The power of the test is given by \[1-\beta = P(1/\hat{\lambda > k | \lambda = \lambda_A} = P(W > 2dk\lambda_A)).\]. The significance level of the test, or Type I error rate, is \(\alpha = P(\hat{\theta}>k | \theta = \theta_0)\). However, one can obtain much simpler, closed-form expressions through a normal approximation. The number of events that we require can be computed using the following function: If a survival distribution estimate is available for the control group, say, from an earlier trial, then we can use that, along with the proportional hazards assumption, to estimate a probability of death without assuming that the survival distribution is exponential. Let \(\tilde{\pi} = \big( \frac{p\pi_0 + (1-p)\pi_1}{\pi_0 \pi_1} \big)^{-1} = \big( \frac{p}{\pi_1} + \frac{1-p}{\pi_0} \big)^{-1}\), then \(\tilde{\pi}\) is a weighted harmonic mean of \(\pi_0\) and \(\pi_1\), and thus may be viewed as an average probability of death across the control and treatment groups. logical indicating if the hash-version of Following is the function to compute \(d\) as described above: This gives us the number of deaths needed to achieve the specified power, but not the total number of patients. Let us use the reparameterization \(\theta = \log(\mu) = -\log(\lambda)\). Why? of nonzero weights must be at least size in this case. We also need to enter the standard deviation, unlike R where we calculated the effect size separately. statistic based on its the confidence interval. Let \(\Phi(z_\alpha) = 1-\alpha\), then \(z_\alpha = \frac{k-\theta_0}{1/\sqrt{d}}\) and hence \(k = \theta_0 + \frac{z_\alpha}{\sqrt{d}}\). By now, samplesize considers the sample takes place from 1:x. Of the four variables that go into the sample size calculation, the variance of the responses can be the most difficult to determine. This allows for determining the number of deaths (or events) required to meet the power and other design specifications. a non-negative integer giving the number of items to choose. This form of the number of patients required can be interpreted as the number of deaths \(d = \frac{(z_\alpha+z_\beta)^2}{\delta^2p(1-p)}\) divided by the probability of death \(\tilde{\pi}\). Thus, we would need to enroll \(n=148\) patients, \(n_1=n_2=74\) patients in each arm, to meet the design specifications. In this lecture we will do some hands-on examples of power and sample size calculations in survival analysis using R. Note: This lecture is designed based on several resources. Non-integer positive numerical values of n or x will be However, we can calculate an expected effect size, given a desired uplift. It is allowed to ask for size = 0 samples with n = 0 or See Annals of surgery, 258(6).↩, Find statistical considerations for a study where the outcome is a time to failure: http://hedwig.mgh.harvard.edu/sample_size/time_to_event/para_time.html↩, \(T_i \stackrel{iid}{\sim} Exp(\lambda) = Gamma(1, \lambda)\), \(L(\lambda) = \lambda^d e^{-\lambda V}\), \(V = \sum\limits_i t_i \sim Gamma(d, \lambda)\), \[T_{(2)} - T_{(1)} \sim Exp((n - 1)\lambda),\], \[T_{(j)} - T_{(j-1)} \sim Exp((n - j + 1)\lambda),\], \(U_j = (n - j + 1)(T_{(j)} - T_{(j-1)})\), \(U_j \stackrel{iid}{\sim} Exp(\lambda)\), \(V = \sum\limits_j u_j \sim Gamma(d, \lambda)\), \(\bar{X} \sim N(\frac{1}{\lambda},\frac{1}{n\lambda^2})\), \(log \bar{X} \sim N(\log(\lambda),\frac{1}{n})\), \(X_i \stackrel{iid}{\sim} Exp(\lambda_1)\), \(Y_i \stackrel{iid}{\sim} Exp(\lambda_2)\), \(\log\big(\frac{\bar{Y}}{\bar{X}}\big) \sim N\big(\Delta,\frac{1}{n_1}+\frac{1}{n_2}\big)\), \(\Delta = \frac{\mu_A}{\mu_0} = \frac{\lambda_0}{\lambda_A}\), \(\alpha = P(\hat{\theta}>k | \theta = \theta_0)\), \(Z = \frac{\hat{\theta}-\theta}{1/\sqrt{d}}\), \(\alpha = P(Z>\frac{k-\theta_0}{1/\sqrt{d}})\), \(z_\alpha = \frac{k-\theta_0}{1/\sqrt{d}}\), \(k = \theta_0 + \frac{z_\alpha}{\sqrt{d}}\), \[ 1-\beta = P(\hat{\theta} > k | \theta = \theta_A) = P\Big(Z>\frac{k-\theta_A}{1/\sqrt{d}}\Big)\], \[z_{1-\beta} = -z_\beta = \sqrt{d}(k-\theta_A) = \sqrt{d}\Big(\theta_0 + \frac{z_\alpha}{\sqrt{d}}-\theta_A\Big)\], \[ \Rightarrow d = \frac{(z_\beta+z_\alpha)^2}{(\theta_A+\theta_0)^2} = \frac{(z_\beta+z_\alpha)^2}{(\log \Delta)^2}.