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Proportions
CHAPTER 17
Proportions INTRODUCTION Previous chapters dealing with categories sometimes used the proportions within each group. It may be easier to work with proportions rather than absolute numbers. A proportion is the relationship of a part to the whole and can be given as a fraction of 1 or as a percentage. A ratio is the relationship of one part to another part. If out of 100 patients with a myocardial infarction 30 of them die, the proportion who die is 30/100 ¼ 0.3 or 30%, whereas the ratio of those who die to those who survive is 30/70 ¼ 0.43 or 43%.
PROPORTIONS AND BINOMIAL THEOREM If there are 37 deaths in 192 people with a certain disease, the number who survive is 192–37¼ 155. The proportion of deaths ( p) is determined by number with the attribute/ total number ¼ 37/192 ¼ 0.1927. The proportion that survives is 155/192 ¼ 0.8073. This is the value 1–p, also symbolized by q (Chapter 16). In repeated samples from this population, the value of p would vary; the sample value p is a point estimate of the population value π. The distribution of p is approximately normal as long as np and nq > 9. The standard deviation of such a binomial is: σ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pð1 pÞ ¼ pq
The standard error of the sample proportion p that estimates the population proportion π is equivalent to the standard error of a mean value and is estimated by dividing the rffiffiffiffiffi pq standard deviation by √ n. Therefore σ ¼ . In the previous example, n rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:1927 0:8073 ¼ 0:02486: σp ¼ 192 The 95% confidence limits of π are p 1.96 0.02486 ¼ 0.1440 to 0.2414. Because the formula for the standard error of a proportion comes from the approximation of the discrete binomial distribution to the normal distribution, we need a continuity correction. Test the null hypothesis by
Basic Biostatistics for Medical and Biomedical Practitioners https://doi.org/10.1016/B978-0-12-817084-7.00017-6
© 2019 Elsevier Inc. All rights reserved.
263
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Basic Biostatistics for Medical and Biomedical Practitioners
z¼
|p π| sep
1 2n :
The effect of the correction is to make the difference slightly smaller, and for large sample sizes the correction is unimportant.
CONFIDENCE LIMITS The observed proportion is a point estimate. To determine its confidence limits, use the rffiffiffiffiffi pq . This method yields Wald method, with the 95% limits determined from p zα=2 N inaccurate results if N is small or p is close to 0 or 1. A simple way to obtain more accurate limits is to use the Agresti-Coull adjustment by adding 2 successes and 2 failures to the X +2 observed counts. Thus pa ¼ , and the adjusted value of p (pa) is substituted for the N +4 value of p in the Wald formula before. rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 37 0:1927 0:8703 If p ¼ ¼ 0:1927, then the Wald limits are 0:1927 1:96 ¼ 192 192 0:1927 0:0579 ¼ 0:1348 to 0:2506. With the adjustment, the adjusted value of p is 37 + 2 ¼ 0:1990, and the adjusted Wald limits are p¼ 192 + 4 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:1990 0:8010 ¼ 0:1990 0:0565 ¼ 0:1425 to 0:2555: 0:1990 1:96 192 Confidence limits may be obtained online from http://www.causascientia.org/math_ stat/ProportionCI.html (a Bayesian calculator), http://www.graphpad.com/quickcalcs/ ConfInterval2.cfm, http://www.sample-size.net/confidence-interval-proportion/, and http://vassarstats.net/prop1.html. Problem 17.1 Nineteen out of 113 (16.8%) men have a serum cholesterol <5.0 mmol/L. What are the 95% and 99% confidence limits for this proportion?
One use for confidence limits is to determine the upper 95% limit for a proportion if zero events occur. If a surgeon operates on 10 patients without a death, what is the upper 95% limit of deaths? It can be determined easily by the rule of 3 (Hanley and Lippman-Hand, 1983; van Belle, 2002). The upper 95% limit for the proportion is determined by 3/N. The upper 95% mortality proportion for the surgical procedure is 3/10 or 0.3 (or 30%). If there were no deaths in 50 operations the upper 95% limit would be 3/50 ¼ 0.06 or 6%. (This assumes that the operations were similar in all respects.)
