資工系 Lecture#11-2 p.1 生物統計學 96-1 2008/01/02 生物統計學 11-2:SPSS for Descriptive Statistics & Testing 01/02/2008 O. Descriptive Statistics Exercise 3.12 (a), (b), (c)。(p. 63)- data set – unicef.dat Percentages of low birth weight infants – weighing less than 2500 gram. Variables – nation, lowbwt, life60, life92 (a) Compute the mean and the median. (b) Compute the 5% trimmed mean. (c) For this data set, which number would you prefer as a measure of central tendency? Result of SPSS: Descriptive Statistics N Minimum percent low birth weight 111 Valid N (listwise) 111 Maximum 4 50 Mean Std. Deviation 11.96 6.524 Case Processing Summary Valid N percent low birth weight 111 Percent 77.1% Cases Missing N Percent 33 22.9% Total N 144 Percent 100.0% De scri ptives percent low birt h weight Mean 95% Confidenc e Int erval for Mean 5% Trimmed Mean Median Variance St d. Deviat ion Minimum Maximum Range Int erquartile Range Sk ewness Kurtos is Lower Bound Upper Bound St atist ic 11.96 10.74 St d. E rror .619 13.19 11.38 10.00 42.562 6.524 4 50 46 8 2.283 9.941 .229 .455 資工系 Lecture#11-2 p.2 生物統計學 96-1 2008/01/02 His to g ra m 50 Fre q u e n c y 40 30 20 10 Me a n = 11.96 S td. De v. = 6.524 N = 111 0 0 10 20 30 40 50 p e rc e n t lo w b irth we ig h t percent low birth weight Stem-and-Leaf Plot Frequency Stem & Leaf .00 0 . 9.00 0 . 444455555 21.00 0 . 666666666677777777777 17.00 0 . 88888888999999999 18.00 1 . 000000000111111111 6.00 1 . 222333 13.00 1 . 4444455555555 12.00 1 . 666666677777 3.00 1 . 899 7.00 2 . 0000011 1.00 2 . 3 2.00 2 . 55 2.00 Extremes (>=33) Stem width: 10 50 Each leaf: 1 case(s) 10 40 56 30 20 10 0 pe rce nt low birth we ight 資工系 Lecture#11-2 p.3 生物統計學 96-1 2008/01/02 Exercise 3.15 (a), (b), (c)。(p. 64)- data set – lowbwt.dat 100 low birth weight infants in two hospitals in Boston, MA. Variables – sbp (systolic blood pressure), sex (gender) (a) Construct a pair of box plots for the SBP – one for boys and one for girls. (b) Compute the mean and s.d. of SBP for males and for females. Compare them. (c) Compute the C.V. of SBP for each gender. Compare them. (CVf – 24%, CVm – 25%) Case Processing Summary Valid sys tolic blood press ure gender Female Male N Percent 100.0% 100.0% 56 44 Cases Mis sing N Percent 0 .0% 0 .0% 16 80 s ys to lic b lo o d p re s s u re 41 60 40 46 20 Fe ma le Ma le gender His to g ra m for s e x= Fe ma le 30 25 Fre q u e n c y 20 15 10 5 Me a n = 46.46 S td. De v. = 11.145 N = 56 0 10 20 30 40 50 s ys to lic b lo o d p re s s u re 60 70 Total N 56 44 Percent 100.0% 100.0% 資工系 Lecture#11-2 p.4 生物統計學 96-1 2008/01/02 His to g ra m for s e x= Ma le 20 Fre q u e n c y 15 10 5 Me a n = 47.86 S td. De v. = 11.806 N = 44 0 20 30 40 50 60 70 80 90 s ys to lic b lo o d p re s s u re Report sys tolic blood press ure gender Female Male Total Mean 46.46 47.86 47.08 N 56 44 100 Median 47.00 46.00 47.00 Minimum 19 27 19 Maximum 67 87 87 Std. Deviation 11.145 11.806 11.403 資工系 Lecture#11-2 p.5 生物統計學 96-1 2008/01/02 I. Hypothesis Testing for One Sample Exercise 10.16 (a), (b) , (c)。