\], \[W = \frac{2d\lambda}{\hat{\lambda}} \sim \chi^2_{2d},\], \(\alpha = P(1/\hat{\lambda} > k |\lambda = \lambda_0) = P(W > 2dk\lambda_0)\), \(k = \frac{\chi^2_{2d,\alpha}}{2d\lambda_0}\), \[1-\beta = P(1/\hat{\lambda > k | \lambda = \lambda_A} = P(W > 2dk\lambda_A)).\], \(\chi^2_{2d,1-\beta} = 2dk\lambda_A \Rightarrow \chi^2_{2d,1-\beta} = \frac{\chi^2_{2d,\alpha}\lambda_A}{\lambda_0}\), \(\Delta = \frac{\lambda_0}{\lambda_A} = \frac{\chi^2_{2d,\alpha}}{\chi^2_{2d,1-\beta}}\), \(LR(d) = \frac{\chi^2_{2d,\alpha}}{\chi^2_{2d,1-\beta}} - \Delta\), \[ \pi = \int\limits_0^a \frac{1}{a} [1-S_\lambda(a+f-t)] dt.\], \[ \pi = 1 - \frac{1}{a\lambda}[e^{-\lambda f} - e^{-\lambda (a+f)}].\], \(\delta = \log \Delta = \log \lambda_0 - \log \lambda_1\), \[\sigma^2 = var(\hat{\delta}) = \frac{1}{E(d_0)} + \frac{1}{E(d_1)} = \frac{1}{n_o\pi_0} + \frac{1}{n_1 \pi_1} = \frac{1}{np(1-p)} \times \frac{p\pi_0 + (1-p)\pi_1}{\pi_0 \pi_1},\], \(\tilde{\pi} = \big( \frac{p\pi_0 + (1-p)\pi_1}{\pi_0 \pi_1} \big)^{-1} = \big( \frac{p}{\pi_1} + \frac{1-p}{\pi_0} \big)^{-1}\), \(\alpha = P(\hat{\delta}>k | \delta = 0) = P(Z > k/\sigma)\), \[1-\beta = P(\hat{\delta}>k|\delta = \delta_A) = P\Big(Z > \frac{k-\delta_A}{\sigma}\Big)\], \(z_{1-\beta} = -z_\beta = \frac{k-\delta_A}{\sigma}\), \(\sigma = \frac{\delta}{z_\alpha+z_\beta}\), \(\sigma^2 = \frac{1}{np(1-p)}\tilde{\pi}^{-1} = \frac{\delta^2}{(z_\alpha+z_\beta)^2}\), \[n = \frac{(z_\alpha+z_\beta)^2}{\delta^2p(1-p)\tilde{\pi}}.\], \(d = \frac{(z_\alpha+z_\beta)^2}{\delta^2p(1-p)}\), \(v_{0i} = var(d_{0i}) = \frac{n_{0i}n_{1i}d_i(n_i - d_i)}{n_i^2(n_i-1)}\), \[v_{0i} \approx \frac{n_{0i}n_{1i}d_i(n_i - d_i)}{n_i^2(n_i-1)} \approx \frac{n_{0i}n_{1i}d_i}{n_i^2} \approx p(1-p)d_i.\], \(V_0 = var(U_0) \approx p(1-p)\sum d_i = p(1-p)d.\), \[d = \frac{(z_{\alpha/2} + z_\beta)^2}{p(1-p)\delta^2}.\], \(\hat{S}_1(t) = [\hat{S}_0(t)]^{1/\Delta}\), \(\hat{S}(t) = p\hat{S}_0(t) + (1-p)\hat{S}_1(t)\), \[\pi = \int\limits_0^a \frac{1}{a} [1-\hat{S}(a+f-t)] dt = 1-\frac{1}{a} \int\limits_f^{a+f} \hat{S}(t) dt\], \(\pi_t = 1-\frac{1}{4} \{ \hat{S}(a+f) + 2\hat{S}(\frac{a}{2} + f) + \hat{S}(f) \}\), \(\pi_s = 1-\frac{1}{6} \{ \hat{S}(a+f) + 4\hat{S}(\frac{a}{2} + f) + \hat{S}(f) \}\), \[\pi_r = \sum\limits_{t_{(i)}:fk | \delta = 0) = P(Z > k/\sigma)\). All the user needs to do is pass some baseline numbers into some functions I have created and they can determine their sample size requirements and experiment duration on an ad-hoc basis. The analyses account for the effects of potential confounder variables that are also included in the multivariate logistic model. the algorithm should be used. notably the change of sample() results with R version 3.6.0. The input for the function is: n – sample size in each group; p1 – the underlying proportion in group 1 (between 0 and 1) p2 – the underlying proportion in group 2 (between 0 and 1) The simple answer is that neither program is using the above formula. Sample () function in R, generates a sample of the specified size from the data set or elements, either with or without replacement.

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