Chapter 14: 1-6, 9, 12; Chapter 15: 8 Solutions When is it appropriate to use the normal approximation to the binomial distribution?

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1 Chapter 14: 1-6, 9, 1; Chapter 15: 8 Solutions 14-1 When is it appropriate to use the normal approximation to the binomial distribution? The usual recommendation is that the approximation is good if np > 5 and ( p) n 1 > Describe the function of a continuity correction When is it most useful? The continuity correction is used to make the probabilities evaluated under a normal approximation to the binomial closer to the actual binomial probabilities It is used when the binomial parameter, p, is known This means that the continuity correction is good to use if you are trying to evaluate a probability when p is known When testing hypotheses, we often need to evaluate the p-value A closer estimate of the p-value can be obtained from the normal approximation to the binomial distribution using the continuity correction When evaluating confidence intervals for the parameter p, simulation study results suggest that the continuity correction should not be used As an example, the P( X 4 p 5, n 10) using Table A1 is given by The normal approximation to the binomial is that X N μ np, σ np 1 p As a result, standardizing X, the z-statistic is given by ( ( )) X μ X np Z σ np( 1 p) Without the continuity correction, we would approximate this probability by 4 5 P( X 4) P σ 5( 075) 4 5 P Z 5( 075) P( Z 1095) With the continuity correction, be540-binomial-ch14-15v1-solutionsdoc 1

2 45 5 P( X 45) P σ 5( 075) 45 5 P Z 5( 075) P( Z 14606) The normal approximation with the continuity correction comes closer to the actual probability from the Binomial distribution If we evaluate a probability greater than or equal to a given value, we subtract 05 for the continuity correction Thus, the P( X 4 p 5, n 10) using Table A1 is given by 041 Without the continuity correction, we would approximate this probability by 4 5 P( X 4) P σ 5( 075) 4 5 P Z 5( 075) P( Z 1095) 0138 With the continuity correction, 35 5 P( X 35) P σ 5( 075) 35 5 P Z 5( 075) P( Z 07303) 033 Once again, with the continuity correction, the approximation is closer 14-3) What factors affect the length of a confidence interval for a proportion? Explain The factors are the sample size, the degree of confidence (α level), and the estimate of the proportion This is true since the length is proportional to z np( 1 p) ˆ ˆ α ) When your are working with a difference in proportions, why does the estimated standard error used in the construction of a confidence interval differ from that used in a hypothesis test? be540-binomial-ch14-15v1-solutionsdoc

3 A confidence interval for a difference in proportions (means) is constructed assuming that there may be a true non-zero difference, meaning that the parameters for each group may actually differ If the parameters (p) differ, they have different variances and the different estimates should be used A hypothesis test of equal proportions is constructed under the assumption that the parameters in the two groups are equal If this is true, a common estimate of the parameter using all the data is better than two separate estimates of the parameter 145) Suppose that you select a random sample of 40 children from the population of newborn infants in Mexico The probability that a child in this population weights at most 500 grams is 015 a) For the sample of size 40, what is the probability that four or fewer of the infants weigh at most 500 grams? Compute the exact binomial probability We interpret the statement weigh at most 500 grams to mean weigh less than 500 grams In order to compute this probability exactly, we need to evaluate the probability that there are 0 infants, 1 infant, infants, 3 infants, or 4 infants that weigh less than 500 grams These probabilities are given by n x n x P( X x) p ( 1 p) x 40 x ( 015 ) ( 085 ) 40 x x We use a spreadsheet to culate these probabilities Notice that 1, 40, ( 39) ( 39)( 38) , ( 39)( 38)( 37) 9880, and ( ) 4 3 ( )( 4) n p x n-x n choose x p to x (1-p) to (n-x) Prob(Xx) Source: be540c14xls Sum 0633 b) Using the normal approximation to the binomial distribution, estimate the probability that four or fewer of the children weight at most 500 grams be540-binomial-ch14-15v1-solutionsdoc 3

4 We use the continuity correction to evaluate this, where X N μ np, σ np 1 p σ np 1 p 51 so that ( ( )) and μ ( ) while ( ) σ P( X 45) P σ 58 P( Z 664) 055 c) Do these two methods provide consistent results? The results are pretty close 14-6 A study was conducted to investigate the relationship between maternal smoking during pregnancy and the presence of congenital malformations in the child Among children who suffer from an abnormality other than Down s syndrome or an oral cleft, 38% have mothers who smoked during pregnancy This proportion is homogeneous for children with various types of defects a) If you were to select repeated samples of size 5 from this population, what could you say about the distribution of sample proportions? List three properties I would expect the distribution of sample proportions of children with smoking mothers to be binomially distributed with mean 038 The properties would be the usual binomial properties ( possible outcomes (ie smoke, don t smoke); n trials independent; the probability of smoking is constant for each trial) b) Among the samples of size 5, what fraction has a sample proportion of 045 or higher? With a sample proportion of 45, there are x115 smokers In order to have a sample proportion greater than 45, we would need to have 1 or more smokers The question can be stated as P( X 1 p 38, n 5 )? We will use the continuity correction in approximating this probability Note that np 8 and σ 5( 038)( 1 038) Using a continuity correction, P( X 115) P σ 38 P( Z 1385) 0084 c) What fraction has a sample proportion of 00 or lower? With a sample proportion of 0, there are x5 smokers In order to have a sample proportion less than or equal to 0, we would need to have 5 or fewer smokers The question P X 5 p 38, n 5? We will use the continuity correction in can be stated as ( ) be540-binomial-ch14-15v1-solutionsdoc 4

