Unit 6 Estimation Week #10 - Practice Problems SOLUTIONS

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1 PubHlth 540 Itroductory Biostatistics Page of 7 Uit 6 Estimatio Week #0 - Practice Problems SOLUTIONS. A etomologist samples a field for egg masses of a harmful isect by placig a yardsquare frame at radom locatios ad carefully examiig the groud withi the frame. A simple radom sample of 75 locatios selected from a couty s pasture lad foud egg masses i 3 locatios. Compute a 95 cofidece iterval estimate of all possible locatios that are ifested. Aswer: (.0876,.590) The settig is estimatio of a biomial proportio π. I this exercise, the umber of trials is N75. Sice this is sufficietly large, we ca obtai a cofidece iterval usig ˆ π ± ( z- α/ ) SEˆ ( ˆ π ) usig the stadard error formula (3) that appears o page 63. Thus, the calculatios are X 3 X.733 N 75 ˆ π X.733 X-X ˆ ( ) (.733)(.867) SE.0437 N 75 z- α / z ˆ π ± z SE ˆ ˆ π.733 ± (.96)(.0437) (.0876,.590) ( ) ( ) - α/

2 PubHlth 540 Itroductory Biostatistics Page of 7. Alzheimers disease has a poorer progosis whe it is diagosed at a relatively youg age. Suppose we wat to estimate the age at which the disease was first diagosed usig a 90% cofidece iterval. Uder the assumptio that the distributio of age at diagosis is ormal, if the populatio variace is 85, how large a sample size is required if we wat a cofidece iterval that is 0 years wide? Aswer: 0 Recall: The 90% cofidece iterval is give 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 )SE(X) - X + (z )SE(X) ()(z ).95 SE(X) ()(z.95) Substitutig width 0 allows us to write 0 ()(z.95) 5 (z.95) (z.95) 5 (z 5.95) (.645) (85) Roudig up yields the required sample size of 0.

3 PubHlth 540 Itroductory Biostatistics Page 3 of 7 3. The Natioal Health ad Nutritio Examiatio Survey of give the followig data o serum cholesterol levels i US males. Group Age, years Populatio Mea, μ Populatio Stadard Deviatio, Suppose the distributio of serum cholesterol is ormal i each age group. If you draw simple radom samples of size 50 from each of the two groups, what is the probability that the differece betwee the two sample meas (Group mea Group mea) will be more tha 5? Aswer:.57 O page 46 of the lecture otes, at the bottom of the page, we lear that (X-X ) is distributed Normal[ (μ-μ ), ( + ) ] Let X Average amog age 0-4. It is distributed Normal(μ 80, X Average amog age It is distributed Normal(μ 99, 43 ) X X 50 ) Thus, Y(X X ). is distributed Normal with μ (μ -μ ) 9 Y X X Y Now we use the z-score method that we leared i Uit 5, Normal Distributio, ad i particular the z-score stadardizatio that is foud o page 9 uder (3), we have that Probability group mea group mea will be more tha 5 Pr[Y > 5] Y-μ Y 5-9 Pr > Y 9.95 Pr[Normal(0,) > 0.65].57

