆ
ˆ
p (1 − p ) / n
0.05
ˆ
p = X / n = 12 / 250 = 0.048 ,
ˆ
ˆ
∆ = z0.05 p(1 − p ) / n = 0.0222
ˆ
ˆ
p − ∆ = 0.048 − 0.0222 = 0.0258 ; p + ∆ = 0.048 + 0.0222 = 0.0702
• Answer: The 90% CI for the proportion p of defective items
produced by the manufacturer is [0.0258; 0.0702]
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13. CI FOR TWO NORMAL MEANS
- VARIANCES ARE KNOWN The Key:
The (1- α) CI for the normal mean µ1− µ2 is:
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14. Example 6:
A random of 20 specimens of cold-rolled steel yield a sample average strength
30.2 , and a sample of 25 galvanized steel specimens gave a sample average
strength 35.1. Assume these two strength distributions are normal with σ 1 =
4.0 and σ 2 = 5.0. Find the 99% CI for the difference in mean µ1- µ2.
Solution: Problem: CI for the difference in mean µ1- µ2
of two normal random variables, known variances
Apply
σ
σ
2
[ (X1 − X 2 ) − Δ , (X1 − X 2 ) + Δ ] , where Δ = z α
2
1
n1
2
+
2
n2
1-  = 0.99,  = 0.01, /2 = 0.005 z /2 = z 0.005 = 2.576
x1 = 30.2,
x 2 = 35.1 , σ 1 = 4.0, σ 2 = 5.0, n1 = 20, n2 = 25
Calculate
2
Δ =zα
2
2
σ1
σ2
4.0 2 5.0 2
+
= (2.576)
+
= 3.456
n1
n2
20
25
(X1 − X 2 ) − Δ = (30.2 − 35.1) − 3.456 = −8.356, (X1 − X 2 ) + Δ = (30.2 − 35.1) + 3.456 = −1.444
Answer: The 99% CI for the difference in mean µ1- µ2 of
cold-rolled steel strength is [-8.356, -1.444]
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15. CI FOR ANY TWO MEANS
- VARIANCES ARE UNKNOWN –
LARGE SAMPLES
 The Key:
The (1- α) CI for the normal mean µ1− µ2 is:
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16. Example 7:
60 pieces of each type of thread are tested under similar conditions. Type A
thread had an average tensile strength of 86.7 kgs with a SD of 6.15 kgs, while
type B had an average tensile strength of 77.8 kgs with a SD of 5.82 kgs. Find
the 90% CI for the difference in mean µA- µB.
Solution: Problem: CI for the difference in mean µ1- µ2 of two
random variables, unknown variances, large samples
Apply [ (X − X ) − Δ , (X − X ) + Δ ] , where Δ = z S + S
2
A
B
A
B
2
A
α
2
B
nA
nB
1-  = 0.90,  = 0.10, /2 = 0.05 z /2 = z 0.05 = 1.645
x A = 86.7,
x B = 77.8 , sA = 6.15, sB = 5.82, nA = 60, nB = 60
Calculate
2
Δ =zα
2
2
SA
SB
6.15 2 5.82 2
+
= (1.645)
+
= 1.798
nA
nB
60
60
(X A − X B ) − Δ = (86.7 − 77.8) − 1.798 = 7.102, (X A − X B ) + Δ = (86.7 − 77.8) + 1.798 = 10.698
Answer: The 90% CI for the difference in mean µA- µB of
cold-rolled steel strength is [7.102, 10.698]
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17. CI FOR ANY TWO MEANS
- VARIANCES ARE UNKNOWN –
SMALL SAMPLES
2
σ1 = σ 2
The Key:
2
( X1 − X 2 ) −(μ 1 −μ 2 )
T=
~ t(n1 + n 2 − 2)
S p 1/n 1 +1/n 2
2
Sp =
2
S 1 (n1 −1) + S 2 (n 2 −1)
, n 1 , n 2 < 30
n1 +n 2 − 2
The (1- α) CI for the normal mean µ1− µ2 is:
[ (X1 − X 2 ) − Δ , (X1 − X 2 ) + Δ ] , where Δ = t α ,n1 + n 2 − 2 S p
2
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1 1
+
n1 n 2
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18. CI FOR ANY TWO MEANS
- VARIANCES ARE UNKNOWN –
SMALL SAMPLES
2
σ1 ≠ σ 2
 The Key:
T=
( X1 − X 2 ) − (μ 1 − μ 2 )
2
~ t(v)
2
S1 /n 1 + S 2 /n 2
2
2
2
(S1 /n1 + S 2 /n 2 ) 2
v=
, n1 , n 2 < 30
2
2
2
2
(S1 /n1 ) /(n1 − 1) + (S 2 /n 2 ) /(n 2 − 1)
The (1- α) CI for the normal mean µ1− µ2 is:
2
[ (X1 − X 2 ) − Δ , (X1 − X 2 ) + Δ ] , where Δ = t α , v
2
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S1 S 2
+
n1 n 2
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19. CI FOR VARIANCE
OF NORMAL R V
 Key:
 Therefore the (1- α) CI for σ2 is:
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20. Example 8:
A hardware manufacturer produces 10-mm bolts. The diameters
(mm) of a sample of 12 bolts, give a sample variance 0.0026 &
sample mean 9.975. Find a 90% CI for σ2 and for µ. Suppose the
bold diameter is normally distributed.
Solution:
Problem: CI for σ 2 of bolt diameters. The diameters are
Normally S 2 (n −1) S 2 
 ( n −1) Distributed

,
• Apply  2
2
•
•
•
•


χα ,n −1
2
χ1− α ,n −1 

2
Given Sample Data: n = 12, S2 = 0.0026
1 -  = 0.90,  = 0.10,  /2 = 0.05, 1 -  /2 = 0.95
X 20.05,11 = 19.675;
X 20.95,11 = 4.575
Calculations: (n − 1) S 2 / χ α2, n − 1 = 0.00145 , (n − 1) S 2 / χ12− α , n − 1 = 0.00625
Answer: The 90% CI for the proportion variance of bolt
2
2
diameters is [0.00145; 0.00625] mm2
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21. Problem: CI for the mean µ of diameters, X. X: Normally Distributed
Variance is unknown
• Apply [ X −∆, X +∆] , where ∆=t ( S / n )
• Given: size n = 12, Standard Deviation, S = 0.0510, x = 9.975 mm
• 1 -  = 0.90,  = 0.10,  / 2 = 0.05, t = 1.796
• Calculate:
α , n−
1
2
0.05, 11
∆ = t α ,n −1 ( S / n ) = (1.796)(0.0510 / 12 ) = 0.0264 , x = 9.975
2
x − ∆ = 9.975 − 0.0264 = 9.949 ;
• Answer: The 90% CI for µ is
x + ∆ = 10.001
[9.949, 10.001] mm.
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22. SUMMARY OF CHAPTER 6
CONFIDENCE INTERVALS
 NOTION: (1 - α) CONFIDENCE INTERVAL
 CI FOR NORMAL POPULATION MEAN µ
 CI FOR ANY POPULATION MEAN :
LARGE SAMPLE
 CI FOR PROPORTION (OR PROBABILITY) p
 CI FOR NORMAL TWO POPULATION MEANS
 CI FOR NORMAL POPULATION VARIANCE σ 2
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