epi.sscc | R Documentation |

Calculates the sample size, power or minimum detectable odds ratio for an unmatched or matched case-control study.

```
epi.sscc(N = NA, OR, p1 = NA, p0, n, power, r = 1,
phi.coef = 0, design = 1, sided.test = 2, nfractional = FALSE,
conf.level = 0.95, method = "unmatched", fleiss = FALSE)
```

`N` |
scalar, the total number of subjects eligible for inclusion in the study. If |

`OR` |
scalar, the expected study odds ratio. |

`p1` |
scalar, the prevalence of exposure amongst the cases. |

`p0` |
scalar, the prevalence of exposure amongst the controls. |

`n` |
scalar, the total number of subjects in the study (i.e., the number of cases plus the number of controls). |

`power` |
scalar, the required study power. |

`r` |
scalar, the number in the control group divided by the number in the case group. |

`phi.coef` |
scalar, the correlation between case and control exposures for matched pairs. Ignored when |

`design` |
scalar, the design effect. |

`sided.test` |
use a one- or two-sided test? Use a two-sided test if you wish to evaluate whether or not the odds of exposure in cases is greater than or less than the odds of exposure in controls. Use a one-sided test to evaluate whether or not the odds of exposure in cases is greater than the odds of exposure in controls. |

`nfractional` |
logical, return fractional sample size. |

`conf.level` |
scalar, the level of confidence in the computed result. |

`method` |
a character string defining the method to be used. Options are |

`fleiss` |
logical, indicating whether or not the Fleiss correction should be applied. This argument is ignored when |

This function implements the methodology described by Dupont (1988). A detailed description of sample size calculations for case-control studies (with numerous worked examples, some of them reproduced below) is provided by Woodward (2014), pp. 295 - 329.

A value for `p1`

is only required if Fleiss correction is used. For this reason the default for `p1`

is set to `NA`

.

The correlation between case and control exposures for matched pairs (`phi.coef`

) can be estimated from previous studies using Equation (6.2) from Fleiss et al. 2003, p. 98. In the function `epi.2by2`

the variable `phi.coef`

is included with the output for each of the uncorrected chi-squared tests. This value can be used for argument `phi.coef`

in `epi.sscc`

.

The methodology described in Woodward (2014), pp. 295 - 329 uses the proportion of discordant pairs to to determine the sample size for a matched case-control study. Note that the proportion of discordant pairs is likely to vary considerably between different studies since it depends not only on the correlation between case and control exposures but also on the exposure prevalence and the odds ratio. In contrast, estimates of `phi.coef`

should be more stable between similar studies.

When no estimate of `phi.coef`

is available, investigators may prefer to perform their power calculations under the assumption that `phi.coef`

equals, say, 0.2 rather than make the questionable independence assumption required by most other methods.

A finite population correction factor is applied to the sample size estimates when a value for `N`

is provided.

Power calculations used in this function are based on the uncorrected chi-square (as opposed to Fisher's exact) test.

A list containing the following:

`n.total` |
the total number of subjects required to estimate the specified odds ratio at the desired level of confidence and power (i.e., the number of cases plus the number of controls). |

`n.case` |
the total number of case subjects required to estimate the specified odds ratio at the desired level of confidence and power. |

`n.control` |
the total number of control subjects required to estimate the specified odds ratio at the desired level of confidence and power. |

`power` |
the power of the study given the number of study subjects, the specified odds ratio and the desired level of confidence. |

`OR` |
the expected detectable odds ratio given the number of study subjects, the desired power and desired level of confidence. |

The power of a study is its ability to demonstrate the presence of an association, given that an association actually exists.

See the documentation for `epi.sscohortc`

for an example using the `design`

facility implemented in this function.

Dupont WD (1988) Power calculations for matched case-control studies. Biometrics 44: 1157 - 1168.

Fleiss JL, Levin B, Paik MC (2003). Statistical Methods for Rates and Proportions. John Wiley and Sons, New York.

Kelsey JL, Thompson WD, Evans AS (1986). Methods in Observational Epidemiology. Oxford University Press, London, pp. 254 - 284.

