Errors in Hypothesis Testing: Type I and Type II Errors, Study notes of Mathematical Statistics

The concepts of type i and type ii errors in hypothesis testing. It explains what these errors mean, how to calculate their probabilities, and the impact of changing the rejection region on the error types. The document also includes examples using normal distribution and r programming.

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Stat 312: Lecture 12
Errors in hypothesis testing
Moo K. Chung
Oct 20, 2004
Type I error: you reject H0when H0is true.
P(Type I error) = P(reject H0|H0true) = α. The
resulting αis called the significance level of the test
and the corresponding test is called a level αtest.
We will use test procedures that give αless than a
specified level (0.05 or 0.01 usually).
I believe that dogs are as smart as people. Assume
IQ of a dog follows XiN(µ, 102). IQ of 10 dogs
are measured: 30, 25, 70, 110, 40, 80, 50, 60, 100,
60. We want to test if dogs are as smart as people by
testing
H0:µ= 100 vs. H1:µ < 100.
(1) Let ¯
Xbe a test statistic and R= (−∞,90) to
be a rejection region. Let’s compute the probabil-
ity of making Type I error based on this testing
procedure. Under the assumption H0is true,
XiN(100,102).
Under this condition, ¯
XN(100,10) and
α=P(¯
X < 90).
>help(pnorm)
...
Usage:
... pnorm(q, mean=0, sd=1,...)
...
> pnorm(90,100,sqrt(10))
[1] 0.0007827011
By using this test procedure, it is highly unlikely
to make Type I error.
(2) Let’s see what happens when we change the test
precedure. When R= (−∞,95) with the same
test statistic ¯
X,
α=P(¯
X < 95).
> pnorm(95,100,sqrt(10))
[1] 0.05692315
70 90 110
0.0 0.4 0.8
a
pnorm(a, 100, sqrt(10))
70 90 110
0.0 0.4 0.8
a
1 − pnorm(a, 90, sqrt(10))
Figure 1: Errors in performing hypothesis testing
based on test procedure if ¯
X < 90, reject H0:µ=
100. Left figure: type I error, right figure: type II error.
(3) When R= (−∞,99),α=P(¯
X < 99).
> pnorm(99,100,sqrt(10))
[1] 0.3759148
The test procedure based on rejecting H0if ¯
X <
99 is producing large Type I error. Why?
Type II error: you do not reject H0when H0is
false.
β=P(Type II error)
=P(do not reject H0|H0false )
= 1 P(reject H0|H1true)
With the fixed test procedure of rejecting H0if ¯
X <
90, let us compute β. When µ= 90 is true,
> 1-pnorm(90,90,sqrt(10))
[1] 0.5
> 1-pnorm(90,95,sqrt(10))
pf2

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Stat 312: Lecture 12

Errors in hypothesis testing

Moo K. Chung

[email protected]

Oct 20, 2004

Type I error : you reject H 0 when H 0 is true. P (Type I error) = P (reject H 0 |H 0 true) = α. The resulting α is called the significance level of the test and the corresponding test is called a level α test. We will use test procedures that give α less than a specified level (0.05 or 0.01 usually).

I believe that dogs are as smart as people. Assume IQ of a dog follows Xi ∼ N (μ, 102 ). IQ of 10 dogs are measured: 30, 25, 70, 110, 40, 80, 50, 60, 100,

  1. We want to test if dogs are as smart as people by testing

H 0 : μ = 100 vs. H 1 : μ < 100.

(1) Let X¯ be a test statistic and R = (−∞, 90) to be a rejection region. Let’s compute the probabil- ity of making Type I error based on this testing procedure. Under the assumption H 0 is true,

Xi ∼ N (100, 102 ).

Under this condition, X¯ ∼ N (100, 10) and α = P ( X <¯ 90).

help(pnorm) ... Usage: ... pnorm(q, mean=0, sd=1,...) ... pnorm(90,100,sqrt(10)) [1] 0.

By using this test procedure, it is highly unlikely to make Type I error.

(2) Let’s see what happens when we change the test precedure. When R = (−∞, 95) with the same test statistic X¯, α = P ( X <¯ 95).

pnorm(95,100,sqrt(10)) [1] 0.

70 90 110

a

pnorm(a, 100, sqrt(10))

70 90 110

a

1 − pnorm(a, 90, sqrt(10))

Figure 1: Errors in performing hypothesis testing based on test procedure if X <¯ 90 , reject H 0 : μ =

  1. Left figure: type I error, right figure: type II error.

(3) When R = (−∞, 99), α = P ( X <¯ 99).

pnorm(99,100,sqrt(10)) [1] 0.

The test procedure based on rejecting H 0 if X <¯ 99 is producing large Type I error. Why?

Type II error : you do not reject H 0 when H 0 is false.

β = P (Type II error) = P (do not reject H 0 |H 0 false ) = 1 − P (reject H 0 |H 1 true)

With the fixed test procedure of rejecting H 0 if X <¯ 90 , let us compute β. When μ = 90 is true,

1-pnorm(90,90,sqrt(10)) [1] 0. 1-pnorm(90,95,sqrt(10))

[1] 0.

1-pnorm(90,99,sqrt(10)) [1] 0.

Now let us see how β changes when we change the rejection region. Assume μ = 90 is true.

a<- -20: a<- a + 90 %Type II error --> right figure plot(a,1-pnorm(a,90,sqrt(10)),type=’l’) %Type I error --> left figure plot(a,pnorm(a,100,sqrt(10)),type=’l’)

Review problems Exercise 8.1., 8.3., 8.5.