540
Introductory PubHlth Biostatistics
Page 1 of 7
Unit 6 Estimation Week #10 - Practice Problems SOLUTIONS
1. An entomologist samples a field for egg masses of a harmful insect by placing a yardsquare frame at random locations and carefully examining the ground within the frame. A simple random sample of 75 locations selected from a county's pasture land found egg masses in 13 locations. Compute a 95 confidence interval estimate of all possible locations that are infested. Answer: (.0876, .2590) Solution: The setting is estimation of a binomial proportion . In this exercise, the number of trials is N=75. Since this is sufficiently large, we can obtain a confidence interval using ^ ^ ^ ( z1- /2 ) SE ( ) using the standard error formula "(3)" that appears on page 63. Thus, the calculations are
X 13 = = .1733 N 75 ^ = X=.1733 X=
^ SE=
X (1-X ) N
=
(.1733)(.8267 ) = .0437
75
z1- /2 = z.975 = 1.96 ^ ^ ^ ( z1- /2 ) SE ( ) = .1733 (1.96)(.0437) = (.0876, .2590)
Wk10_solutions.doc
PubHlth 540
Introductory Biostatistics
Page 2 of 7
2. Alzheimers' disease has a poorer prognosis when it is diagnosed at a relatively young age. Suppose we want to estimate the age at which the disease was first diagnosed using a 90% confidence interval. Under the assumption that the distribution of age at diagnosis is normal, if the population variance is 2=85, how large a sample size is required if we want a confidence interval that is 10 years wide? Answer: n=10 Solution: Recall: The 90% confidence interval is given by X ( z.95 ) SE(X) , width = [upper limit of CI] - [lower limit of CI] = X+(z.95 )SE(X) - X-(z.95 )SE(X)
= X+(z.95 )SE(X) - X + (z.95 )SE(X) = (2)(z.95 )SE(X) = (2)(z.95 ) n
Substituting width = 10 allows us to write 10= (2)(z.95 ) n 5= (z.95 ) n (z ) n = .95 5
(z ) n= .95 5
2
(1.645) 2 (85) n= = 9.2005 25
Rounding up yields the required sample size of 10.
Wk10_solutions.doc
PubHlth 540
Introductory Biostatistics
Page 3 of 7
3. The National Health and Nutrition Examination Survey of 1975-1980 give the following data on serum cholesterol levels in US males. Age, years 20-24 25-34 Population Mean, 180 199 Population Standard Deviation, 43 49
Group 1 2
Suppose the distribution of serum cholesterol is normal in each age group. If you draw simple random samples of size 50 from each of the two groups, what is the probability that the difference between the two sample means (Group 2 mean Group 1 mean) will be more than 25?
Answer: .257 Solution: On page 46 of the lecture notes, at the bottom of the page, we learn that
2 2 1 2 (X 2 -X1 ) is distributed Normal[ ( 2 -1 ), ( + ) ] n1 n 2
2 Let X1 = Average among age 20-24. It is distributed Normal(1=180, X1 = 2 X 2 = Average among age 25-34. It is distributed Normal(1=199, X2 =
12
n1
2 2
= =
432 ) 50 492 ) 50
n2
Thus, Y=(X 2 - X1 ) . is distributed Normal with
Y = ( X2 - X2 ) = 19
2 2 432 492 2 Y = 1 + 2 = + = 85 n1 n 2 50 50
Now we use the z-score method that we learned in Unit 5, Normal Distribution, and in particular the z-score standardization that is found on page 19 under "(3)", we have that Probability group 2 mean group 1 mean will be more than 25 = Pr[Y > 25]
Y- Y 25-19 =Pr > Y 9.2195 =Pr[Normal(0,1) > 0.65] = .257
Wk10_solutions.doc
PubHlth 540
Introductory Biostatistics
Page 4 of 7
4.
