Section 8.
1: Hypotheses and Testing Procedure
Concepts and Formulae: Suppose there is a claim or a statement. The null hypothesis is an answer in which the claim or the statement is true, and denoted by H0. The alternative hypothesis is an answer in which the claim or the statement is NOT true, and denoted by Ha (HA or H1). If we conclude H0, we say we accept H0 or fail to reject H0; if we conclude Ha, we say we reject H0 and conclude Ha.
A testing procedure is specied by the following steps: Choose a test statistic. A test statistic is a function of data which makes the decision about the rejection or acceptance of H0. A rejection range is a set in which we reject H0 if the test statistic is in the set and accept H0 if the test statistic is outside of the set. Let T be the test statistic and R be the rejection range. Then, we conclude H0 Ha if T R if T R
In general, a test can be summarized into the following table. Truth H0 Correct Type I Error
Conclude H0 Ha
Ha Type II Error Correct
Type I error rate (probability) is dened by P (Conclude Ha|Truth is H0) and type II error rate (probability) is dened by P (Conclude H0|Truth is Ha). We want to make both probabilities small. However, this is usually impossible in real applications. Thus classically, we only control type I error probabilities. The signicance level denoted by is the maximum of type I error probabilities. Therefore, if we choose = 0.05, we guarantee the type I error probability is less than or equal to 0.05.
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First example of Section 8.1: Examples 8.1 and 8.3 on textbook. Suppose old rate of no visible damage is 25%. An experiment took 20 samples and want to know whether the rate increases with a new method. Null hypothesis is H0 : rate does not increase versus the alternative Ha : rate increases. Under H0, the count X of visible damage follows Bin(20, 0.25). Under Ha, X Bin(20, p) with p > 0.25. Then, we can write H0 : p = 0.25 Ha : p > 0.25. Assume the rejection range is R = {8, 9, 10, 11, , 20}.
Then, the type I error probability is P (X 8|p = 0.25) =1 B(7; 20, 0.25) =0.102. The signicance level = 0.102. The type II error probability is P (X < 8|p > 0.25) = B(7; 20, p) which is a function of p. Suppose we change the testing problem as H0 : p 0.25 Ha : p > 0.25. Then, the type I error probability is 1 B(7; 20, p) for p 0.25 and the type II error probability is B(7; 20, p) for p > 0.25. In this case, we still have the signicance level = 0.102.
0.10
0.08
Type II Error Probability 0.00 0.05 0.10 0.15 0.20 0.25
Type I Error Probability
0.06
0.04
0.00
0.02
0.0
0.2
0.4
0.6
0.8
0.4
0.6 True Proportion
0.8
1.0
True Proportion
Graph of Type I and Type II error probabilities The signicance level is the maximum of type I error probabilities, which gives = 0.102.
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Second example of Section 8.1: Examples 8.2 and 8.4 on textbook. Assume we choose 25 samples from N (, 81) and suppose we test H0 : = 75 Ha : < 75. Suppose we use R = X < 70.8, i.e., we conclude H0 Ha if if X 70.8 X < 70.8
Then, the type I error is P (X 70.8 if = 75) 70.8 75 =( ) 81/25 =(2.33) = 0.01. Then, the signicance level is = 0.01 and type II error probability is 70.8 ). 1.8 for < 75, which is a function of for < 75. For example, we have () = P (X 70.8) = 1 ( (72) = 1 (0.67) = 0.7486.
0.010
0.008
Type II Error Probability 75 76 77 78 79 80
Type I Error Probability
0.006
0.004
0.002
0.000
0.4 70
0.5
0.6
0.7
0.8
0.9
1.0
71
72
73
74
75
True Expected value
True Expected value
Graph of Type I and Type II error probabilities The signicance level is the maximum of type I error probabilities, which gives = 0.01.
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