Nur Antwort d ist richtig –> Wenn der Effekt, also die Differenz kleiner ist, dann braucht es bei gleichem alpha-level und gleicher Power mehr Personen:
power.prop.test(p1 = 0.26, p2 = 0.11, sig.level = 0.01, power = 0.9)
##
## Two-sample comparison of proportions power calculation
##
## n = 196.9295
## p1 = 0.26
## p2 = 0.11
## sig.level = 0.01
## power = 0.9
## alternative = two.sided
##
## NOTE: n is number in *each* group
power.prop.test(p1 = 0.26, p2 = 0.16, sig.level = 0.01, power = 0.9)
##
## Two-sample comparison of proportions power calculation
##
## n = 491.2201
## p1 = 0.26
## p2 = 0.16
## sig.level = 0.01
## power = 0.9
## alternative = two.sided
##
## NOTE: n is number in *each* group
power.prop.test(p1 = 0.40, p2 = 0.25, sig.level = 0.01, power = 0.9)
##
## Two-sample comparison of proportions power calculation
##
## n = 287.6656
## p1 = 0.4
## p2 = 0.25
## sig.level = 0.01
## power = 0.9
## alternative = two.sided
##
## NOTE: n is number in *each* group
Insgesamt müssten 576 Personen eingeschlossen werden.
power.prop.test(p1 = 0.40, p2 = 0.25, sig.level = 0.05, power = 0.8)
##
## Two-sample comparison of proportions power calculation
##
## n = 151.8689
## p1 = 0.4
## p2 = 0.25
## sig.level = 0.05
## power = 0.8
## alternative = two.sided
##
## NOTE: n is number in *each* group
Insgesamt müssten 304 Personen eingeschlossen werden.
Aus einer Studie haben Sie folgende Angaben:
On the basis of results from RAVEL (the Randomized Study with the Sirolimus-Coated Bx Velocity Balloon-Expandable Stent in the Treatment of Patients with de Novo Native Coronary Artery Lesions) we assumed an incidence of major adverse cardiac events of 6 percent in the sirolimus-stent group and of 12 percent in the paclitaxel-stent group. Enrollment of 1010 patients would provide the study with a statistical power of 90 percent to detect this difference with a two-sided significance level of 0.05.
power.prop.test(p1 = 0.12, p2 = 0.06, sig.level = 0.05, power = 0.9)
##
## Two-sample comparison of proportions power calculation
##
## n = 476.0072
## p1 = 0.12
## p2 = 0.06
## sig.level = 0.05
## power = 0.9
## alternative = two.sided
##
## NOTE: n is number in *each* group
Eigentlich wären nur ca. 954 Personen nötig. Die Autoren rechnen wohl mit einigen Drop-outs, weshalb eine Reserve einberechnet wurde. Dies ist bei Sample-Size Berechnungen nicht selten.
power.prop.test(p1 = 0.12, p2 = 0.06, sig.level = 0.05, power = 0.8)
##
## Two-sample comparison of proportions power calculation
##
## n = 355.9428
## p1 = 0.12
## p2 = 0.06
## sig.level = 0.05
## power = 0.8
## alternative = two.sided
##
## NOTE: n is number in *each* group
Es bräuchte 356 Personen pro Gruppe, für die ganze Studie also 712 Personen. Rechnet man wie oben mit einer Drop-out Rate von ca. 6%, dann würde man 756 Personen einschliessen.
Sie haben folgende Angaben zu einer Studie:
Design, Setting, and Patients: Randomized, double-blind, placebo-controlled trial conducted among 212 patients with peripheral artery disease (mean age, 65.5 [SD, 6.2] years), initiated in May 2008 and completed in August 2011 and conducted at 3 hospitals in Australia.
Intervention: Patients were randomized to receive 10 mg/d of ramipril (n = 106) or matching placebo (n = 106) for 24 weeks.
Main Outcome Measures: Maximum and pain-free walking times were recorded during a standard treadmill test. The Walking Impairment Questionnaire (WIQ) and Short-Form 36 Health Survey (SF-36) were used to assess walking ability and quality of life, respectively.
