Introductory Biostatistics Practice Test III Solutions: Estimation and Hypothesis Testing, Exams of Biostatistics

Download Introductory Biostatistics Practice Test III Solutions: Estimation and Hypothesis Testing and more Exams Biostatistics in PDF only on Docsity! PubHlth 540 Introductory Biostatistics Page 1 of 8 Practice Test III Unit 6 – Estimation Unit 7 – Hypothesis Testing SOLUTIONS 1. a. True or False. A hypothesis test for which the type I error occurs with probability α has probability of type II error equal to (1 - α). Answer: FALSE Solution: A type I error can only occur when the null hypothesis is true and a type II error can only occur when the alternative hypothesis is true. b. True or False. If a one sided test indicates that the null hypothesis can be rejected at the 5% level, then a two sided test performed on the same set of data is necessarily significant at the 5% level. Answer: FALSE Solution: The significance level for a two sided test is the sum of two areas under the curve whereas it is the sum of one area under the curve when the test is one sided. Thus, for a given test statistic value, the two sided p-value will always be larger than the one sided p- value. Accordingly, if a one sided p-value is .05 or close to it, then the two sided p-value cannot necessarily be less than or equal to .05. Practicetest3_solutions.doc PubHlth 540 Introductory Biostatistics Page 2 of 8 c. True or False. For a given sample variance s2 and sample mean X , a 90% confidence interval for an unknown mean μ is narrower than a 99% confidence interval. Answer: TRUE Solution: All other things constant, the confidence coefficient multiplier for a 90% CI is smaller than the confidence coefficient multiplier for a 99% CI. As a result the width of the CI, being [ upper limit – lower limit ] will be narrower. For example, 90% and 99% confidence intervals based on the Normal distribution have multipliers 1.645 and 2.576, respectively. d. True or False. An investigator is performing a t-test for which the assumptions are satisfied could, in the absence of a student’s t-distribution tables, use a Normal(0,1) probability table provided the degrees of freedom is sufficiently large. Answer: TRUE Solution: Consider Type I Error (2 sided) Test Statistic Critical Value of Test Statistic .10 t df=30 1.697 t df=50 1.676 t df=100 1.66 Normal (0,1) 1.645 .01 t df=30 2.75 t df=50 2.678 t df=100 2.626 Normal (0,1) 2.576 Practicetest3_solutions.doc PubHlth 540 Introductory Biostatistics Page 5 of 8 4. Consider testing the null hypothesis that the mean of a normal probability distribution is μ = 98.6. The population standard deviation is known and is equal to σ = 1.5. You draw a sample of size 25 from this distribution and calculate a sample mean. If it is of interest to perform a two sided test of the null hypothesis with a pre-specified type I error equal to 0.02, what values must the sample mean lie between in order for the null hypothesis to be not rejected? Answer: Between 97.9 and 99.29 Solution: (1) For Z ~ Normal(0,1) and desired two sided type I error = .02, obtain the critical region for the Z test. To do this, obtain from a Normal (0,1) calculator that Pr [ -2.326 < Z < +2.326 ] = .98 (2) Identify the null hypothesis distribution of n=25X n=25X ~ Normal ( 2 2 2 NULL X 1.5μ = 98.6 and σ = = n 25 σ ) under null. (3) Solve for critical region values of n=25X by using the standardization formula backwards. Here, it takes the form NULL n=25 NULLX X(-2.326) [σ ] + μ X (+2.326) [σ ] + μ≤ ≤ n=25= (-2.326) [1.5/ 25] + 98.6 X (+2.326) [1.5/ 25] + 98.6≤ ≤ = 97.9, 99.29 Practicetest3_solutions.doc PubHlth 540 Introductory Biostatistics Page 6 of 8 5. A randomized trial of “diet continuation” versus “diet termination” was performed to evaluate whether diet termination is associated with a reduction in normal development among children with PKU. The outcome measured was Bitnet IQ, a measure of mental retardation. The following summary statistics were obtained from complete data on 22 children: Group Diet Continuation Diet Termination Sample size, n 12 10 Sample mean IQ, X 103.6 99.8 Carry out the appropriate statistical hypothesis test to evaluate the study data. In developing your answer you may assume that the observations are distributed normal in both groups. You may also assume that the population standard deviations of a Bitnet IQ are σ = 13 in the diet continuation population and σ = 17 in the diet termination population. Answer: - Hnull: μC = μT Halternative: μT < μC , one sided - Z statistic = 22 T C CT T C T C T C (X -X ) - [0] σσ; where SE (X -X )= + SE (X -X ) n n ~ Normal(0,1) under null - Z 22 2 2 CT T C σ(99.8-103.6) σ 17 130.5793; where 6.56= + + 6.56 n n 10 12 = = − = - pvalue = 0.28 - Do not reject. Conclude these data are insufficient evidence of an effect of diet continuation on Bitnet IQ in children with PKU Practicetest3_solutions.doc PubHlth 540 Introductory Biostatistics Page 7 of 8 6. It has been reported anecdotally that the onsets of menses for women living together (for example, a college dormitory) are more synchronized that those of women living separately. An investigator for the Journal of Irreproducible Results decides to investigate this hypothesis. During the month of December 2008, the dates of mensis onset were recorded for two samples of women. The data for the first sample consisted of mensis onset histories for 10 women who had lived together continuously in a single dormitory at Smith College in Northampton, MA during 2008. The sample variance was calculated to be s2 = 30.3. The data for the second sample included mensis onset histories for 25 women of comparable ages attending the University of Massachusetts/Amherst, all of whom lived in separate dwellings. The sample variance for this sample was s2 = 69.7 Using the appropriate statistical hypothesis test, do these data suggest that mensis onset dates are more synchronized for women living together relative to women living separately? Answer: - Hnull: Halternative: , one sided 2 2separately togetherσ = σ 2 2 separately togetherσ > σ - F statistic = 2 separately 2 together S S ~ F24,9 under null - F 69.7 2.3 30.3 = = - pvalue = 0.09726 - Do not reject. Conclude these data are insufficient evidence of a synchrony of menses among women who live together, relative to women who do not live together. Practicetest3_solutions.doc