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Power of the Test

Power of the Test. Power of a Test, Section 10.7. Type I Error. Tests are set up assuming the null hypothesis is true “ Under the null hypothesis ” We set a level of significance, for example, of 0.05 The risk we are willing to assume of rejecting a correct null hypothesis

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Power of the Test

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  1. Power of the Test Power of a Test, Section 10.7

  2. Type I Error • Tests are set up assuming the null hypothesis is true • “Under the null hypothesis” • We set a level of significance, for example, of 0.05 • The risk we are willing to assume of rejecting a correct null hypothesis • Is this a small Type I error? • We will reject the true hypothesis in 5 out of 100 samples • Costs are associated with making mistakes • If this is too high a risk of rejecting a true hypothesis, what can we do in terms of our test? • Change the decision rule and widen the acceptance region PP 4

  3. Sampling Distribution of Sample Means, Normally Distributed .99 do not reject .95 do not reject 7814.1 -2.58 -1.96 1.96 2.58 Changing the Level of Significance • H0:  = 8000H1:  8000 • When ⍺ = 0.05 • DR: If (-1.96 < Z < 1.96) do not reject the claim • Reject H0 because test statistic = -1.97 • Lower ⍺: Let ⍺ = 0.01 • DR: if (-2.58 < Z < 2.58) do not reject the claim • Do not reject H0:  = 8000 -1.97 Z PP 4

  4. TYPE II Error • Widen the “non- reject” interval • ⍺ (1-⍺) • What if population mean ≠ 8000? • Unknown to statistician, null H is false • Make a second type of error • Type II error, that is, we do not reject a false hypothesis • Probability of not rejecting a false hypothesis =  • Probability of correctly rejecting a false hypothesis = 1 -  • 1 -  is the power of the test • If the null hypothesis is false, by widening the acceptance region, we increase the chance of not rejecting the false hypothesis PP 4

  5. Calculating the Probability of a Type II Error, β • Let • H0: μ≤ 8000 • Statement about mean annual earnings in 1968 • H1: μ > 8000 • Looking for sample evidence that the population mean is greater than 8000 • Given σ = 4191.2 and n = 1976, so the standard error, = 94.28 • Let ⍺ = 0.05, Z.05 = 1.645 • DR: if (Z ≤ 1.645) do not reject PP 4

  6. Sampling Distribution under the null hypothesis 0.05 8000 1.64 Z The DR could beif ( ≤ 8154.63) Do not reject = 8154.63 Calculating the Probability of a Type II Error, β • Ask what is the that has 5% of sample means greater than it? • Solve for the unknown Do not reject ? PP 4

  7. Calculating the Probability of a Type II Error, β • Solve for β • We are not testing a null hypothesis • Must be given a value for the true population mean • Let “true” μ = 8350 • Arbitrary choice • Ask what is the probability of not rejecting the false null hypothesis if the true population mean = 8350 PP 4

  8. Sampling Distribution under the null hypothesis True Sampling Distribution with μ = 8350 1 - β Do not reject 8000 8350 Ask : When do I make a mistake and not reject the false null hypothesis? Answer: When I draw a sample mean from the true distribution that lies within the non-reject region. = βRemember: if ( ≤ 8154.63) Do not reject Calculating the Probability of a Type II Error, β, whenμ = 8350 8154.63 PP 4

  9. Sampling Distribution under the null hypothesis True Sampling Distribution with μ = 8350 .95 Do not reject 8000 8350 = (8154.63 - 8350)/94.28 = -2.07 Calculating the Probability of a Type II Error, β, whenμ = 8350 1 - β 8154.63 0 -2.07 Z P(0 < Z < 2.07) = .4808 β = = .5000 - .4808 = .0192 1 - β = .9808 PP 4

  10. Type II Error • Is this a large or small risk? • It depends on the costs associated with the particular situation • Generally, researchers accept  = 0.05, 0.10, 0.15 • Two points to remember • In order to calculate a Type II error, you must select a value for the “true” population mean • As the acceptance interval widens,  increases • Thus as  decreases,  increases PP 4