Proportions
The corollary to this is to determine how many surgical operations must be observed in order to find at least one death if we know the average mortality. If the mortality of a procedure is 1%, there is a 95% chance of observing 1 death in 3/0.01 ¼ 300 operations. This calculation does not allow for differences in surgical skill or severity of illness.
SAMPLE AND POPULATION PROPORTIONS To compare any observed value of p with the population value π, use the normal distribution curve. πp z¼ σp To determine if the sample value of p ¼ 0.1927 could have come from a population in 0:3 0:1927 which π ¼ 0.3, calculate z ¼ ¼ 4:3162. 0:02486 P ¼ 0.000016 (two-tailed), and we would reject the null hypothesis.
SAMPLE SIZE To compare two proportions, we need to know how many subjects we will need to minimize type I and type II errors. The basic formulas are discussed by Fleiss (1981) and Bland (2015). They are relatively complex and approximate, can be written in several ways, and are best replaced by free online calculation at http://statpages.org/proppowr.html, http://www.stat.ubc.ca/rollin/ stats/ssize/b2.html, http://www.cct.cuhk.edu.hk/stat/proportion/Casagrande.htm, http://www.sample-size.net/confidence-interval-proportion/, and https://www.statstodo. com/SSiz2Props_Pgm.php. A simplified approximation was devised by Lehr (1992) who used the equation n¼
16pq , ðp1 p2 Þ2
where p is the average of p1 and p2, and q is 1-average p. This gives the number in each group for a power of 0.8 (Type II error β ¼¼ 0.2) and a Type I error α ¼ 0.05 (twotailed). If a power of 0.9 is wanted, the constant is changed from 16 to 21. According to Lehr, this estimate is slightly low for Fisher’s exact test and slightly high for a 2 2 chi-square test. As with all sample size calculations, these are estimates, not precise numbers. As an example, consider how many subjects (equal sized groups) are needed to show a difference between a remission rate of 0.7 for one treatment and 0.6 for another. We wish to set α ¼ 0.05 and β ¼ 0.2, that is, power ¼ 0.8. By Lehr’s formula, n¼
16 0:65 0:35 ¼ 364 as the number required in each group. The more formal ð0:7 0:6Þ2
265
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Basic Biostatistics for Medical and Biomedical Practitioners
calculation online gives 356 in each group, or 376 if the continuity correction is used. For a power of 0.9, Lehr’s formula gives 478 in each group, and the online calculation gives 476, or 496 with the continuity correction. The major advantage of using an online calculator is that it allows for unequal group sizes. In estimating the numbers needed, we need some idea of the effect size that we want. Take 0.50 as the null hypothesis, that is, there is no difference between the two groups. Then either select an effect size based on previous work or else try to decide what minimum effect size to detect. For example, suppose the null hypothesis is that the probability of survival of a disease under standard treatment is 0.50, we would probably not be interested in a treatment that changed the proportion to only 0.49. Cohen (1988) classified effect size of the difference from 0.50 as small (<0.05, that is, 0.45–0.55), medium (0.15, that is, 0.35–0.65), and large (0.25, i.e., 0.25–0.75). He pointed out that large differences were rare. For example, in presidential elections in the United States there has never been a division as extreme as 65:35, and even a division of 55:45 would be regarded as a landslide victory. Cohen gives tables, and online calculators are available. As always, it takes a huge number of measurements or counts to detect a small difference. Problem 17.2 In two different populations of men, serum cholesterol concentrations below 5.0 mmol/L occur in 6% and 9%, respectively. What sample size is needed to show that this difference allows us to reject the null hypothesis with α ¼ 0.05 and β ¼ 0.2? (assume equal sized groups)?
