HYPOTHESIS TESTING III: POPULATION PROPORTIONS, ETC.


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1 HYPOTHESIS TESTING III: POPULATION PROPORTIONS, ETC. HYPOTHESIS TESTS OF POPULATION PROPORTIONS Purpose: to determine whether the proportion in the population with some characteristic is or is not equal to some particular value typically, of course, the proportion being tested for is imposed from outside the problem tests involve the probability of getting particular sample results on the basis of an hypothesized value of p, hence involve the probability distribution of p s, the sample proportion on the hypothesis that p has the particular value, say, p 0 p ( 1 p) recall that σ p s = n here of course p is unknown, but since it is assumed to be p 0, then σ ps is assumed to have value σ p s = p ( 1 p0) n 0 In the pvalue approach (too many p s!) we will find probabilities by using NORMDIST with mean = p 0 and σ ps as above. In the critical value approach, the relevant statistic is p s p0 z = σ p s which will be compared to a critical z value found from the z table or from NORMSINV Examples: A direct mail appeal for contributions to a university's alumni must have at least a 10% favorable response to be worthwhile. Before undertaking a fullscale campaign, the fundraisers decide to use a pilot sample of 200 to test the waters. They wish to test their hypothesis at 8% significance. In their sample of 200, 17 respond favorably. What should the fundraisers conclude? 1. we have H 0 : p 0.1 H 1 : p < In proportion problems, the sampling distribution is always normal, and the test statistic is always a z. 3. We ve specified significance level The critical value, for a lower onetailed test, is We will reject H 0 if the calculated z is less than this. Alternatively, we ll reject H 0 if pvalue < By hypothesis, σ ps = (0.1 (1 0.1))/200 = pvalue approach: NORMDIST(17/200,.08,.0212,true) = 0.24 > 0.08, so we fail to reject critical value approach: z = ( ) = > 1.41 so we fail to reject 5. the fundraisers can go ahead with the fullscale appeal hypothesis testing, page 1
2 A veterans advocate claims that male veterans of military service are much more likely to be homeless than nonveterans. The proportion of males who are veterans in the general population is 19%. A survey of 5,233 homeless men finds that 33.6% are veterans. Use this information to test the claim. 1. We should think of the alternative hypothesis first: H 1 : p > 19%, which suggests H 0 : p 19. H 1 : p > 19% 2. p s is normally distributed. 3. Let s use α = σ ps = (19 81 )/5233 = NORMDIST(33.6, 19, , true) = 0 and the null hypothesis can be decisively rejected. 5. Accept the claim. A defense contractor claims that, when properly maintained, the breakdown rate of its trucks under field conditions is less than 5%. To test this claim, army engineers will carefully track the maintenance of 150 trucks and observe their breakdown rate during an extended field exercise. If the null hypothesis is to be tested at α = 0.01, when should the army reject the contractor's claim? H 0 : p 0.05 H 1 : p < 0.05, σ ps = ( /150 = z C = 2.33 for a lower 1% test p sc = ( 2.33) = , so the null hypothesis should be rejected if the breakdown rate is less than 0.86% notice that this is an extremely stringent test: the manufacturer's claim will be accepted only if the null hypothesis is rejected, and that will require a sample breakdown rate of less than 1 per cent. If 3 of the trucks break down during the test, what is the pvalue of the test? p s = = 0.02 NORMDIST(.02,.05, ,true) =.046 Fail to reject: the breakdown rate is at least 5%. hypothesis testing, page 2
3 hypothesis testing, page 3 HYPOTHESIS TESTING IV: ERRORS ERRORS IN HYPOTHESIS TESTING The hypothesis testing procedure does not guarantee that µ or p has or does not have the hypothesized value two kinds of errors may occur: reject a null hypothesis which is true accept a null hypothesis which is false Examples: Hypothesis: My toe is broken. Hypothesis: The bridge is structurally sound. Hypothesis: The prof will ask questions about this topic on the exam. Reject a true null hypothesis filling boxes of Fruit Loops: H 0 : µ 16 oz H 1 : µ < 16 oz. if σ = 0.5 oz, n = 25, α = 0.05, we will reject H 0 whenever X < (z C = z = ( X µ 0 )/σ X ; substitute for z and solve for X) Suppose in fact µ = 16 oz. What proportion of X s will be less than oz.? NORMDIST(15.836, 16, 0.1, TRUE) = 0.05 Any X < will cause us to reject H 0 :, but there is a 5% probability of such X even if H 0 is true Rejecting H 0 is an error that will cause us to stop and adjust our machinery even when it doesn't need it a rejection error or Type I error probability of a rejection error = α or the significance level of the test selecting α sets the probability of a rejection error shutting down machinery unnecessarily imposes costs on firm of lost time and employee waiting. These costs must be weighed against costs of having to honor guarantees or reprocess goods accept a false null hypothesis from above, suppose µ = 15.8, so that H 0 is false. We accept H 0 whenever X > oz. Of course will sometimes be that large: NORMDIST(15.836, 15.8, 0.1, TRUE) = there is 36% probability of accepting X even when it is false This is an error that exposes us to the possibility of having to honor freereplacement guarantees and so on an acceptance error or Type II error the probability of an acceptance error = β β depends on the actual value of µ suppose µ = 15.6: β = NORMDIST(15.836, 15.8, 0.1, TRUE) = ACCEPTANCE AND REJECTION ERRORS INTERACT α and β vary inversely to one another α is chosen as part of the test
4 β depends on the actual value of µ and on the α selected The sketches below illustrate the situation: Figure 1: We are testing H 0 : µ 150 vs. H 1 : µ > 150. We have σ = 60 and n = 9, so that σ X = 20, and we have selected a significance level of 5%, that is, α = We will reject H 0 whenever X 183. If, in fact, µ = 210, we will commit an acceptance error whenever X < 183. This probability is shown as the lightshaded tail in the sketch. In this case the probability of a rejection error is zero because H 0 is in fact false. Figure 2: Changing α will reposition the critical value of Gx and so will change β. Here we continue to suppose the true mean = 210, but selecting α = 0.01, will change X C to 201.6, moving it to the right. Consequently, β increases to Thus we can conclude that choosing α determines the critical value of X in the test and thus determines β increasing α reduces β and viceversa hypothesis testing, page 4
5 Is there a way to reduce both α and β? Increasing the sample size reduces the dispersion in the sampling distribution, reduces, that is, σ X in the example above, let n = 25. Then σ X = 12, and for α = 0.05 X C = = If in fact µ = 210, β = P( X < 169.7). For this probability, z = ( )/12 = 3.36 and β = Clearly, we could reduce α to, say, and still have β < thus, by appropriate choice of sample size, we can control both α and β Designing hypothesis tests consists of balancing three costs and two probabilities: the cost of the test procedure, which increases with sample size; the cost of a rejection error, which has probability α; and the cost of an acceptance error, which has probability β THE POWER OF A TEST Definition: the power of a test = 1 β since β is the probability of accepting a false null hypothesis, power is the probability of rejecting a false null hypothesis hypothesis testing, page 5
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