  11. Sampling Distribution under the null hypothesis True Sampling Distribution with μ = 8350 Do not reject 8000 8350 The DR isif ( ≤ 8219.67) Do not reject As  Decreases from 0.05 to 0.01,  Increases .99 1 - β 8219.67 Let ⍺ = .01, Z.01 = 2.33. The sample mean that is exceeded only 1% of the time is As ⍺↓ β↑ and (1 - β)↓ PP 4

  12. Sampling Distribution under the null hypothesis True Sampling Distribution with μ = 8300 8000 .95 8300 .4382 .5000 Do not reject 8154.63 -1.54 Type II Error and Power of the Test Depends upon How Different the True Mean Is from the Hypothesized Mean • Hypothesis we are testing is false • H0: ≤ 8000 • Let true population mean = 8300 Z = (8154.63 – 8300)/94.28 = -1.54 P(0 < z < 1.54) = .4382  = .5000 - .4382 = .06181 -  = .4382 + .5000 = .9382 PP 4 ⍺ = .05 and if (Sample Mean ≤ 8154.63) Do not reject

  13. True Sampling Distribution with μ = 8100 8000 8100 8154.63 Z Type II Error and Power of the Test Depends upon How Different the True Mean Is from the Hypothesized Mean • Hypothesis we are testing is false • H0: ≤ 8000 • Let true population mean = 8100 Sampling Distribution under the null hypothesis .2190 Do not reject 0 .579 Z = (8154.63 - 8100)/94.28 = 0.579 P(0 < z < .58) = .2190  = 1 - .281 = .7191 -  = .5000 -.2190 = .281 PP 4 ⍺ = .05 and if (Sample Mean ≤ 8154.63) Do not reject

  14. 8300 8200 8500 if ( ≤ 8154.63) Do not reject Type II Error and Power of the Test Depends upon How Different the True Mean Is from the Hypothesized Mean As the true population mean becomes closer to the mean under the null hypothesis, the probability of “accepting” the false null increases; the power of the test declines. Sampling Distribution under the null hypothesis Do not reject 8000 8154.63 PP 4

  15. Creating the Power Curve • For H0:≤ 8000,  = .05, n = 1976,  = 4191.2, one tail test PP 4

  16. Creating the Power Curve TEST 2 TEST 1 PP 4

  17. How Can We Change the Power of the Test? • For a given , n ,  and type of test • The greater the distance between the true mean and 0, the greater is the power of the test • For given , n,  and true mean • A one tailed test is more powerful than a two tail test • One tail test places all of the rejection region in one tail • For given n, , type of test and true mean • The larger the level of significance, , the smaller is , and the larger is the power, 1- • For given , , type of test and true mean • The larger is the sample size, the greater is the power of the test PP 4

  18. Examples of Type I and Type II Error • M&M/Mars claims that at least 20% of the M&M’s in each package are the new blue color. State the two types of error and the consequences. • H0:  .20 There are enough blues • H1:  .20 There are not enough blues, change production • Type I error • We reject the null hypothesis which is true • The company then increases the number of blues although there are already enough • This changes production and is costly • Type II error • We do not reject the null hypothesis which is false • The company changes nothing, but in fact there are not enough blues • Customer may complain • Type I error is more costly • Set  = 0.01 or 0.005 PP 4

  19. Examples of Type I and Type II Error • You are inspecting bolts that are used to fasten engines to airplanes. The mean strength of the bolts must be at least 100 pounds per square inch (psi). You need to decide whether a shipment of these bolts is OK. You can not test all of the bolts because the test destroys the bolt. • H0:  100 psi The shipment is OK • H1:  < 100 psi The shipment is not OK • Type I error • You believe the shipment is not OK when in fact it is good • You send the bolts back to be reworked, although they do not need to be • Type II error • You “accept” that the bolts are OK when they are not • You keep bolts that are not strong enough • Type II error is more serious • Try to minimize  • As ↑  ↓ • Set  = 0.10 PP 4

  20. Do problems in Weier’s • No associated Cengage problems • Section 10.7 Power of a Test, pp. 343-349 Work on problems: 10.79-10.86. • Solutions are on website PP 4

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