Biostatistics 110A/F08: Hypothesis Testing for Means - Assignment #8 Solutions, Assignments of Biostatistics

Download Biostatistics 110A/F08: Hypothesis Testing for Means - Assignment #8 Solutions and more Assignments Biostatistics in PDF only on Docsity! Biostat 110A/F08 Solution key for Assignment #8 1. a. Let μ1= Mean for population 1 and μ2 = Mean for population 2. Our hypotheses are H0 : μ1=μ2 vs. HA : μ1=μ2 Test statistic 1.6 X - X = 25 20 + 25 20 X - X = Z 2121 Critical Region: Reject H0 in favor of HA at α =0.05 if |Z|  z1-α/2 = 1.96 b. d=2/1.6=1.5811. π(2) = Φ(1.5811 - 1.96)+Φ(-1.5811 - 1.96) = Φ(-0.38) + Φ(-3.54)  .352 c. Let n1 = n2 = n. Note that π(2) = π(-2) . π(2) = Φ(n ∙ 2/40-1.96) + Φ (-n ∙ 2/40-1.96)  Φ(n ∙ 2/40-1.96) Now set π(2) = 0.80 and solve for n: n  2/40-1.96 = Φ -1 (0.80) = 0.842 Or, n= (1.96+0.842)  (2/40) = 8.86. Or n=78.5. So take n 1 = n 2 = 79 for total of 158. d. H0 : μ1=μ2 vs. HA : μ1=μ2 . Test statistic: 1.66- = .084.257 122-120 = n 1 + n 1 S X - X =t 21 p 21 Critical Region: Reject H0 in favor of HA at α =0.05 if |t|  t 0.975[48] = 2.01 Hence, we fail to reject H0 at 5%. .10 < P-value < .20 using the table. (Using Stata: .1034) e. Intuitively thinking |μ1-μ2|=4 is further away from Null hypothesis than |μ1-μ2|=2 , which means you have higher power. You can also prove it by direct calculation of power at |μ1-μ2|=4 and compare it with power at |μ1-μ2|=2. 2. (Hint) Note that when n1=n2 we have 2/)S + S( = S 22212p . So the EŜ 's are exactly the same. 3. a. These two methods did not use the same child twice. 20 children were randomly assigned to the two groups, 10 to each group. This makes the statistics calculated from each group independent. b. Assuming σ1 = σ2=σ, we have Sp 2 = 108.39 i.e. Sp =10.411. H0 : μ1=μ2 vs. Ha : μ1=μ2 Test statistic: 1.267- = 10/210.411 X - X =t 21 Critical Region: Reject H0 in favor of HA at 1% if |t|  t 0.995[18] = 2.878. So, we fail to reject H0 at α= 0.01 c. 99% CI for μ1-μ2 is : (80.5-86.4)  2.878 (10.441) (2/10) = (-19.3,7.5) d. CI does include( or cover) 0. e. CI means fail to reject H0, since there is a 1-1 relationship between 99% CI and two-sided test at 1%. f. H0 : μ1=μ2 vs. Ha : μ1=μ2 , Since the sample sizes are equal we have: t 0.995 = t 9, 0.995 Test statistic t = (80.5-86.4)/((152.5/10+64.27/10)) = -1.267 Critical Region: Reject H0 in favor of HA at α = 0.01 if |t|  t 0.995  3.25 Hence fail to reject H0 at 1%. g. 99% CI for μ1-μ2 is : (80.5-86.4)  3.25 (10.411) (2/10) -> (-21, 9.2). CI includes 0; because the same estimate of the standard error is used for the CI and test, there is a perfect relationship between the 99% CI the two-sided test. h. P-value for t-test using equal variance is = 0.2212 P-value for approximate t-test is = 0.243 4. 8.74 = S ; .18- = D D H0 : μD=0 vs. Ha : μD<0 Test Statistic: t=-5.04 Critical Region: Reject H0 in favor of HA at 5% if tt.05[5]= -t.95[5]=-2.015 Hence reject at 5%. (Drug lowers blood pressure) b. H0 : μD=0 vs. HA : μD=0. Test statistic: t= - 5.041 Critical Region: Reject H0 in favor of HA if |t|  t 0.975[5] = 2.571 Hence reject at 5%. (Drug changes blood pressure). c. Must have a strong a priori evidence to use a) otherwise use(b). d. 95% CI is ( -27.17, -8.83); 99% CI is (-32.39, -3.61) 5. Let X= drug measurement and Y= saline measurement. a. H0 : μx=μy vs. Ha : μx=μy Test statistic: t= (28.183 - 24.1)/ (1.66667/6 +7.0/5) = 3.15 , Use t b. H0 : μx=μy vs. Ha : μx>μy Test statistics: t= 3.15, Use t c. 95% CI : Equivalent to test (a) 7. Let S stands for smokers and NS stands for non smokers. μS= Mean blood-lead conc. for children born to smokers. μNS = Mean blood-lead conc. for children born to nonsmokers. a. H0 : μS=μNS vs. Ha : μS>μNS c. The following will be changed : i) Degrees of freedom for t-statistic: df = 50 (almost no change at all) ii) Critical region of the test (but since the df's are almost the same this will be essentially the same too). The following will not change at all (because the sample sizes are the same!) i) test statistic ii) estimated standard error 8.a.Let pD= Proportion of drinkers having hearing loss pND=Proportion of nondrinkers having hearing loss. H0 : pD=pND vs. HA : pD = pND b. We will use large sample test for proportions (two independent samples). The pooled .271 = 20000/))10000(.262+)(10000(.28 = p̂ Test statistic: Z= (.2800-.2620)/(.2710(1-.2710) (1/10000+1/10000)) = 2.8636 Critical Region: Reject H0 in favor of HA at 5% if |Z|1.96 Hence reject H0 at 5%. P-value=.0042 c. It is highly significant statistically. But this does not necessarily imply scientific significance! d. 95% CI for pD - pND is given by: [(.2800-.2620)  1.96 (.2710(1-.2710) (1/10000+1/10000))] = [0.018 .0123201] -> (.0057, .0303)