COMPARING PROPORTIONS To compare two proportions, for example, the survival of patients with a given disease who have two different treatments, let the proportion surviving in group 1 be p1 and that in group 2 be p2. If these two proportions are similar, we conclude that treatment did not affect survival. If they are quite different, then we can ask if the null hypothesis is true. To do this, calculate the standard error of the difference as s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p1 q1 p2 q2 : sp1p2 ¼ n1 n2 Then relate the difference p1–p2 to the standard error of the difference; p1 p2 z ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , p1 q1 p2 q2 n1 n2
Proportions
For example, if treatment 1 gives a survival proportion of 37/192 ¼ 0.1927 and treatment 2 gives a survival of 17/168 ¼ 0.1012, could that difference have occurred by chance? Calculate z as 0:1927 0:1012 0:0915 ffi z ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:0008102 + 0:0005414 0:1927 0:8073 0:1012 0:8988 + 192 168 ¼ 2:4888 Therefore P ¼ 0.0128 (two-sided) and we can reasonably reject the null hypothesis and believe that treatment 2 might be better. The 95% confidence limits for the difference are 0.0915 1.96 0.03676 ¼ 0.05474 to 0.1291. Because this does not include zero, it confirms the results of the z test. The continuity correction can be done by using 1 1 1 |P1 P2 | + 2 n1 n2 Z¼ seðp1p2Þ Sample sizes for adequate power are provided by Cohen (1988). Proportions can be compared at http://www.measuringusability.com/ab-calc.php, https://www.medcalc.org/calc/comparison_of_proportions.php, http://vassarstats.net/ prop2_ind.html, or http://in-silico.net/statistics/ztest/two-proportion and sample size can be calculated at http://www.sample-size/net, and http://www.statstodo.com/ SSiz2Props_Pgm.php. Note: Comparing two proportions is the same as doing a 2 2 chi-square test.
Pooling samples If there are several small samples from a population, with X1 successes in N1 trials, X2 successes in N2 trials, up to Xk successes in Nk trials, then an average proportion of successes can be calculated as P¼
X1 + X2 + …:Xk : N1 + N2 + …Nk
REFERENCES Bland, M., 2015. An Introduction to Medical Statistics. Oxford University Press, Oxford. Cohen, J., 1988. Statistical Power Analysis for Behavioral Sciences. Lawrence Erlbaum Associates, Hillsdale, NJ. Fleiss, J.L., 1981. Statistical Methods for Rates and Proportions. John Wiley & Sons, New York. Hanley, J.A., Lippman-Hand, A., 1983. If nothing goes wrong, is everything alright? JAMA 249, 1743–1745. Lehr, R., 1992. Sixteen s-squared over d-squared: a relation for crude sample size estimates. Stat. Med. 11, 1099–1102. van Belle, G., 2002. Statistical Rules of Thumb. John Wiley & Sons, Inc., New York.
267
Proportions INTRODUCTION Previous chapters dealing with categories sometimes used the proportions within each group. It may be easier to work with proportions rather than absolute numbers. A proportion is the relationship of a part to the whole and can be given as a fraction of 1 or as a percentage. A ratio is the relationship of one part to another part. If out of 100 patients with a myocardial infarction 30 of them die, the proportion who die is 30/100 ¼ 0.3 or 30%, whereas the ratio of those who die to those who survive is 30/70 ¼ 0.43 or 43%.
PROPORTIONS AND BINOMIAL THEOREM If there are 37 deaths in 192 people with a certain disease, the number who survive is 192–37¼ 155. The proportion of deaths ( p) is determined by number with the attribute/ total number ¼ 37/192 ¼ 0.1927. The proportion that survives is 155/192 ¼ 0.8073. This is the value 1–p, also symbolized by q (Chapter 16). In repeated samples from this population, the value of p would vary; the sample value p is a point estimate of the population value π. The distribution of p is approximately normal as long as np and nq > 9. The standard deviation of such a binomial is: σ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pð1 pÞ ¼ pq
The standard error of the sample proportion p that estimates the population proportion π is equivalent to the standard error of a mean value and is estimated by dividing the rffiffiffiffiffi pq standard deviation by √ n. Therefore σ ¼ . In the previous example, n rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:1927 0:8073 ¼ 0:02486: σp ¼ 192 The 95% confidence limits of π are p 1.96 0.02486 ¼ 0.1440 to 0.2414. Because the formula for the standard error of a proportion comes from the approximation of the discrete binomial distribution to the normal distribution, we need a continuity correction. Test the null hypothesis by
Basic Biostatistics for Medical and Biomedical Practitioners https://doi.org/10.1016/B978-0-12-817084-7.00017-6
© 2019 Elsevier Inc. All rights reserved.
263
264
Basic Biostatistics for Medical and Biomedical Practitioners
z¼
|p π| sep
1 2n :
The effect of the correction is to make the difference slightly smaller, and for large sample sizes the correction is unimportant.