The objectives of a study by Kennedy and Bhambhani (1991) were to use physiological measurements to determine the test-retest reliability of the Baltimore Therapeutic Equipment Work Simulator during three simulated tasks performed at light, medium, and heavy work intensities, and to examine the criterion validity of these tasks by comparing them to real tasks performed in a controlled laboratory setting. Subjects were 30 healthy men between the ages of 18 and 35. The investigators reported a standard deviation of s=0.57 for the variable peak oxygen consumption (1/min) during one of the procedures. Assuming normality, compute a 95% confidence interval for the population variance for the oxygen consumption variable.
Answer: (.21, .59) Solution: (n-1) = 29 Lower limit =
S2 = 0.57 2
2 2 1-/2 = .975; DF=29 =45.722 2 2 /2 = .025; DF=29 =16.047
(n-1)S2
2 1- /2;df=(n-1)
=
(29)(0.57 2 ) = .2061 45.722
Upper limit =
(n-1)S2
2 /2;df=(n-1)
(29)(0.57 2 ) = = .5872 16.047
5. The purpose of an investigation by Alahuhta et al (1991) was to evaluate the influence of extradural block for elective caesarian section simultaneously on several and maternal fetal hemodynamic variables and to determine if the block modified fetal myocardial function. The study subjects were eight healthy parturient in gestational weeks 38-42 with uncomplicated singleton pregnancies undergoing elective caesarian section under extradural anesthesia. Among the measurements taken, were maternal diastolic arterial pressure during two stages of the study. The following are the lowest values of this variable at the two stages. Compute a 95% confidence interval for the difference in diastolic blood pressure between the two stages. Patient ID Stage 1 Stage 2 1 70 79 2 87 87 3 72 73 4 70 77 5 73 80 6 66 64 7 63 64 8 57 60
Wk10_solutions.doc
PubHlth 540
Introductory Biostatistics
Page 5 of 7
Answer: (-0.06, +6.6) Solution: Because two measurements are made on each patient, at stages 1 and 2, these data fit the definition of "paired". The analysis focuses on the differences, per the table below: Patient ID, i di = Stage 2-1 1 9 2 0 3 1 4 7 5 7 6 -2 7 1 8 3
The actual calculations required to complete the solution are similar to the example on page 43. In particular, for this exercise, we have
d=3.25 df=(n-1)=7
2 Sd = 15.643
Sd = 3.9551
SE(d)=
Sd 3.9551 = = 1.3983 n 8
t1- /2;df = t .975;7 = 2.365
95% CI for d =d(t .975;DF=7 )SE(d)
= (3.25) (2.365)(1.3983) = (-0.057, 6.557)
6. A possible environmental determinant of lung function in children is the amount of cigarette smoking in the home. To study this question, two groups of children were studied. Group 1 consisted of 23 nonsmoking children aged 5-9 both of whose parents smoke in the home. Group 2 consisted f 20 nonsmoking children aged 5-9 neither of whose parents smoke. The mean (SD) of FEV1 for group 1 is 2.1 L (0.7) and for the Group 2 children, the mean (SD) of FEV1 is 2.3 L (0.4). Under the assumption of normality, construct a 95% confidence interval for the ratio of the variance of the two groups. What is your conclusion regarding the reasonableness of the assumption of equality of variances? Answer: (1.24, 7.37) Solution: The following link is good here. http://www.psychstat.missouristate.edu/pdf/pdfj.htm Play with it using a known row F distribution probability from the back of your text book. Thus, you will discover that this table works with areas under the curve to the right.
Wk10_solutions.doc
PubHlth 540
Introductory Biostatistics
Page 6 of 7
2 S1 = 0.7 2 S2 = 0.42 2
(n1 - 1) = (23 - 1) = 22 (n 2 - 1) = (20 - 1) = 19
Fn1 -1,n 2 -1;1- / 2 = F22,19;.975 = 2.4783
Fn1 -1,n 2 -1; / 2 = F22,19;.025 = 0.4155
th
For example, to obtain the 97.5 percentile, enter .025 as the area under the curve to the right as shown below. Then click on the left arrow key to obtain the val...