 # 1 Power and Sample Size in Testing One Mean. 2 Type I & Type II Error Type I Error: reject the null hypothesis when it is true. The probability of a Type.

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1 Power and Sample Size in Testing One Mean

2 Type I & Type II Error Type I Error: reject the null hypothesis when it is true. The probability of a Type I Error is denoted by . Type II Error: accept the null hypothesis when it is false and the alternative hypothesis is true. The probability of a Type II Error is denoted by .

3 Null hypothesis DecisionTrueFalse Reject H 0 Type I ErrorCorrect Decision  1   Do not reject H 0 Correct Decision Type II Error 1  

4 The serum cholesterol levels for all 20- to 24-year-old males of a certain population was claimed to be normally distributed with a mean of 180 mg/100ml and standard deviation is 46 mg/100ml. A Population

5 Is the mean cholesterol level of the population of 20- to 74-year-olds higher than 180 mg/100ml ? H 0 :  = 180 mg/100ml (or   180 mg/100ml) H a :  > 180 mg/100ml Research Hypothesis

6 Power of the Test

7 The power of the statistical test is the ability of the study as designed to distinguish between the hypothesized value and some specific alternative value. That is, the power is the probability of rejecting the null hypothesis if the null hypothesis is false. Power = P(reject H 0 | H 0 is false) Power = 1   The power is usually calculated given a simple alternative hypothesis: H a :  = 211 mg/100ml Power = P(reject H 0 | H a is true)

8 0 1.645 Rejection region z Right tail area is 0.05 Assuming that we use a sample of size 25 and test the hypothesis with a level of significant  = 0.05, what would be the  and the power of the test, i.e. 1   ? If we take the critical value approach, the critical value would be 1.645. At the level of significant  = 0.05, we would reject H 0 if z  1.645.

9 If a sample of size 25 is selected, what would be the critical value in terms of the mean cholesterol level? 0 1.645 Rejection region z Right tail area is 0.05 180 ? Rejection region 195.1

10 If the alternative hypothesis H a :  =  a = 211 mg/100ml, what would be the probability of NOT rejecting H 0, i.e.  ? H 0 :   = 180 v.s. H a :   = 211 That is to say “how powerful can this test detect a 31 mg/100ml increase in average cholesterol level?”  = ? 180 195.1 211 Power of the test when sample size is 25: Power = 1 –  = 1 – 0.042 = 0.958. 0.042

11 Assuming that we use a sample of size 100 and test the hypothesis with a level of significant  = 0.05, what would be the  and the power of the test, i.e. 1   ? 0 1.645 Rejection region z Right tail area is 0.05 180 ? Rejection region 187.6

12 If we assume the actual population mean is 211 mg/100ml for H a, what is the probability of accepting H 0, i.e.  ?  = ? 180 187.6 211  =.05 Power of the test when sample size is 100: Power = 1 –   ? Recall that the power was.958 when sample size was 25.  0.00 1 – 0  1

13 180 211 n = 25 187.6 n = 100 195.1

14 Sample Size Estimation

15 H 0 :  = 180 mg/100ml (or   180 mg/100ml) v.s. H A :  > 180 mg/100ml, to perform a test at the power level of 0.95, with the level of significance of 0.01, how large a sample do we need? (Power of 0.95 means we want to risk a 5% chance of failing to reject the null hypothesis if the true mean is as large as 211 mg/100ml. Or, to say, we want to have a 95% chance of rejecting the null hypothesis if the true mean is as large as 211 mg/100ml.)

16 180 211  = 0.05  =.01

17 If  =.01, then the critical value in z-score would be 2.32. The critical value in average cholesterol level would be If the true mean is 211 mg/100ml, with a power of 0.95, i.e.  =.05, we would accept the null hypothesis when the sample average is less than  = 0.05 Let the unknown sample size to be n.  =.01 118 211

18 Set the two equations equal to each other: Solve for n: So, the sample size needed is 35. 118 211

19 Formula: The estimated sample size for a one- sided test with level of significance  and a power of 1   to detect a difference of  a –  0 is,

20 For a two-sided test, the formula is

21 Example: For testing hypothesis that H 0 :  =  0 = 70 v.s. H a :   70 (Two-sided Test) with a level of significance of  =.05. Find the sample size so that one can have a power of 1   =.90 to reject the null hypothesis if the actual mean is  a = 80. The standard deviation, , is approximately equal 15.

22  =.05 1   =.90   =.10 The sample size needed is 24. z  /2 = z.025 = 1.96 z  = z.1 = 1.28 

23 Example: For testing hypothesis that H 0 :  =  0 = 70 v.s. H a :  < 70 (One-sided Test) with a level of significance of  =.05. Find the sample size so that one can have a power of 1   .95 to reject the null hypothesis if the actual mean is  a  60 (or the actual difference in means is 10). The standard deviation, , is approximately equal 15.

24 Example: For testing hypothesis that H 0 :  =  0 = 70 v.s. H a :   70 (Two-sided Test) with a level of significance of  =.05. Find the sample size so that one can have a power of 1   .95 to reject the null hypothesis if the actual mean is  a  75. The standard deviation, , is approximately equal 15.

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