errors in hypothesis testing
DESCRIPTION
Errors in Hypothesis Testing. 2 TYPES OF ERRORS. TRUE CASE H A is true H A is false WE Accept H A SAY Do not Accept H A. TYPE I ERROR. CORRECT. PROB = α. TYPE II ERROR. CORRECT. PROB = β. α is set by the decision maker. β varies and depends on: - PowerPoint PPT PresentationTRANSCRIPT
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Errors inErrors in
Hypothesis TestingHypothesis Testing
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TRUE CASETRUE CASE
HAis true HAis false
WEWE Accept HA
SAYSAY Do not Accept HA
2 TYPES OF ERRORS
CORRECT
CORRECT
TYPE I
ERROR
TYPE II
ERROR
PROB = α
PROB = β
α is set by the decision maker
β varies and depends on:
(1) α; (2) n; (3) the true value of
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Relationship Between and is the Probability of making a Type II error
– i.e. the probability of not concluding HA is true when it is
depends on the true value of • The closer the true value of is to its hypothesized value,
the more likely we are of not concluding that HA is true -- i.e. is large (closer to 1)
is calculated BEFOREBEFORE a sample is taken– We do not use the results of a sample to calculate
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• Example: If we take a sample of n = 49, with = 4.2, “What is the probability we will get a sample from which we would not conclude > 25 when really = 25.5?” (Use = .05)
REWRITE REJECTION REGION IN TERMS OF
CALCULATING
x
The Hypothesis Test The Hypothesis Test
987.2549
2.4645.125x
645.1
494.2
25xz if HAccept A
or if
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CALCULATING (cont’d)
• So when = 25.5,– If we get an > 25.987, we will correctly conclude
that > 25– If we get an < 25.987 we will not conclude that
> 25 even though really = 25.5
x
x
That’s aTYPE II ERROR!!
P(Making this error) =
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CALCULATING (cont’d)
x
49
x
ββ
• So what is P(not getting an > 25.987 when really = 25.5? That is P(getting an < 25.987)?Calculate z = (25.987 - 25.5)/(4.2/ ) .81
• is the area to the leftleft of .81 for a “>” test• P(Z < .81) = .7910.7910
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“>” TestDetermining When = 25.5
0 .81 Z
X25.5 25.987
ACCEPT HA
RIGHT!
DO NOTACCEPT HA
WRONGProb = =.7910
.7910.7910
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What is When = 27?
x
49
x
ββ
This shows that the further the true value of is from the
hypothesized value of , the smallersmaller the value of β; that is we
are less likely to NOT conclude that HA is true (and it is!)
• So what is P(not getting an > 25.987 when really = 27? That is P(getting an < 25.987)?Calculate z = (25.987 - 27)/(4.2/ ) -1.69
• is the area to the leftleft of -1.69 for a “>” test• P(Z < -1.69) = .0455.0455
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“>” TestDetermining When = 27
-1.69 0 Z
X25.987 27
ACCEPT HA
RIGHT!
DO NOTACCEPT HA
WRONGProb = =.0455
.0455.0455
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for “<” Tests
• For n = 49, = 4.2, “What is the probability of not concluding that < 27, when really is 25.5? (With = .05)
• This time is the area to the rightright of x
013.2649
2.4645.127x
645.1
494.2
27xz if HAccept A
or if
The Hypothesis Test The Hypothesis Test
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What is When = 25.5?
x
49
x
ββ
• So what is P(not getting an < 26.013 when really = 25.5? That is P(getting an > 26.013)?Calculate z = (26.013 – 25.5)/(4.2/ ) .86
• is the area to the rightright of .86 for a “<” test• P(Z > .86) = 1 - .8051 = .1949.1949
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“<” TESTDetermining When = 25.5
0 .86 Z
X25.5 26.013
ACCEPT HA
RIGHT!
