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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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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,
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 =
(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-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.