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Power and Sample Size in Testing One Mean

Power and Sample Size in Testing One Mean. 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  .

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Power and Sample Size in Testing One Mean

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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. 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

  4. Is the mean cholesterol level of the population of 20- to 74-year-olds higher than 180 mg/100ml ? Research Hypothesis H0:  = 180 mg/100ml (or  180 mg/100ml) Ha:  > 180 mg/100ml

  5. Power of the Test

  6. Power of the Test 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 H0 | H0 is false) Power = 1  The power is usually calculated given a simple alternative hypothesis: Ha: m = 211mg/100ml Power = P(reject H0 | Ha is true)

  7. Rejection region Right tail area is 0.05 z 0 1.645 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, wewould reject H0 if z  1.645.

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

  9. 180 195.1 211  = ? If the alternative hypothesisHa: m = ma = 211 mg/100ml,what would be the probability of NOT rejectingH0, i.e.?H0: m = m0 = 180 v.s. Ha: m = ma = 211That is to say “how powerful can this test detect a 31 mg/100ml increase in average cholesterol level?” Power of the test when sample size is 25: Power = 1 –  = 1 – 0.042 = 0.958. 0.042

  10. Rejection region Right tail area is 0.05 z 0 1.645 Rejection region 180 ? 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 ? 187.6

  11. 180 187.6 211  0.00  = ?  = .05 1 – 0  1 If we assume the actual population mean is 211 mg/100ml forHa, what is the probability of acceptingH0, i.e.  ? Power of the test when sample size is 100: Power = 1 –  ? Recall that the power was .958 when sample size was 25.

  12. n = 100 187.6 195.1 n = 25 180 211

  13. Sample Size Estimation

  14. H0:  = 180 mg/100ml (or  180 mg/100ml) v.s.HA:  > 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.)

  15. 180 211  = .01  = 0.05

  16. 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 118 211  = 0.05  = .01 Let the unknown sample size to ben.

  17. 118 211 Set the two equations equal to each other: Solve forn: So, the sample size needed is35.

  18. Formula: The estimated sample size for a one-sided test with level of significance  and a power of 1  to detect a difference of ma – m0 is,

  19. For a two-sided test, the formula is

  20. Example: For testing hypothesis that H0: m = m0 = 70 v.s. Ha: m 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 isma= 80. The standard deviation, s, is approximately equal 15.

  21. z/2 = z.025 = 1.96 z= z.1 = 1.28  = .05 1  = .90  = .10 The sample size needed is24.

  22. Example: For testing hypothesis that H0: m = m0 = 70 v.s. Ha: m< 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 isma= 60 (or the actual difference in means is 10). The standard deviation, s, is approximately equal 15.

  23. Example: For testing hypothesis that H0: m = m0 = 70 v.s. Ha: m 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 isma= 75. The standard deviation, s, is approximately equal 15.

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