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MTH 4452 Test #2 - Solutions Fall 2007 Pat Rossi Name Instructions. Show CLEARLY how you arrive at your answers 1. In a study, 31 patients with an inflammation and swelling of the Achilles tendon (Achilles Tendinopathy) were examined and found to have Achilles tendons with a mean diameter of 9.8 mm and a standard deviation of 1.95 mm. The mean diameter for Achilles tendons in the general population is 5.97 mm. Is there sufficient evidence to indicate that the average diameter of Achilles tendons of patients with Achilles Tendinopathy is greater than 5.97 mm? Test at the 5% level of significance. First of all, we suspect that the average diameter of Achilles tendons of patients with Achilles Tendinopathy is greater than 5.97 mm. If we consider only those people with Achilles Tendinopathy, and let µ be the mean of the population of people with Achilles Tendinopathy, then our hypotheses are: H0 : µ = µ0 = 5.97 Ha : µ > µ0 = 5.97 Next, since the size of the sample, n = 31, is greater than 30, we can assume that the sampling distribution for the sample means is normal with mean µ and standard 0 . deviation (standard error) σ x̄ = √σn ≈ √sn . Our test statistic is z = x̄−µ σ x̄ Also, since we are testing at the 5% level of significance, α = 0.05. Our test is a one-tailed test, so our rejection region is as shown below: Rejection Region Area = 0.05 z" = 1.645 Note that zα = 1.645. Observe: z = x̄−µ0 σ x̄ ³ ´ = 10.936 = 9.8−5.97 1.95 √ 31 Since z > zα = 1.645, z is in the rejection region, and therefore, we reject H0 and accept Ha . There is sufficient evidence to indicate that the average diameter of Achilles tendons of patients with Achilles Tendinopathy is greater than 5.97 mm. 2. The addition of MMT to gasoline has caused concern about human exposure to Mn. In a study of ambient air concentrations of fine Mn, samples of air were taken in California and at National Park sites. The information is given below. Mean Standard Deviation Number of Sites National Parks California 0.94 2.8 1.2 2.8 36 26 (a) At the 5% significance level, test to see if there is sufficient evidence that the mean concentrations of Mn differ at the two types of sites. (Even though n2 < 30, use the large sample z-test.) Since we are testing to see if the mean concentrations of Mn differ, this will be a two-tailed test. H0 : (µ1 − µ2 ) = D0 = 0. Ha : (µ1 − µ2 ) 6= D0 = 0. We are testing at the 5% significance level, so α = 0.05. Since this is a two-tailed test, we are interested in α 2 = 0.025. Our critical values are ±z α2 = ±1.96. Our rejection region is shown below: Area = 0.025 Area = 0.025 !z"/2 = !1.96 z"/2 = 1.96 Rejection Region 2 Our test statistic is z = 0.58441 Observe: z = (µ1 −µ2 )−D0 σ x̄1 −x̄2 (µ1 −µ2 )−D0 , where σ x̄1 −x̄2 = (0.94−2.8)−0 0.58441 σ x̄1 −x̄2 = r s21 n1 + s22 n2 = r (1.2)2 36 + (2.8)2 26 = = −3.1827 Since z = −3.1827 is in the rejection region, we reject H0 and accept Ha . We conclude that there is sufficient evidence that the mean concentrations of Mn differ at the two types of sites. (b) What is the p-value of this test? Since this is a two-tailed test, the p-value is P (|z| > 3.1827) . (i.e. The probability of getting a z-value that is farther from the mean than −3.1827.) Note: the area of the tails corresponding to z = ±3.1827 is 2 · (0.0007) = 0.0014. The p-value is 0.0014. 3. A random sample of 1400 observations from a binomial population produced x = 529 successes. (a) If your research hypothesis is that p differs from 0.4, then what hypotheses should you test? H0 : p = p0 = 0.4 Ha : p 6= p0 = 0.4 Note: This is a two-tailed test. (b) Calculate the test statistic and its p-value. At the 1% level of significance, use the p-value to determine if there is sufficient evidence to indicate that p is different from 0.4. First, note that np0 = (1400) (0.4) = 560 > 5, and nq0 = (1400) (0.6) = 840 > 5. Thus, we have the right to assume that the sampling distribution for pˆ is normal. pˆ = x n = 529 1400 = 0.378 The test statistic is z = z= p̂−p0 σ p̂ = 0.378−0.4 0.0131 p̂−p0 , σ p̂ where σ p̂ = q p0 q0 n = q (0.4)(0.6) 1400 = 0.0131 = −1.6794 Since this is a two-tailed test, the p-value is P (|z| > 1.6794) . (i.e., The probability of getting a z-value that is farther from the mean than −1.6794.) Note: the area of the tails corresponding to z = ±1.6794 is 2 · (0.0465) = 0.093 The p-value is 0.093. At the 1% level of significance, α = 0.01 3 Since the p-value is greater than α = 0.01, we do not reject H0 . There is not sufficient evidence to indicate that p is different from 0.4. 4. An experiment was conducted to test the effect of a new drug on a viral infection. The infection was induced in 100 mice, and the mice were randomly split into two groups of 50. The first group, the control group, received no treatment for the infection. The second received the drug. After a 30 day period, the proportions of survivors, p1 and p2 , in the two groups were found to be 0.36 and 0.60, respectively. (a) Is there sufficient evidence, at the 5% significance level, to indicate that the drug is effective in treating the viral infection? H0 : (p1 − p2 ) = D0 = 0 Ha : (p1 − p2 ) < D0 = 0 (Our test is a one-tailed test.) At the 5% significance level, α = 0.05. The rejection region is shown below: Area = 0.05 !z" = !1.645 Rejection Region For the sake of being thorough, note that n1 p1 = 18; n1 q1 = 32; n2 p2 = 30; n2 q2 = 20. Since all of these are greater than or equal to 5, we can assume normality. First note, that since the hypothesis is centered about zero, we will use a pooled estimator for p̂ , when computing sp̂1 −p̂2 . p̂ = x1 +x2 n1 +n2 = 18+30 50+50 σ p̂1 −p̂2 ≈ sp̂1 −p̂2 = = 0.48 q p̂q̂ n1 + Our test statistic is z = p̂q̂ n2 = q (0.48)(0.52) 50 (p1 −p2 )−D0 σp̂1 −p̂2 = + (0.48)(0.52) 50 (0.36−0.60)−0 0.09992 = 0.09992 = −2.4019 Since z = −2.4019 is in the rejection region, we reject H0 and accept Ha . 4 There is sufficient evidence to indicate that the drug is effective in treating the viral infection. 5