CONFIDENCE LIMITS The observed proportion is a point estimate. To determine its confidence limits, use the rffiffiffiffiffi pq . This method yields Wald method, with the 95% limits determined from p zα=2 N inaccurate results if N is small or p is close to 0 or 1. A simple way to obtain more accurate limits is to use the Agresti-Coull adjustment by adding 2 successes and 2 failures to the X +2 observed counts. Thus pa ¼ , and the adjusted value of p (pa) is substituted for the N +4 value of p in the Wald formula before. rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 37 0:1927 0:8703 If p ¼ ¼ 0:1927, then the Wald limits are 0:1927 1:96 ¼ 192 192 0:1927 0:0579 ¼ 0:1348 to 0:2506. With the adjustment, the adjusted value of p is 37 + 2 ¼ 0:1990, and the adjusted Wald limits are p¼ 192 + 4 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:1990 0:8010 ¼ 0:1990 0:0565 ¼ 0:1425 to 0:2555: 0:1990 1:96 192 Confidence limits may be obtained online from http://www.causascientia.org/math_ stat/ProportionCI.html (a Bayesian calculator), http://www.graphpad.com/quickcalcs/ ConfInterval2.cfm, http://www.sample-size.net/confidence-interval-proportion/, and http://vassarstats.net/prop1.html. Problem 17.1 Nineteen out of 113 (16.8%) men have a serum cholesterol <5.0 mmol/L. What are the 95% and 99% confidence limits for this proportion?
One use for confidence limits is to determine the upper 95% limit for a proportion if zero events occur. If a surgeon operates on 10 patients without a death, what is the upper 95% limit of deaths? It can be determined easily by the rule of 3 (Hanley and Lippman-Hand, 1983; van Belle, 2002). The upper 95% limit for the proportion is determined by 3/N. The upper 95% mortality proportion for the surgical procedure is 3/10 or 0.3 (or 30%). If there were no deaths in 50 operations the upper 95% limit would be 3/50 ¼ 0.06 or 6%. (This assumes that the operations were similar in all respects.)
Proportions
The corollary to this is to determine how many surgical operations must be observed in order to find at least one death if we know the average mortality. If the mortality of a procedure is 1%, there is a 95% chance of observing 1 death in 3/0.01 ¼ 300 operations. This calculation does not allow for differences in surgical skill or severity of illness.
SAMPLE AND POPULATION PROPORTIONS To compare any observed value of p with the population value π, use the normal distribution curve. πp z¼ σp To determine if the sample value of p ¼ 0.1927 could have come from a population in 0:3 0:1927 which π ¼ 0.3, calculate z ¼ ¼ 4:3162. 0:02486 P ¼ 0.000016 (two-tailed), and we would reject the null hypothesis.
SAMPLE SIZE To compare two proportions, we need to know how many subjects we will need to minimize type I and type II errors. The basic formulas are discussed by Fleiss (1981) and Bland (2015). They are relatively complex and approximate, can be written in several ways, and are best replaced by free online calculation at http://statpages.org/proppowr.html, http://www.stat.ubc.ca/rollin/ stats/ssize/b2.html, http://www.cct.cuhk.edu.hk/stat/proportion/Casagrande.htm, http://www.sample-size.net/confidence-interval-proportion/, and https://www.statstodo. com/SSiz2Props_Pgm.php. A simplified approximation was devised by Lehr (1992) who used the equation n¼
16pq , ðp1 p2 Þ2
where p is the average of p1 and p2, and q is 1-average p. This gives the number in each group for a power of 0.8 (Type II error β ¼¼ 0.2) and a Type I error α ¼ 0.05 (twotailed). If a power of 0.9 is wanted, the constant is changed from 16 to 21. According to Lehr, this estimate is slightly low for Fisher’s exact test and slightly high for a 2 2 chi-square test. As with all sample size calculations, these are estimates, not precise numbers. As an example, consider how many subjects (equal sized groups) are needed to show a difference between a remission rate of 0.7 for one treatment and 0.6 for another. We wish to set α ¼ 0.05 and β ¼ 0.2, that is, power ¼ 0.8. By Lehr’s formula, n¼
16 0:65 0:35 ¼ 364 as the number required in each group. The more formal ð0:7 0:6Þ2
265
266
Basic Biostatistics for Medical and Biomedical Practitioners
