A2.6 Sample size calculations for case-control studies

A2.6 PRACTICAL: R

Estimating sample size for case-control studies

You have now been asked to help another research group with their power calculations. They want to conduct a case-control study of sterilization (tubal ligation) and ovarian cancer. You could use parameters from a recent paper (reference below). Parameters of interest might include the proportion of cases or controls with tubal ligation, and the odds ratio.

Tubal ligation, hysterectomy and epithelial ovarian cancer in the New England Case-Control Study. Rice MS, Murphy MA, Vitonis AF, Cramer DW, Titus LJ, Tworoger SS, Terry KL. Int J Cancer. 2013 Nov 15;133(10):2415-21. doi: 10.1002/ijc.28249. Epub 2013 Jul 9.

Looking at the paper, you could extract the following relevant information needed to estimate your sample size:

Proportion of controls with tubal ligation: 18.5%.
Proportion of cases with tubal ligation: 12.8%.
Odds ratio for the association between tubal ligation and ovarian cancer: 0.82.
This study used 2,265 cases and 2,333 controls.
Note: you probably won’t need to use all of these parameters.

Below is a plot showing how the sample size would change, depending on the odds ratio (assuming 18% of controls have had tubal ligation, and 90% power, and allowing the odds ratio to vary between 0.6 and 0.9):

Sample size for case-control studies plot

If you want to use a test of proportions to assess how large your sample needs to be for an expected odds ratio of the association between tubal ligation and ovarian cancer, you need to have two proportions for the command. Above we assume that 18% of controls have had tubal ligation, so we then can use this formula to work background from an expected odds ratio to see the other proportion:

p2 = (OR*p1 )/(1 +p1 (OR-1))

Question A2.6a: Looking at the plot, how many people do you need to recruit into your study if your odds ratio is less than 0.8 and you want to have power of at least 80%?
Question A2.6b: Now use a test of two proportions to assess how large your sample needs to be if the odds ratio of tubal ligation and ovarian cancer is 0.85. You assume 18% of the controls have had tubal ligation and you want 90% power. You need to have two proportions to run the command, so use the formula presented above.

Answer

Answer A2.6a: If the odds ratio in this scenario is ≤ 0.8, (i.e. an effect size of 20% -40%) then you can reach a power ≥ 0.8 with a sample size of around 5000 people. If the true odds ratio is closer to 0.9 (i.e. a 10% effect size), then you would require a much larger sample size.

Answer A2.6b:

We substitute a proportion of 18% and an OR of 0.85 into the formula:

p2 = (OR*p1 )/(1 +p1 (OR-1))

p2= (0.85*0.18)/(1+0.18(0.85-1)

Using R like a calculator, we get:

> (0.85*0.18)/(1+0.18*(0.85-1))

[1] 0.1572456

> power11<-pwr.2p.test(h=ES.h(p1=0.18, p2=0.1572), power=0.9, sig.level=0.05)

> power11

Difference of proportion power calculation for binomial distribution (arcsine transformation)

h = 0.06092931

n = 5660.744

sig.level = 0.05

power = 0.9

alternative = two.sided

NOTE: same sample sizes

You need 5661 cases and 5661 controls, so about 11,322 participants in total, to obtain 90% power if your OR of tubal ligation with ovarian cancer is 0.85.

A2.6 PRACTICAL: Stata

Estimating sample size for case-control studies.

You have now been asked to help another research group with their power calculations. They want to conduct a case-control study of sterilization (tubal ligation) and ovarian cancer. You could use parameters from a recent paper (reference below). Parameters of interest might include the proportion of cases or controls with tubal ligation, and the odds ratio.

Tubal ligation, hysterectomy and epithelial ovarian cancer in the New England Case-Control Study. Rice MS, Murphy MA, Vitonis AF, Cramer DW, Titus LJ, Tworoger SS, Terry KL. Int J Cancer. 2013 Nov 15;133(10):2415-21. doi: 10.1002/ijc.28249. Epub 2013 Jul 9.

Looking at the paper, you could extract the following relevant information needed to estimate your sample size:

- Proportion of controls with tubal ligation: 18.5%.
- Proportion of cases with tubal ligation: 12.8%.
- Odds ratio for the association between tubal ligation and ovarian cancer: 0.82.
- This study used 2,265 cases and 2,333 controls.

Note: you probably won’t need to use all of these parameters.