5 approximating this probability Note that np 8 and σ 5( 038)( 1 038) Using a continuity correction, 55 8 P( X 55) P σ 38 P( Z 1133) 019 d) What value of p cuts off the lower 10% of the distribution? This question can be interpreted as asking what sample proportions, ˆp would fall in the lower 10% of the distribution of all possible values In order to determine this, we need to work backwards The lower 10% corresponds to a 18 P Z z, so that ( ) x 8 Using this value of z, we need to determine a value of x such that 18 or x Rel that with the continuity correction, this value of x corresponds to a count of 465 This means that sample proportions p ˆ 4/5 16 would be in the lower 10%, while sample proportions p ˆ 5/5 would be above 10% of the sampling distribution 14-9 In NY City, a study was conducted to evaluate whether any information that is available at the time of birth can be used to identify children with special educational needs In a random sample of 45 third-graders enrolled in the special education program of the public school system, 4 have mothers who have had more than 1 years of schooling a) Construct a 90% confidence interval for the population proportion of children with special educational needs whose mothers have had more than 1 years of schooling We use a normal approximation to the confidence interval: pˆ( 1 pˆ) pˆ ± z95 n ( 091) ± ± 054 (0034,014) b) In 1980, % of all third-graders enrolled in the NY city public school system had mothers who had had more than 1 years of schooling Suppose you wish to know whether this proportion is the same for children n the special education program What are the null and alternative hypotheses? H : 0 0 p H : p 0 a be540-binomial-ch14-15v1-solutionsdoc 5

6 c, d Conduct the test at the 005 level of significance What do you conclude? 4 p ˆ z 137 0( 078) We reject the null hypothesis if z > 196 Since this is true, we reject the null hypothesis e) If the true population proportion of children with special educational needs whose mother have had more than 1 years of schooling is as low as 010, you want to risk only a 5% chance of failing to reject the null hypothesis If you are conducting a two-sided test at the 005 level of significance, how large a sample would be required? We use the sample size formula given on page 331 to solve this problem The formula is given by z ( ) ( ) α/ p0 1 p0 zβ p1 1 p 1 n p1 p0 In this problem, p 0 0, p 1 01, α β 005, z α / 196, z β 1645, so that ( ) ( ) n so select a sample of 119 subjects Suppose you are interested in investigating the factors that affect the prevalence of tuberculosis among intravenous drug users In a group of 97 individuals who admit to sharing needles, 47% had a positive tuberculin skin test result; among 161 drug users who deny sharing needles, 174% had a positive test result a) Assuming that the population proportions of positive skin test results are in fact equal, estimate their common value, p npˆ pˆ n npˆ n ( ) ( ) b, c) Test the null hypothesis that the proportions of intravenous drug users who have a positive tuberculin skin test results are identi for those who share needles and those who do not State your conclusion be540-binomial-ch14-15v1-solutionsdoc 6

7 H : p p 0 1 H : p p a 1 We reject the null hypothesis if 196 z ˆ ( ) ( 1 ) ( 1 ) pˆ1 p p1 p pˆ pˆ pˆ pˆ n1 n (8) (8) z > Since this is not true, we fail to reject the null hypothesis, and conclude that the current data is insufficient to conclude that there is a difference in proportions d) Construct a 95% confidence interval for the true difference in proportions ( 1 ) ( 1 ) pˆ pˆ pˆ pˆ pˆ ˆ 1 p ± zα / n n ( ) ( ) ± ± ± 0104 ( 0031,0177) 158 The following data come from a study designed to investigate drinking problems among college students In 1983, a group of students were asked whether they had ever driven an automobile while drinking In 1987, after the legal drinking age was raised, a different group of college students were asked the same question Year Year Drove while Drinking Total yes no Total be540-binomial-ch14-15v1-solutionsdoc 7

8 a,b) Use a the chi-square test to evaluate the null hypothesis that the population proportions of students who drove while drinking are the same in the two endar years What are your conclusions? Ho: Population proportions driving while drinking are homogenous between the years Ha: Population proportions are not homogeneous between years Year Year Drove while Drinking Total yes no Total Prop(Yes Expected Year Year Drove while Drinking Total yes no Total check sum (Obs-Exp)/Exp Year Year Drove while Drinking Total yes no Total Chisq check sum We compare the chi-square statistic, χ 5535 with the criti value from a chi-square distribution (equal to 384 when α 005 with 1 degree of freedom) see Table A8 We conclude that the population proportions are not homogeneous between years c) Again test the null hypothesis that the proportions of students who drove while drinking are identi for the two endar years using the test in Section 146 H : p p 0 1 H : p p a 1 be540-binomial-ch14-15v1-solutionsdoc 8

9 We reject the null hypothesis if 196 In fact, ( ) z χ z ˆ ( ) ( 1 ) ( 1 ) pˆ1 p p1 p pˆ pˆ pˆ pˆ n1 n (58) 4(58) z > Since this is true, we reject the null hypothesis d) Construct a 95% confidence interval for the true difference in population proportions pˆ pˆ pˆ pˆ pˆ1 pˆ ± zα / n1 n 047( 053) 037( 063) 01± ± ± 006 (0074,016) ( 1 ) ( 1 ) 1 1 be540-binomial-ch14-15v1-solutionsdoc 9

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