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Hypothesis Testing: Type II Error and Power

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**Actually True Actually False**

Type I and Type II Error Revisited NULL HYPOTHESIS Actually True Actually False 1-a Type II error b Type I error a 1-b Fail to Reject DECISION Either type error is undesirable and we would like both a and b to be small. How do we control these?

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**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 error a also represents the level of significance of the decision rule or test You, as the investigator, select this level

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**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 error 1-b represents the POWER of a test: The probability of rejecting a false null hypothesis The value of b depends on a specific alternative hypothesis b can be decreased (power increased) by increasing sample size

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**Computing Power of a Test**

Example: Suppose we have test of a mean with Ho: mo = vs Ha: mo 100 s = 10 n = 25 a = .05 If 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?

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**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 )

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**Suppose, in fact, that ma = 105.**

We will reject Ho if x is greater than or x is less than 96.08 Let’s look at these decision points relative to our specific alternative. Suppose, in fact, that ma = 105. Distribution based on Ha 96.08 103.92 ma=105

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ma=105 96.08 103.92 z

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**b note: a is fixed in advance by the investigator b depends on**

the sample size se = (s / n) the specific alternative, ma we assume that the variance s2 holds for both the null and alternative distributions a/2 a/2 b ma 105 m0 100 (se) = 96.08 (se) =

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**Again, looking at our specific alternative: ma = 105**

b: area where we fail to reject Ho even though Ha is correct a/2: area where we reject Ho for Ha – Good! a/2 ma 105 m0 100 (se) = 96.08 (se) =

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We define power as 1-b power = Pr(rejecting Ho | Ha is true) In our example, power = 1-b = 1 – = .8315 That is, with a = .05 a sample size of n=25 a true mean of ma= 105, the power to reject the null hypothesis (mo=100) is %.

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**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 67 We have n=25 and we know s = 3. To test this hypothesis we establish our critical region. a/2 a/2 ? ?

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**Here, we reject Ho, at the a=.05 level when:**

or a/2: Rejection region a/2: Rejection region

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**Now, select a specific alternative to compute b: Let Ha1: ma=67.5**

“fail-to-reject” region based on H0 z – or Power = 1-b = 13%

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**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 for a = .05, n=25, mo = 67, s = 3 ma zlower zupper b Power =1-b 68.5 - 4.47 - .53 .29 .71 68 - 3.36 0.30 .62 .38 67.5 - 2.80 1.13 .87 .13 67 - 1.96 1.96 .95 .05 66.5 - 1.13 2.80 66 - 0.30 3.36 65.5 +0.53 4.47 mo

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**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 = a 1.00 The farther the alternative is from m0, the greater the power. 0.75 1 - b 0.50 0.25 0.00 65 66 67 68 69 m0 ma

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**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 67 But we will now use n=100. We establish our critical region – now with sx= s / n = 3/10 = .3 a/2 a/2 ? ?

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**With n=100, we reject Ho, at the a=.05 level when:**

or a/2: Rejection region a/2: Rejection region

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**Again, select a specific alternative to compute b: Let Ha: ma=67.5**

“fail-to-reject” region based on H0 z – or Power = 1-b = 38%

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**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 for a = .05, n=100, mo = 67, s = 3 ma zlower zupper b Power =1-b 68.5 - 6.97 - 3.04 .00 1.00 68 - 5.30 - 1.37 .09 .91 67.5 - 3.63 0.30 .62 .38 67 - 1.96 1.96 .95 .05 66.5 - 0.30 3.63 66 1.37 5.30 65.5 3.04 6.97 mo

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**1 - b ma Power Curves: Power (1-b) vs. ma for n=25, 100**

a = .05, mo = 67 0.00 0.25 0.50 0.75 1.00 – n = 100 – n = 25 1 - b For the same alternative ma, greater n gives greater power. 65 66 67 68 69 ma

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**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 Ha even for alternatives that are close to the value of mo.

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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-ma The sample size, n The standard deviation, s

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**Using Minitab to estimate Sample Size: **

Stat Power and Sample Size 1-Sample Z Sample size (to specify several, separate with a space) Difference between mo and ma ( to specify several, separate with a space) 2-sided test s

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Power and Sample Size 1-Sample Z Test Testing mean = null (versus not = null) Calculating power for mean = null + difference Alpha = Assumed standard deviation = 10 Sample Difference Size Power

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