Woodward M (2014). Epidemiology Study Design and Data Analysis. Chapman & Hall/CRC, New York, pp. 295 - 329.

```
## EXAMPLE 1 (from Woodward 2014 Example 8.17 p. 318):
## A case-control study of the relationship between smoking and CHD is
## planned. A sample of men with newly diagnosed CHD will be compared for
## smoking status with a sample of controls. Assuming an equal number of
## cases and controls, how many study subject are required to detect an
## odds ratio of 2.0 with 0.90 power using a two-sided 0.05 test? Previous
## surveys have shown that around 0.30 of males without CHD are smokers.
epi.sscc(N = NA, OR = 2.0, p1 = NA, p0 = 0.30, n = NA, power = 0.90, r = 1,
phi.coef = 0, design = 1, sided.test = 2, nfractional = FALSE,
conf.level = 0.95, method = "unmatched", fleiss = FALSE)
## A total of 376 men need to be sampled: 188 cases and 188 controls.
## EXAMPLE 2 (from Woodward 2014 Example 8.18 p. 320):
## Suppose we wish to determine the power to detect an odds ratio of 2.0
## using a two-sided 0.05 test when 188 cases and 940 controls
## are available (that is, the ratio of controls to cases is 5:1). Assume
## the prevalence of smoking in males without CHD is 0.30.
n <- 188 + 940
epi.sscc(N = NA, OR = 2.0, p1 = NA, p0 = 0.30, n = n, power = NA, r = 5,
phi.coef = 0, design = 1, sided.test = 2, nfractional = FALSE,
conf.level = 0.95, method = "unmatched", fleiss = TRUE)
## The power of this study, with the given sample size allocation is 0.99.
## EXAMPLE 3:
## The following statement appeared in a study proposal to identify risk
## factors for campylobacteriosis in humans:
## `We will prospectively recruit 300 culture-confirmed Campylobacter cases
## reported under the Public Health Act. We will then recruit one control per
## case from general practices of the enrolled cases, using frequency matching
## by age and sex. With exposure levels of 10% (thought to be realistic
## given past foodborne disease case control studies) this sample size
## will provide 80% power to detect an odds ratio of 2 at the 5% alpha
## level.'
## Confirm the statement that 300 case subjects will provide 80% power in
## this study.
epi.sscc(N = NA, OR = 2.0, p1 = NA, p0 = 0.10, n = 600, power = NA, r = 1,
phi.coef = 0.01, design = 1, sided.test = 2, nfractional = FALSE,
conf.level = 0.95, method = "matched", fleiss = TRUE)
## If the true odds ratio for Campylobacter in exposed subjects relative to
## unexposed subjects is 2.0 we will be able to reject the null hypothesis
## that this odds ratio equals 1 with probability (power) 0.826. The Type I
# error probability associated with this test of this null hypothesis is 0.05.
## EXAMPLE 4:
## We wish to conduct a case-control study to assess whether bladder cancer
## may be associated with past exposure to cigarette smoking. Cases will be
## patients with bladder cancer and controls will be patients hospitalised
## for injury. It is assumed that 20% of controls will be smokers or past
## smokers, and we wish to detect an odds ratio of 2 with power 90%.
## Three controls will be recruited for every case. How many subjects need
## to be enrolled in the study?
epi.sscc(N = NA, OR = 2.0, p1 = NA, p0 = 0.20, n = NA, power = 0.90, r = 3,
phi.coef = 0, design = 1, sided.test = 2, nfractional = FALSE,
conf.level = 0.95, method = "unmatched", fleiss = FALSE)
## A total of 620 subjects need to be enrolled in the study: 155 cases and
## 465 controls.
## An alternative is to conduct a matched case-control study rather than the
## unmatched design outlined above. One case will be matched to one control
## and the correlation between case and control exposures for matched pairs
## (phi.coef) is estimated to be 0.01 (low). Using the same assumptions as
## those described above, how many study subjects will be required?
epi.sscc(N = NA, OR = 2.0, p1 = NA, p0 = 0.20, n = NA, power = 0.90, r = 1,
phi.coef = 0.01, design = 1, sided.test = 2, nfractional = FALSE,
conf.level = 0.95, method = "matched", fleiss = FALSE)
## A total of 456 subjects need to be enrolled in the study: 228 cases and
## 228 controls.
## EXAMPLE 5:
## Code to reproduce the isograph shown in Figure 2 in Dupont (1988):
r <- 1
p0 = seq(from = 0.05, to = 0.95, length = 50)
OR <- seq(from = 1.05, to = 6, length = 100)
dat.df05 <- expand.grid(p0 = p0, OR = OR)
dat.df05$n.total <- NA
for(i in 1:nrow(dat.df05)){
dat.df05$n.total[i] <- epi.sscc(N = NA, OR = dat.df05$OR[i], p1 = NA,
p0 = dat.df05$p0[i], n = NA, power = 0.80, r = 1,
phi.coef = 0, design = 1, sided.test = 2, nfractional = FALSE,
conf.level = 0.95, method = "unmatched", fleiss = FALSE)$n.total
}
grid.n <- matrix(dat.df05$n.total, nrow = length(p0))
breaks <- c(22:30,32,34,36,40,45,50,55,60,70,80,90,100,125,150,175,
200,300,500,1000)
par(mar = c(5,5,0,5), bty = "n")
contour(x = p0, y = OR, z = log10(grid.n), add = FALSE, levels = log10(breaks),
labels = breaks, xlim = c(0,1), ylim = c(1,6), las = 1, method = "flattest",
xlab = 'Proportion of controls exposed', ylab = "Minimum OR to detect")
## Not run:
## The same plot using ggplot2:
library(ggplot2)
ggplot(data = dat.df05, aes(x = p0, y = OR, z = n.total)) +
theme_bw() +
geom_contour(aes(colour = ..level..), breaks = breaks) +
scale_x_continuous(limits = c(0,1), name = "Proportion of controls exposed") +
scale_y_continuous(limits = c(1,6), name = "Minimum OR to detect")
## End(Not run)
## EXAMPLE 6 (from Dupont 1988, p. 1164):
## A matched case control study is to be carried out to quantify the
## association between exposure A and an outcome B. Assume the prevalence
## of exposure in controls is 0.60 and the correlation between case and
## control exposures for matched pairs (phi.coef) is 0.20 (moderate). Assuming
## an equal number of cases and controls, how many subjects need to be
## enrolled into the study to detect an odds ratio of 3.0 with 0.80 power
## using a two-sided 0.05 test?
epi.sscc(N = NA, OR = 3.0, p1 = NA, p0 = 0.60, n = NA, power = 0.80, r = 1,
phi.coef = 0.2, design = 1, sided.test = 2, nfractional = FALSE,
conf.level = 0.95, method = "matched", fleiss = FALSE)
## A total of 162 subjects need to be enrolled in the study: 81 cases and
## 81 controls.
## How many cases and controls are required if we select three
## controls per case?
epi.sscc(N = NA, OR = 3.0, p1 = NA, p0 = 0.60, n = NA, power = 0.80, r = 3,
phi.coef = 0.2, design = 1, sided.test = 2, nfractional = FALSE,
conf.level = 0.95, method = "matched", fleiss = FALSE)
## A total of 204 subjects need to be enrolled in the study: 51 cases and
## 153 controls.
```

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