2 Actually True Actually False Type I and Type II Error RevisitedNULL HYPOTHESISActually True Actually False1-aType II errorbType I errora1-bFail toRejectDECISIONEither type error is undesirable and we would like both a and b to be small.How do we control these?
3 A Type I error, or an a-error is made when a true hypothesis is rejected. The letter “a” (alpha) is used to denote the probability related to a type I errora also represents the level of significance of the decision rule or testYou, as the investigator, select this level
4 A Type II error, or an b-error is made when a false hypothesis is NOT rejected. The letter “b” (beta) is used to denote the probability related to a type II error1-b represents the POWER of a test:The probability of rejecting a false null hypothesisThe value of b depends on a specific alternative hypothesisb can be decreased (power increased) byincreasing sample size
5 Computing Power of a Test Example: Suppose we have test of a mean withHo: mo = vs Ha: mo 100s = 10n = 25a = .05If the true mean is in fact m = 105,what is b, the probability of failing to reject Ho when we should ?What is the power (1-b) of our test to reject Ho when we should reject it?
6 In this example, the standard error is s/n = 10/5=2, so that: mo=100(2) = 96.08(2) =We will reject Ho if (x 96.08) or if (x )
7 Suppose, in fact, that ma = 105. We will reject Hoif x is greater thanor x is less than 96.08Let’s look at these decision points relative to our specific alternative.Suppose, in fact, that ma = 105.Distribution based on Ha96.08103.92ma=105
9 b note: a is fixed in advance by the investigator b depends on the sample size se = (s / n)the specific alternative, mawe assume that the variance s2 holds for both the null and alternative distributionsa/2a/2bma105m0100(se) = 96.08(se) =
10 Again, looking at our specific alternative: ma = 105 b: area where we fail to reject Ho even though Ha is correcta/2: area where we reject Ho for Ha – Good!a/2ma105m0100(se) = 96.08(se) =
11 We define power as 1-bpower = Pr(rejecting Ho | Ha is true)In our example,power = 1-b = 1 – = .8315That is,with a = .05a sample size of n=25a true mean of ma= 105,the power to reject the null hypothesis (mo=100) is %.
12 To test this hypothesis we establish our critical region. Example 2:Suppose we want to test, at the a = .05 level, the following hypothesis:Ho: m = 67 vs. Ha: m 67We have n=25 and we know s = 3.To test this hypothesis we establish our critical region.a/2a/2? ?
13 Here, we reject Ho, at the a=.05 level when: ora/2:Rejection regiona/2:Rejection region
14 Now, select a specific alternative to compute b: Let Ha1: ma=67.5 “fail-to-reject” region based on H0z–or Power = 1-b = 13%
15 Type II Error (b) and Power of Test for Now look at the same thing for different values of ma:Type II Error (b) and Power of Test fora = .05, n=25, mo = 67, s = 3mazlowerzupperbPower=1-b68.5- 4.47- .53.29.7168- 3.3126.96.36.19967.5- 2.801.13.87.1367- 1.961.96.95.0566.5- 1.132.8066- 0.303.3665.5+0.534.47mo
16 ma Let us plot Power (1-b) vs. alternative mean (µa). This plot will be called the power curve.Note: at ma= mo 1-b = a1.00The farther the alternative is from m0, the greater the power.0.751 - b0.500.250.006566676869m0ma
17 We establish our critical region – now with sx= s / n = 3/10 = .3 Suppose we want to test, the same hypothesis, still at the a = .05 level, s = 3 :Ho: m = 67 vs. Ha: m 67But we will now use n=100.We establish our critical region – now withsx= s / n = 3/10 = .3a/2a/2? ?
18 With n=100, we reject Ho, at the a=.05 level when: ora/2:Rejection regiona/2:Rejection region
19 Again, select a specific alternative to compute b: Let Ha: ma=67.5 “fail-to-reject” region based on H0z–or Power = 1-b = 38%
20 Type II Error (b) and Power of Test for Now look at the same thing for different values of ma:Type II Error (b) and Power of Test fora = .05, n=100, mo = 67, s = 3mazlowerzupperbPower=1-b68.5- 6.97- 3.04.001.0068- 5.30- 1.37.09.9167.5- 3.6188.8.131.5267- 1.961.96.95.0566.5- 0.303.63661.375.3065.53.046.97mo
21 1 - b ma Power Curves: Power (1-b) vs. ma for n=25, 100 a = .05, mo = 670.000.250.500.751.00– n = 100– n = 251 - bFor the same alternative ma, greater n gives greater power.6566676869ma
22 Clearly, the larger sample size has resulted in a more powerful test.However, the increase in power required an additional 75 observations.In all cases a = .05.Greater power means:we have a greater chance of rejecting Ho in favor of Haeven for alternatives that are close to the value of mo.
23 We will revisit our discussion of power when we discuss sample size in the context of hypothesis testing.Minitab allows you to compute power of a test for a specific alternative:You must supply:The difference between the null and a specific alternative mean: m0-maThe sample size, nThe standard deviation, s
24 Using Minitab to estimate Sample Size: Stat Power and Sample Size 1-Sample ZSample size (to specify several, separate with a space)Difference between mo and ma ( to specify several, separate with a space)2-sided tests
25 Power and Sample Size1-Sample Z TestTesting mean = null (versus not = null)Calculating power for mean = null + differenceAlpha = Assumed standard deviation = 10SampleDifference Size Power