To detect a 120-second change in walking time (assuming an SD of 300 seconds) and a 65-second change in pain-free walking time (assuming an SD of 161 seconds) with ramipril, 100 patients per group would provide a power of 80% at an α of .05. A 2-sided P value less than .05 was deemed significant.
Überprüfen Sie die Berechnungen oben in R.
power.t.test(delta=120, sd=300, sig.level = 0.05, power = 0.8)
##
## Two-sample t test power calculation
##
## n = 99.08057
## delta = 120
## sd = 300
## sig.level = 0.05
## power = 0.8
## alternative = two.sided
##
## NOTE: n is number in *each* group
power.t.test(delta=65, sd=161, sig.level = 0.05, power = 0.8)
##
## Two-sample t test power calculation
##
## n = 97.27782
## delta = 65
## sd = 161
## sig.level = 0.05
## power = 0.8
## alternative = two.sided
##
## NOTE: n is number in *each* group
Laden Sie den Datensatz “bpstudy.csv”. Die Variable BPX ist der Blutdruck in der Interventionsgruppe und die Variable BPC der Blutdruck in der Kontrollgruppe. Ist der Unterschied der beiden Gruppenmittelwerte statistisch signifikant?
Anmerkung: Das ist ein didaktisches Beispiel. In der Realität macht eine nachträgliche Power-Berechnung wenig Sinn
library(rio)
<- import("../../Data/bpstudy.csv")
data t.test(data$BPX, data$BPC, paired = FALSE)
##
## Welch Two Sample t-test
##
## data: data$BPX and data$BPC
## t = -13.763, df = 37.842, p-value = 2.61e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -20.64801 -15.35199
## sample estimates:
## mean of x mean of y
## 131 149
<- mean(data$BPX) - mean(data$BPC)
mean_diff <- sqrt(sd(data$BPX)^2 + sd(data$BPC)^2)
sd_diff power.t.test(n = 20, delta = mean_diff, sd = sd_diff, sig.level = 0.05)
##
## Two-sample t test power calculation
##
## n = 20
## delta = 18
## sd = 5.848977
## sig.level = 0.05
## power = 1
## alternative = two.sided
##
## NOTE: n is number in *each* group
Der p-Wert ist sehr klein. Folglich ist die Power (beinahe) bei 1.
Ein spezifischer genetischer Marker \(x\) ist bei 10% der gesunden Menschen zu finden. Eine Pilot-Fall-Kontrollstudie hat gezeigt, dass die Odds für diesen Marker bei Menschen mit der Krankheit \(y\) doppelt so gross sind wie bei gesunden Menschen (Odds Ratio = 2).
Tipp: Man braucht die Formeln zum Umrechnen von Proportionen in Odds und umgekehrt.
Zuerst muss die Proportion mit dem Marker bei den Erkrankten berechnet werden:
<- 2
OR <- 0.1
p_controls <- p_controls/(1-p_controls)
odds_controls <- odds_controls * OR
odds_cases <- odds_cases/(1+odds_cases)
p_cases p_cases
## [1] 0.1818182
Danach kann man den power.prop.test() durchführen.
power.prop.test(p1 = p_controls, p2 = p_cases, sig.level = 0.05, power = 0.8)
##
## Two-sample comparison of proportions power calculation
##
## n = 282.686
## p1 = 0.1
## p2 = 0.1818182
## sig.level = 0.05
## power = 0.8
## alternative = two.sided
##
## NOTE: n is number in *each* group
Demnach braucht es 283 Personen pro Gruppe.
Zuerst muss die Proportion mit dem Marker bei den Erkrankten berechnet werden:
<- 2
OR <- 0.05
p_controls <- p_controls/(1-p_controls)
odds_controls <- odds_controls * OR
odds_cases <- odds_cases/(1+odds_cases)
p_cases p_cases
## [1] 0.0952381
power.prop.test(p1 = p_controls, p2 = p_cases, sig.level = 0.05, power = 0.8)
##
## Two-sample comparison of proportions power calculation
##
## n = 515.3997
## p1 = 0.05
## p2 = 0.0952381
## sig.level = 0.05
## power = 0.8
## alternative = two.sided
##
## NOTE: n is number in *each* group