.8051.8051
.1949.1949
DO NOTACCEPT HA
WRONGProb = =.1949
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for “” Tests• For n = 49, = 4.2, “What is the probability of not
concluding that 26, when really is 25.5? (With = .05)
• This time is the area in the middlein the middle between the two critical values of
x
176.2749
2.496.126x
or 824.2449
2.496.126x
1.96or 1.96
494.2
26xz if HAccept A
or if
The Hypothesis TestThe Hypothesis Test
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What is When = 25.5?
x
x
49
ββ
49
• So what is P(not getting an < 24.824 or > 27.176 when really = 25.5? That is P(24.824 < < 27.176)?Calculate z’s = (24.824 – 25.5)/(4.2/ ) -1.13
and = (27.176 – 25.5)/(4.2/ ) 2.79
• is the area in betweenin between -1.13 and 2.79 for a “” test
• P(Z < 2.79) = .9974.9974 • P(Z < -1.13) = .1292.1292
P(-1.13 < Z < 2.79 = .9974.9974 - .1292.1292 = .8682.8682
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“” TESTDetermining When = 25.5
-1.13 0 2.79 Z
X24.824 25.5 27.176
DO NOTACCEPT HA
WRONGProb = =.9974 –
.1292 =.8682
ACCEPT HA
RIGHT!
.9974.1292
.8682
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The Power of a Test = 1 -
is the Probability of making a Type II error– i.e. the probability of not concluding HA is true when
it is depends on the true value of and sample
size, n• The Power of the test for a particular value of
is defined to be the probability of concluding HA is true when it is -- i.e. 1 - 1 -
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Power Curve Characteristics
• The power increases with:– Sample Size, n– The distance the true value of μ is from the
hypothesized value of μ
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Power Curves For HA: μ 26With n = 25 and n = 49
n = 49
n = 25
α = .05
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Calculating Using Excel“> Tests”
Suppose H0 is = 25; = 4.2, n = 49, = .05
““>” TESTS>” TESTS: HA: > 25 and we want when the true value of = 25.5
1) Calculate the criticalcritical x-bar x-bar value = 25 + NORMSINV(.95)*(4.2/SQRT(49))
2) Calculate z = (criticalcritical x-bar x-bar -25.5)/ (4.2/SQRT(49))
3) Calculate the the probability of getting a z- value < than this critical z value: -- this is this is =NORMSDIST(z)NORMSDIST(z)
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Calculating Using Excel“< Tests”
Suppose H0 is = 27; = 4.2, n = 49, = .05
““< TESTS”:< TESTS”: HA: < 27 and we want when the true value of = 25.5
1) Calculate the critical x-barcritical x-bar value = 27 - NORMSINV(.95)*(4.2/SQRT(49))
2) Calculate z = (criticalcritical x-bar x-bar -25.5)/ (4.2/SQRT(49))
3) Calculate the the probability of getting a z- value > than the critical value: -- this is this is
=1-NORMSDIST(z)
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Calculating Using Excel“ Tests”
Suppose H0 is = 26; = 4.2, n = 49, = .05
TESTS:TESTS: HA: 26 and we want when the true value of = 25.5
1) Calculate the critical uppercritical upper x-barx-barUU value and the
lower criticallower critical x-barx-barLL value
= 26 - NORMSINV(.975)*(4.2/SQRT(49)) (x-barx-barLL)
= 26 + NORMSINV(.975)*(4.2/SQRT(49)) (x-barx-barUU)
2) Calculate zU = (x-barx-barUU -25.5)/ (4.2/SQRT(49)) and zL = (x-x-
barbarLL -25.5)/ (4.2/SQRT(49))
3) Calculate the the probability of getting an z- value in between zL and zU - this is this is =NORMSDIST(zU) - NORMSDIST(zL)
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β for “>” Tests
=B3+NORMSINV(1-B2)*(B5/SQRT(B6))
=(B8-B7)/(B5/SQRT(B6))
=NORMSDIST(B9)
=1-B10
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β for “<” Tests
=B3-NORMSINV(1-B2)*(B5/SQRT(B6))
=(B8-B7)/(B5/SQRT(B6))
=1-NORMSDIST(B9)
=1-B10
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β for “” Tests =B3-NORMSINV(1-B2/2)*(B5/SQRT(B6))
=B3+NORMSINV(1-B2/2)*(B5/SQRT(B6))
=(B8-B7)/(B5/SQRT(B6))
=(B9-B7)/(B5/SQRT(B6))
=NORMSDIST(B11)-NORMSDIST(B10)
=1-B12
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REVIEW
• Type I and Type II Errors = Prob (Type I error) = Prob (Type II error) -- depends on , n and α
• How to calculate for:– “>” Tests– “<” Tests– “” Tests
• Power of a Test at = 1- • How to calculate using EXCEL