(p. 282)- data set – heart.dat Infant development study on two scores of index - PDI (Psychomotor Development Index), MDI (Mental Development Index); variables: trtment (0 - ca; 1 - lf), pdi, mdi. (a) At α = 0.05, is the mean PDI score equal to 100, the mean score for healthy infants? (b) Test that the mean MDI score for the analogous test of hypothesis. (a) H0 :μ= 100 vs H1 :μ≠ 100, the p-value = 0.000. We reject H0 :μ= 100 at α = 0.05 level. In fact, the mean PDI is less than 100, and the 95% CI is [100 – 7.84, 100 – 2.60]. Result of SPSS: One-Sample Statistics N ps ychomotor development index Mean 143 Std. Deviation Std. Error Mean 15.851 1.326 94.78 One-Sample Test Test Value = 100 t ps ychomotor development index df -3.936 Sig. (2-tailed) Mean Difference .000 -5.217 142 95% Confidence Interval of the Difference Lower Upper -7.84 -2.60 (b) H0 :μ= 100 vs H1 :μ≠ 100, the p-value = 0.000. We reject H0 :μ= 100 at α = 0.05 level. The mean MDI is higher than 100, and the 95% CI is [100 + 2.17, 100 + 7.31]. One-Sample Statistics N mental development index Mean 144 104.74 Std. Deviation Std. Error Mean 15.604 1.300 One-Sample Test Test Value = 100 t mental development index 3.642 df 143 Sig. (2-tailed) Mean Difference .000 4.736 95% Confidence Interval of the Difference Lower Upper 2.17 7.31 資工系 Lecture#11-2 p.6 生物統計學 96-1 2008/01/02 II. Hypothesis Testing for Matched or Paired Samples Example 1: Starting salaries for environmental jobs, in dollars, for 5 positions requiring no degree are as follows: Position Salary 1 2 3 4 5 32,000 25,000 54,000 40,000 45,000 The starting salaries for the same position for persons holding a degree in geography or computer science are: Position Salary 1 2 3 4 5 32,000 26,000 55,000 43,000 49,000 Result of SPSS: Paired Samples Statistics Pair 1 Mean 39200.00 41000.00 salary1 salary2 N 5 5 Std. Deviation 11256.109 11937.336 Std. Error Mean 5033.885 5338.539 Pa ired Sa mples Correlations N Pair 1 salary1 & s alary2 5 Correlation .992 Sig. .001 Pa ired Sa mpl es Test Paired Differenc es Pair 1 Mean salary1 - s alary 2 -1800.000 test statistic td = St d. Deviat ion 1643.168 St d. Error Mean 734.847 95% Confidenc e Int erval of t he Difference Lower Upper -3840.262 240.262 xd d = 1800/(1643/ 5 ) = 2.45, sd / n The p-value = 0.0352 = 0.070/2 t -2. 449 df = n – 1 = 4 (one-sided p-value = two-sided /2) df 4 Sig. (2-tailed) .070 資工系 Lecture#11-2 p.7 生物統計學 96-1 2008/01/02 Example 2: The study of males with coronary artery disease. For each of the 63 men, calculate the percent decrease in time to angina when he is exposed to carbon monoxide minus the percent decrease when he is exposed to unadulterated air. (exposure to air vs exposure to co4%) Result of SPSS: Paired Samples Statistics Pair 1 100*(air1-air2)/air1 (@4co1-@4co2)/@4co1 Mean .957895 7.586281 N Std. Deviation 17.4402749 16.8956695 63 63 Std. Error Mean 2.1972681 2.1286543 Paired Samples Correlations N Pair 1 100*(air1-air2)/air1 & (@4co1-@4co2)/@4co1 Correlation 63 Sig. .302 .016 Paired Samples Test Paired Differences Mean 100*(air1-air2)/air1 (@4co1-@4co2)/@4co1 test stat. td = Std. Error Mean 20.2911395 2.5564433 -6.62839 -11.7386 -1.51813 xd d = (-6.63 – 0)/(20.29/ 63 ) = -2.59, sd / n The p-value = 0.006 = 0.012/2 t -2.593 df Sig. (2-tailed) 62 df = n – 1 = 62 (one-sided p-value = two-sided /2) (The t62, 0.0125 = -2.299, t62, 0.005 = -2.660 for one-sided t-test, so 0.005 < p-value < 0.0125) 40 35 30 Fre q u e n c y Pair 1 Std. Deviation 95% Confidence Interval of the Difference Lower Upper 25 20 15 10 5 Me a n = -6 .6 2 8 4 S td . De v. = 2 0 .2 9 1 1 4 N = 63 0 -8 0 .0 0 -6 0 .0 0 -4 0 .0 0 -2 0 .0 0 0 .0 0 d iff1 -d iff2 2 0 .0 0 4 0 .0 0 6 0 .0 0 .012 資工系 Lecture#11-2 p.8 生物統計學 96-1 2008/01/02 Further Application Example 2-1: The study of males with coronary artery disease. (exposure to air vs exposure to co2%; there was one missing data, so n = 62) Result of SPSS: Paired Samples Statistics Pair 1 Mean .9254 5.8738 100*(air1-air2)/air1 (@2co1-@2co2)/@2co1 N 62 62 Std. Deviation 17.58073 14.18603 Std. Error Mean 2.23275 1.80163 Paired Samples Correlations N Pair 1 100*(air1-air2)/air1 & (@2co1-@2co2)/@2co1 Correlation 62 .296 Sig. .020 Paired Samples Test Paired Differences Mean Pair 1 100*(air1-air2)/air1 (@2co1-@2co2)/@2co1 test stat. td = -4.94833 Std. Deviation Std. Error Mean 19.04708 2.41898 95% Confidence Interval of the Difference Lower Upper -9.78538 -.11128 xd d = (-4.95 – 0)/(19.05/ 62 ) = -2.59, sd / n The p-value = 0.0225 = 0.045/2 t -2.046 df = n – 1 = 61 (one-sided p-value = two-sided /2) df Sig. (2-tailed) 61 .045 資工系 Lecture#11-2 p.9 生物統計學 96-1 2008/01/02 III. Hypo. Testing for Independent Groups: Two Population Means Exercise 11.14 (a), (b)。(p. 281)- data set – lowbwt.dat 100 low birth weight infants in two hospitals in Boston, MA. Variables – sbp (systolic blood pressure), sex (gender) (a) A histogram of SBP for this sample. Is it approximately normally distributed? (b) At α = 0.05, is the mean SBP for boys equal to the mean for girls? (a) Yes, the values appear to be fairly symmetric, and could be considered to be approximately normally distributed. 50 Frequency 40 30 20 10 Me a n = 4 7 .0 8 S td . De v. = 1 1 .4 0 3 N = 100 0 0 20 40 60 80 100 s y s t o lic b lo o d p r e s s u r e (b) H0 :μf =μm vs H1 :μf ≠μm , the p-value = 0.545 for equal variance assumed. We are unable to reject H0 :μf =μm at the 0.05 level of significance. Result of SPSS: Group Statistics sys tolic blood press ure gender Female Male N 56 44 Mean 46.46 47.86 Std. Deviation 11.145 11.806 Std. Error Mean 1.489 1.780 Independent Samples Test Levene's Test for Equality of Variances F systolic blood press ure Equal variances as sumed Equal variances not ass umed .079 Sig. .779 t-test for Equality of Means t df Sig. (2-tailed) Mean Difference Std. Error Difference 95% Confidence Interval of the Difference Lower Upper -.607 98 .545 -1.399 2.305 -5.973 3.174 -.603 89.858 .548 -1.399 2.321 -6.010 3.211 資工系 Lecture#11-2 p.10 生物統計學 96-1 2008/01/02 Exercise 11.15 (a), (b) , (c)。(p. 282)- data set – heart.dat Infant development study on two scores of index - PDI (Psychomotor Development Index), MDI (Mental Development Index); variables: trtment (0 - ca; 1 - lf), pdi, mdi. (a) At α = 0.05, is the mean PDI score for circulatory arrest group (ca) equal to the mean for low-flow bypass group (lf)? What is the p-value? (b) Test that the mean MDI scores are identical for the two treatment groups. (a) H0 :μca =μlf vs H1 :μca ≠μlf , the p-value = 0.027 for equal variance assumed. We reject H0 :μca =μlf at α = 0.05 level; mean PDI is higher for the low-flow group. Result of SPSS: Group Statistics ps ychomotor development index treatment group CA LF N Mean 91.92 97.77 73 70 Std. Deviation 16.488 14.685 Std. Error Mean 1.930 1.755 Independent Samples Test Levene's Test for Equality of Variances F ps ychomotor development index Equal variances as sumed Equal variances not ass umed 2.815 t-test for Equality of Means Sig. t .096 df Sig. (2-tailed) Mean Difference Std. Error Difference 95% Confidence Interval of the Difference Lower Upper -2.239 141 .027 -5.854 2.615 -11.023 -.684 -2.244 140.247 .026 -5.854 2.609 -11.011 -.696 (b) H0 :μca =μlf vs H1 :μca ≠μlf , the p-value = 0.214 for equal variance assumed. Group Statistics mental development index treatment group CA LF N Mean 103.16 106.40 74 70 Std. Deviation 16.465 14.573 Std. Error Mean 1.914 1.742 Independent Samples Test Levene's Test for Equality of Variances F mental development