For a table showing how the sample size would change, depending on the odds ratio (assuming 18% of controls have had tubal ligation, and 90% power, and allowing the odds ratio to vary between 0.6 and 0.9):

power twoproportions 0.18, alpha(0.05) effect(oratio) power(0.9)
oratio(0.6,0.7,0.8,0.9) table

Estimated sample sizes for a two-sample proportions test
Pearson’s chi-squared test
Ho: p2 = p1 versus Ha: p2 != p1

+————————————————————————-+
| alpha power N N1 N2 delta p1 p2 oratio |
|————————————————————————-|
| .05 .9 1308 654 654 .6 .18 .1164 .6 |
| .05 .9 2530 1265 1265 .7 .18 .1332 .7 |
| .05 .9 6162 3081 3081 .8 .18 .1494 .8 |
| .05 .9 26552 13276 13276 .9 .18 .165 .9 |
+————————————————————————-+

For a graph illustrating how the power and sample size would change with different odds ratios (again assuming 18% of controls have had tubal ligation, and allowing the odds ratio to vary from 0.6 to 0.9, and the sample size to vary from 1000 to 15,000):

*– Basic plot;
power twoproportions 0.18, alpha(0.05) effect(oratio) ///
oratio(0.6,0.7,0.8,0.9) n(1000(1000)15000) graph

*– Changing the graph using plot options:
power twoproportions 0.18, alpha(0.05) effect(oratio) ///
oratio(0.6,0.7,0.8,0.9) n(1000(1000)15000) ///
graph(plot1opts(msymbol(O) lpattern(solid)) ///
plot2opts(msymbol(D) lpattern(dash)) ///
plot3opts(msymbol(S) lpattern(dot)) ///
plot4opts(msymbol(T) lpattern(longdash_dot)) ///
graphregion(color(white)) xlabel(, nogrid) ylabel(, nogrid))

sample size for case-control studies plot

Question A2.6: Looking at the plot, how many people do you need to recruit into your study if your odds ratio is less than 0.8 and you want to have power of at least 80%?

Answer

If the odds ratio in this scenario is ≤ 0.8, then you can reach a power ≥ 0.8 with a sample size of around 5000 people. If the true odds ratio is closer to 0.9, then you would require a much larger sample size.

A2.6 PRACTICAL: SPSS

Estimating sample size for case-control studies

You have now been asked to help another research group with their power calculations. They want to conduct a case-control study of sterilization (tubal ligation) and ovarian cancer. You could use parameters from a recent paper (reference below). Parameters of interest might include the proportion of cases or controls with tubal ligation, and the odds ratio.

Tubal ligation, hysterectomy and epithelial ovarian cancer in the New England Case-Control Study. Rice MS, Murphy MA, Vitonis AF, Cramer DW, Titus LJ, Tworoger SS, Terry KL. Int J Cancer. 2013 Nov 15;133(10):2415-21. doi: 10.1002/ijc.28249. Epub 2013 Jul 9.

Looking at the paper, you could extract the following relevant information needed to estimate your sample size:

Proportion of controls with tubal ligation: 18.5%.
Proportion of cases with tubal ligation: 12.8%.
Odds ratio for the association between tubal ligation and ovarian cancer: 0.82.
This study used 2,265 cases and 2,333 controls.
Note: you probably won’t need to use all of these parameters.

Below is a plot showing how the sample size would change, depending on the odds ratio (assuming 18% of controls have had tubal ligation, and 90% power, and allowing the odds ratio to vary between 0.6 and 0.9):

If you want to use a test of proportions to assess how large your sample needs to be for an expected odds ratio of the association between tubal ligation and ovarian cancer, you need to have two proportions for the command. Above we assume that 18% of controls have had tubal ligation, so we then can use this formula to work background from an expected odds ratio to see the other proportion:

p2 = (OR*p1 )/(1 +p1 (OR-1))

i, Looking at the plot, how many people do you need to recruit into your study if your odds ratio is less than 0.8 and you want to have power of at least 80%?

ii. Now use the Independent Samples Binomial Test as before to assess how large your sample needs to be if the odds ratio of tubal ligation and ovarian cancer is 0.85. You assume 18% of the controls have had tubal ligation and you want 90% power. You need to have two proportions to run the command, so use the formula presented above.

Answer

i. If the odds ratio in this scenario is ≤ 0.8, (i.e. an effect size of 20% -40%) then you can reach a power ≥ 0.8 with a sample size of around 5000 people. If the true odds ratio is closer to 0.9 (i.e. a 10% effect size), then you would require a much larger sample size.

ii. Firstly we substitute a proportion of 18% and an OR of 0.85 into the formula to work out the second proportion:

p2 = (OR*p1 )/(1 +p1 (OR-1))

= (0.85*0.18)/(1+0.18(0.85-1)

= (0.85*0.18)/(1+0.18*(0.85-1))

= 0.1572456

Then input the proportions into the Independent Samples Binomial Test with a power of 0.9 to calculate the required sample size.

You need 5665 cases and 5665 controls, so about 11,330 participants in total, to obtain 90% power if your OR of tubal ligation with ovarian cancer is 0.85.

FoSSA: Fundamentals of Statistical Software & Analysis

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A2.6 Sample size calculations for case-control studies

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Estimating sample size for case-control studies.

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FoSSA: Fundamentals of Statistical Software & Analysis

Course Information

Participants 1660

A2.6 Sample size calculations for case-control studies

Learning Outcomes

Estimating sample size for case-control studies.

Live Bootcamps

Training & Consulting

Free Courses

Get In Touch

TERMS OF SERVICE

PRIVACY POLICY

Join Our Mailing List

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