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type I error.
type
P(Rej H0H0 iis true) = α
s 2) Fail to Reject H0 when Ha iis true at some
s
value. This is called a type II error.
value.
P(Fail to Rej H0Ha iis true at some value) = β
s
to Rej
G. Baker, Department of Statistics
G.
University of South Carolina; Slide 69
University
69 Avg Life of Light Bulb  Type I Error
Life
H0: µ < 2000
Ha: µ > 2000 Z Fail to reject H0. Assumes H0
is true. α = Probability that
we will reject Ho
when Ho is true.
G. Baker, Department of Statistics
G.
University of South Carolina; Slide 70
University
70 Type I and Type II Errors
H0: µ = 2000 β = Probability we will
fail to reject Ho when
Ha is true at µ = 2200 What if µ = 2200 α = Probability that
we will reject Ho
when Ho is true. G. Baker, Department of Statistics
G.
University of South Carolina; Slide 71
University
71 How can we control the size of β?
How
The value of α.
Location of our point of interest.
Location
Sample size.
Sample G. Baker, Department of Statistics
G.
University of South Carolina; Slide 72
University
72 Calculating β
Calculating
If µ = 2200, what is the probability of a type II
2200,
error?
error?
Given: α = 0.05 and we are assuming
µ = 2000. We will also assume we know σ =
2000.
216.
216. P( Z > 1.645) = 0.05
y − 2000
1.645 =
→ y = 2091
216 / 15
G. Baker, Department of Statistics
G.
University of South Carolina; Slide 73
University
73 Calculating β
Calculating
H0: µ = 2000 Fail to Reject Ho What if µ = 2200 2091 Reject Ho P( y < 2091  µ = 2200) = β
G. Baker, Department of Statistics
G.
University of South Carolina; Slide 74
University
74 Calculating β
Calculating
β = P( y < 2091  µ = 2200)
2091 − 2200 β = P z < = P( z < −1.9544) = 0.0254
216 / 15 β = P(Fail to Reject H 0  µ = 2200) = 0.0254
G. Baker, Department of Statistics
G.
University of South Carolina; Slide 75
University
75 α, β and Power
α = P(Reject H0µ = 2000) = 0.05
β = P(Fail to Rej H0 µ = 2200) = 0.0254
P(Fail
Rej
We say that the power of this test at
of
µ = 2200 is 1 – 0.0254 = 0.9746
2200
Power = 1 –β
Power = P(Rej H0µ is at some Ha level)
P(Rej G. Baker, Department of Statistics
G.
University of South Carolina; Slide 76
University
76 Plastic Injection Molding
A plastic injection molding process for a
part that has a critical width dimension
historically follows a normal distribution.
historically
A recent sample of n = 4 yielded a sample
mean of 101.4 and sample standard
deviation of 8.
deviation
Does this data support the statement:
“The true average width is greater than
The
95.”?
95.
G. Baker, Department of Statistics
G.
University of South Carolina; Slide 77
University
77 Plastic Injection Molding
Confidence Interval Approach
95% confidence interval on µ: y ± t df =3, 0.025 s
n 8
101.4 ± 3.182
= 101.4 ± 12.728
4 (93.56,109.24)
G. Baker, Department of Statistics
G.
University of South Carolina; Slide 78
University
78 Plastic Injection Molding
Hypothesis Test Approach
H0: α = 0.05 Ha:
Test statistics is
pvalue =
Conclusion:
G. Baker, Department of Statistics
G.
University of South Carolina; Slide 79
University
79...
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This note was uploaded on 01/12/2014 for the course STA 509 taught by Professor Wang during the Fall '13 term at South Carolina.
 Fall '13
 Wang
 Statistics

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