index Equal variances as sumed Equal variances not ass umed .404 Sig. .526 t-test for Equality of Means t df Sig. (2-tailed) Mean Difference Std. Error Difference 95% Confidence Interval of the Difference Lower Upper -1.247 142 .214 -3.238 2.597 -8.371 1.895 -1.251 141.386 .213 -3.238 2.588 -8.354 1.878 資工系 Lecture#11-2 p.11 生物統計學 96-1 2008/01/02 Example 3 in Lecture#11 Source: http://archpedi.ama-assn.org/cgi/content/abstract/143/8/969 American J. of Diseases of Children Vol. 143 No. 8, August 1989 or Archives of Pediatrics and Adolescent Medicine Vol. 143 No. 8, August 1989 Effect of pancreatic enzyme supplements on iron absorption W. T. Zempsky, B. J. Rosenstein, J. A. Carroll and F. A. Oski Department of Pediatrics, Johns Hopkins University School of Medicine, Baltimore, Md. Iron deficiency has been reported in one third of patients with cystic fibrosis. There are data that suggest that iron absorption is increased with exocrine pancreatic deficiency and that administration of pancreatic enzymes may impair oral iron absorption. We compared oral iron absorption over a 3-hour period in the presence and absence of exogenous pancreatic enzymes in 13 stable young-adult patients with cystic fibrosis and 9 age-matched control patients. Although none of the patients with cystic fibrosis had a hemoglobin level less than 119 g/L, serum ferritin levels were less than 25 micrograms/L in 5 of the 13 patients, and the mean corpuscular volume was significantly lower in the patient group (86.1 +/- 2.7 vs 90.9 +/- 5 fL). Baseline mean serum iron levels were higher in controls (18.9 +/- 5.9 mumol/L) than in patients (11.9 +/- 6.3 mumol/L). There was no difference in iron absorption in the absence of exogenous pancreatic enzymes. Significant impairment of iron absorption was detected in both patients with cystic fibrosis and controls after administration of a preparation of pancreatic enzymes. There was an inverse relationship between iron stores, as measured by serum ferritin, and iron absorption. These findings suggest that long-term consumption of pancreatic enzymes by patients with cystic fibrosis may contribute to iron deficiency. (no data) 資工系 Lecture#11-2 p.12 生物統計學 96-1 2008/01/02 Example 4 in Lecture#11: http://www.nlm.nih.gov/databases/alerts/hypertension.html Results of the NHLBI and NIA clinical trial of the efficacy of antihypertensive treatment for isolated systolic hypertension in those aged 60 years and older National Heart, Lung, and Blood Institute (NHLBI) National Institute on Aging (NIA) June 26, 1991 Abstract: NIH's National Heart, Lung, and Blood Institute and the National Institute on Aging have announced a treatment that reduces the complications of isolated systolic hypertension. This form of high blood pressure, which affects more than 3 million people aged 60 years and above, can be treated with a stepped-care regimen of low dose chlorthalidone and atenolol with a resultant reduction of stroke (fatal plus non-fatal). The Systolic Hypertension in the Elderly Program (SHEP) was a 5-year double-blind randomized trial that assessed the ability of antihypertensive drug treatment to reduce risk of nonfatal plus fatal (total) stroke in men and women age 60 and above with isolated systolic hypertension. Persons with average systolic blood pressure 160-219 mm Hg and average diastolic blood pressure 90 mmHg were eligible. SHEP randomized 4,736 people to either active treatment (2,365 persons) or placebo (2,371). The two treatment groups were comparable at