calculation online gives 356 in each group, or 376 if the continuity correction is used. For a power of 0.9, Lehr’s formula gives 478 in each group, and the online calculation gives 476, or 496 with the continuity correction. The major advantage of using an online calculator is that it allows for unequal group sizes. In estimating the numbers needed, we need some idea of the effect size that we want. Take 0.50 as the null hypothesis, that is, there is no difference between the two groups. Then either select an effect size based on previous work or else try to decide what minimum effect size to detect. For example, suppose the null hypothesis is that the probability of survival of a disease under standard treatment is 0.50, we would probably not be interested in a treatment that changed the proportion to only 0.49. Cohen (1988) classified effect size of the difference from 0.50 as small (<0.05, that is, 0.45–0.55), medium (0.15, that is, 0.35–0.65), and large (0.25, i.e., 0.25–0.75). He pointed out that large differences were rare. For example, in presidential elections in the United States there has never been a division as extreme as 65:35, and even a division of 55:45 would be regarded as a landslide victory. Cohen gives tables, and online calculators are available. As always, it takes a huge number of measurements or counts to detect a small difference. Problem 17.2 In two different populations of men, serum cholesterol concentrations below 5.0 mmol/L occur in 6% and 9%, respectively. What sample size is needed to show that this difference allows us to reject the null hypothesis with α ¼ 0.05 and β ¼ 0.2? (assume equal sized groups)?
COMPARING PROPORTIONS To compare two proportions, for example, the survival of patients with a given disease who have two different treatments, let the proportion surviving in group 1 be p1 and that in group 2 be p2. If these two proportions are similar, we conclude that treatment did not affect survival. If they are quite different, then we can ask if the null hypothesis is true. To do this, calculate the standard error of the difference as s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p1 q1 p2 q2 : sp1p2 ¼ n1 n2 Then relate the difference p1–p2 to the standard error of the difference; p1 p2 z ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , p1 q1 p2 q2 n1 n2
Proportions
For example, if treatment 1 gives a survival proportion of 37/192 ¼ 0.1927 and treatment 2 gives a survival of 17/168 ¼ 0.1012, could that difference have occurred by chance? Calculate z as 0:1927 0:1012 0:0915 ffi z ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:0008102 + 0:0005414 0:1927 0:8073 0:1012 0:8988 + 192 168 ¼ 2:4888 Therefore P ¼ 0.0128 (two-sided) and we can reasonably reject the null hypothesis and believe that treatment 2 might be better. The 95% confidence limits for the difference are 0.0915 1.96 0.03676 ¼ 0.05474 to 0.1291. Because this does not include zero, it confirms the results of the z test. The continuity correction can be done by using 1 1 1 |P1 P2 | + 2 n1 n2 Z¼ seðp1p2Þ Sample sizes for adequate power are provided by Cohen (1988). Proportions can be compared at http://www.measuringusability.com/ab-calc.php, https://www.medcalc.org/calc/comparison_of_proportions.php, http://vassarstats.net/ prop2_ind.html, or http://in-silico.net/statistics/ztest/two-proportion and sample size can be calculated at http://www.sample-size/net, and http://www.statstodo.com/ SSiz2Props_Pgm.php. Note: Comparing two proportions is the same as doing a 2 2 chi-square test.
Pooling samples If there are several small samples from a population, with X1 successes in N1 trials, X2 successes in N2 trials, up to Xk successes in Nk trials, then an average proportion of successes can be calculated as P¼
X1 + X2 + …:Xk : N1 + N2 + …Nk
REFERENCES Bland, M., 2015. An Introduction to Medical Statistics. Oxford University Press, Oxford. Cohen, J., 1988. Statistical Power Analysis for Behavioral Sciences. Lawrence Erlbaum Associates, Hillsdale, NJ. Fleiss, J.L., 1981. Statistical Methods for Rates and Proportions. John Wiley & Sons, New York. Hanley, J.A., Lippman-Hand, A., 1983. If nothing goes wrong, is everything alright? JAMA 249, 1743–1745. Lehr, R., 1992. Sixteen s-squared over d-squared: a relation for crude sample size estimates. Stat. Med. 11, 1099–1102. van Belle, G., 2002. Statistical Rules of Thumb. John Wiley & Sons, Inc., New York.
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