4 PubHlth 540 Itroductory Biostatistics Page 4 of 7 4. The objectives of a study by Keedy ad Bhambhai (99) were to use physiological measuremets to determie the test-retest reliability of the Baltimore Therapeutic Equipmet Work Simulator durig three simulated tasks performed at light, medium, ad heavy work itesities, ad to examie the criterio validity of these tasks by comparig them to real tasks performed i a cotrolled laboratory settig. Subjects were 30 healthy me betwee the ages of 8 ad 35. The ivestigators reported a stadard deviatio of s0.57 for the variable peak oxyge cosumptio (/mi) durig oe of the procedures. Assumig ormality, compute a 95% cofidece iterval for the populatio variace for the oxyge cosumptio variable. Aswer: (.,.59) (-) 9 S 0.57 χ χ α/.975; DF9 χ α/ χ.05; DF Lower limit (-)S χ - α /;df(-) (9)(0.57 ) Upper limit (-)S χ α /;df(-) (9)(0.57 ) The purpose of a ivestigatio by Alahuhta et al (99) was to evaluate the ifluece of extradural block for elective caesaria sectio simultaeously o several materal ad fetal hemodyamic variables ad to determie if the block modified fetal myocardial fuctio. The study subjects were eight healthy parturiet i gestatioal weeks 38-4 with ucomplicated sigleto pregacies udergoig elective caesaria sectio uder extradural aesthesia. Amog the measuremets take, were materal diastolic arterial pressure durig two stages of the study. The followig are the lowest values of this variable at the two stages. Compute a 95% cofidece iterval for the differece i diastolic blood pressure betwee the two stages. Patiet ID Stage Stage

5 PubHlth 540 Itroductory Biostatistics Page 5 of 7 Aswer: (-0.06, +6.6) Because two measuremets are made o each patiet, at stages ad, these data fit the defiitio of paired. The aalysis focuses o the differeces, per the table below: Patiet ID, i d i Stage The actual calculatios required to complete the solutio are similar to the example o page 43. I particular, for this exercise, we have d3.5 Sd Sd d S SE(d) df(-)7 t- α /;df t.975; % CI for μ d±(t )SE(d) d.975;df7 6. A possible evirometal determiat of lug fuctio i childre is the amout of cigarette smokig i the home. To study this questio, two groups of childre were studied. Group cosisted of 3 osmokig childre aged 5-9 both of whose parets smoke i the home. Group cosisted f 0 osmokig childre aged 5-9 either of whose parets smoke. The mea (SD) of FEV for group is. L (0.7) ad for the Group childre, the mea (SD) of FEV is.3 L (0.4). Uder the assumptio of ormality, costruct a 95% cofidece iterval for the ratio of the variace of the two groups. What is your coclusio regardig the reasoableess of the assumptio of equality of variaces? Aswer: (.4, 7.37) The followig lik is good here. (3.5) ± (.365)(.3983) (-0.057, 6.557) Play with it usig a kow row F distributio probability from the back of your text book. Thus, you will discover that this table works with areas uder the curve to the right.

6 PubHlth 540 Itroductory Biostatistics Page 6 of 7 S 0.7 S 0.4 ( ) (3 ) ( ) (0 ) 9 F, ; α / F,9; F, ; α / F,9; For example, to obtai the 97.5 th percetile, eter.05 as the area uder the curve to the right as show below. The click o the left arrow key to obtai the value of the percetile. You should see the followig ad thus obtai 97.5 th percetile value.478. S 0.7 Lower limit.357 F ; ;( α /) S S 0.7 Upper limit F ; ; S / α

7 PubHlth 540 Itroductory Biostatistics Page 7 of 7 Sice the cofidece iterval has lower limit equal to.357, a umber that is above, these data are ot cosistet with the assumptio of equal variaces. (Logic if the variaces are equal, the their ratio is equal to. It the follows that, if the cofidece iterval for the ratio does ot iclude, the the data are ot cosistet with the assumptio of equal variaces). 7. For the same data i problem #6 ad drawig upo your aswer to #6 (regardig the reasoableess of equality of variaces), compute a 95% cofidece iterval for the true mea differece i FEV betwee 5-9 year old childre whose parets smoke ad comparable childre whose parets do ot smoke. Aswer: (-0.55, +0.5) This solutio assumes that the variaces are Uequal because the cofidece iterval obtaied for #6 does ot iclude a ratio of variaces value. Therefore, the correct stadard error formula to use is Solutio 3 o page 48 of the otes. X X S S SEˆ X X S S f S S t t.08 -α/;f.975;df36 ( ) ( ) ˆ.975;DF36 ( ) 95%CI X -X ± t SE X -X (-0.)±(.08)(0.7) ( 0.547, )