baseline. Mean baseline age was 71.6 years; 47.5% were white women, 38.6% white men, 9.3% black women, and 4.6% black men. Mean systolic blood pressure was 170 mmHg: mean diastolic blood pressure, 77 mm Hg. For the active treatment group, a stepped-care regimen was used which included chlorthalidone 12.5 or 25 mg/day, and as needed, addition of atenolol 25 or 50 mg/day or reserpine, 0.05 or 0.10 mg/day. Treatment goal was to reduce systolic blood pressure by at least 20 mm Hg from baseline and to below 160 mm Hg with minimal amounts of study medication. Five-year average systolic blood pressures was 155 mmHg for the placebo group and 143 mm Hg for the active treatment group; diastolic blood pressure values were 72 mmHg and 68 mm Hg respectively. Goal blood pressure was achieved by about 70% of the actively treated group and about 35% of the placebo group during the trial. Average follow-up was 4.5 years. Five-year incidence of total stroke was 5.2/100 for active treatment and 8.2/100 for placebo. The relative risk as computed by proportional hazards regression was 0.64, 95% confidence interval 0.51 to 0.82, p = 0.0003. The results were similar across age and race-sex subgroups. 資工系 Lecture#11-2 p.13 生物統計學 96-1 2008/01/02 For SHEP secondary end points, rates were again lower for active treatment than placebo, e.g., clinical nonfatal myocardial infarction plus coronary death relative risk was 0.73 (95% confidence interval 0.57 - 0.94); combined major cardiovascular events, relative risk is 0.68 (95% confidence interval 0.58 - 0.79); deaths from all causes, relative risk 0.87 (95% confidence interval 0.73 -1.05). Occurrence of adverse effects was of low order, and findings from behavioral evaluation (including dementia and depression) were similar throughout the trial for the two treatment groups. Two additional secondary questions were stated at the time of trial planning as subgroup hypotheses. First, would treatment of ISH reduce the frequency of total stroke (fatal and nonfatal) similarly in those receiving and not receiving antihypertensive medication at initial contact? Second, would treatment of ISH reduce the incidence of sudden cardiac death or of coronary plus nonfatal myocardial infarction similarly in those free of baseline ECG abnormalities and in those with such abnormalities? Reduction in stroke incidence for those on and not on medication at initial contact was similar during the trial. The reduction in nonfatal myocardial infarction plus coronary death was greater in those with ECG abnormalities at baseline than in those without such abnormalities. The results for sudden death and for nonfatal myocardial infarction plus coronary death were similar in the two groups. Conclusion: In persons aged 60 years and over with isolated systolic hypertension antihypertensive stepped-care drug treatment with low dose chlorthalidone as Step 1 medication reduces incidence of total stroke by 36% with 5-year absolute benefit estimated at 30/1,000 participants. Major cardiovascular events are also reduced, with 5-year absolute benefit estimated as 55/1000 participants. Implications: The SHEP trial conclusively demonstrates that ISH can be effectively and safely treated with a stepped-care regimen including low dose diuretic, chlorthalidone, and low dose beta-blocker, atenolol. These are low cost medications which should be considered as the first choice of treatment for ISH. Other medication regimens have not been tested for efficacy in the treatment of ISH. (no data)