
# 1 Introduction

A feeding study served to authorize the MON863 maize, a genetically modified organism (GMO) developed by the Monsanto company, by the European and American authorities. It included male and female rats. For each sex, one group was fed with GMOs in the equilibrated diet, and one with the closest control regimen without GMOs.

We are interested in the weight of the rats after a period of 14 weeks.

ratWeight <- read.csv("ratWeight.csv")
data <- subset(ratWeight, week==14)
head(data)
##        id week weight  regime gender dosage
## 14 B38602   14  514.9 Control   Male    11%
## 28 B38603   14  505.0 Control   Male    11%
## 42 B38604   14  545.1 Control   Male    11%
## 56 B38605   14  596.6 Control   Male    11%
## 70 B38606   14  516.8 Control   Male    11%
## 84 B38607   14  518.1 Control   Male    11%

The data per gender and regime is displayed below

library(ggplot2)
theme_set(theme_bw())
ggplot(data=data) + geom_point(aes(x=weight,y=as.numeric(regime),colour=regime, shape=gender)) +
ylab(NULL) + scale_y_continuous(breaks=NULL, limits=c(-5,10)) + xlab("weight (g)") 

The following table provides the mean weight in each group

aggregate(weight ~ regime+gender, data=data, FUN= "mean" )
##    regime gender   weight
## 1 Control Female 278.2825
## 2     GMO Female 287.3225
## 3 Control   Male 513.7077
## 4     GMO   Male 498.7359

Our main objective is to detect some possible effect of the diet on the weight. More precisly, we would like to know if the differences observed in the data are due to random fluctuations in sampling or to differences in diet.

# 2 Student’s t-test

## 2.1 One sample t-test

Before considering the problem of comparing two groups, let us start looking at the weight of the male rats only:

ggplot(data=subset(data,gender=="Male")) + geom_point(aes(x=weight,y=0), colour="red") +
ylab(NULL) + scale_y_continuous(breaks=NULL) + xlab("weight (g)")

Let $$x_1, x_2, x_n$$ the weights of the $$n$$ male rats. We will assume that the $$x_i$$’s are independent and normally distributed with mean $$\mu$$ and variance $$\sigma^2$$:

$x_i \iid {\cal N}(\mu \ , \ \sigma^2)$

### 2.1.1 One sided test

We want to test

$H_0: \ \mu \leq \mu_0" \quad \text{versus} \quad H_1: \ \mu > \mu_0"$

Function t.test can be used for performing this test:

x <- data[datagender=="Male","weight"] mu0 <- 500 t.test(x, alternative="greater", mu=mu0) ## ## One Sample t-test ## ## data: x ## t = 1.2708, df = 77, p-value = 0.1038 ## alternative hypothesis: true mean is greater than 500 ## 95 percent confidence interval: ## 498.0706 Inf ## sample estimates: ## mean of x ## 506.2218 Let us see what these outputs are and how they are computed. Let $$\bar{x} = n^{-1}\sum_{i=1}^n x_i$$ be the empirical mean of the data. $\bar{x} \sim {\cal N}(\mu \ , \ \frac{\sigma^2}{n})$ Then, \begin{aligned} \frac{\sqrt{n}(\bar{x} - \mu)}{\sigma} \ & \sim \ {\cal N}(0 \ , \ 1) \\ \frac{\sqrt{n}(\bar{x} - \mu)}{s} \ & \sim \ t_{n-1} \end{aligned} where $s^2 = \frac{1}{n-1}\sum_{i=1}^n (x_i - \bar{x})^2$ is the empirical variance of $$(x_i)$$. The statistic used for the test should be a function of the data whose distribution under $$H_0$$ is known, and whose expected behavior under $$H_1$$ allows one to define a rejection region (or critical region) for the null hypothesis. Here, the test statistic is $T_{\rm stat} = \frac{(\bar{x} - \mu_0)}{s/\sqrt{n}}$ which follows a $$t$$-distribution with $$n-1$$ degrees of freedom when $$\mu=\mu_0$$. $$\bar{x}$$ is expected to be less than or equal to $$\mu_0$$ under the null hypothesis, and greater than $$\mu_0$$ under the alternative hypothesis, Hence, $$T_{\rm stat}$$ is expected to be less than or equal to 0 under $$H_0$$ and greater than 0 under $$H_1$$. We then reject the null hypothesis $$H_0$$ if $$T_{\rm stat}$$ is greater than some threshold $$q$$. Such decision rule may lead to two kinds of error: • The type I error is the incorrect rejection of null hypothesis when it is true, • The type II error is the failure to reject the null hypothesis when it is false. The type I error rate or significance level is therefore the probability of rejecting the null hypothesis given that it is true. In our case, for a given significance level $$\alpha$$, we will reject $$H_0$$ if $$T_{\rm stat} > qt_{1-\alpha,n-1}$$, where $$qt_{1-\alpha,n-1}$$ is the quantile of order $$1-\alpha$$ for a $$t$$-distribution with $$n-1$$ degrees of freedom. Indeed, by definition, \begin{aligned} \prob{\text{reject } H_0 \ | \ H_0 \ \text{true}} &= {\mathbb P}(T_{\rm stat} > qt_{1-\alpha,n-1} \ | \ \mu \leq \mu_0) \\ & \leq {\mathbb P}(T_{\rm stat} > qt_{1-\alpha,n-1} \ | \ \mu = \mu_0) \\ & \leq {\mathbb P}(t_{n-1} > qt_{1-\alpha,n-1}) \\ & \leq \alpha \end{aligned} alpha <- 0.05 x.mean <- mean(x) x.sd <- sd(x) n <- length(x) df <- n-1 t.stat <- sqrt(n)*(x.mean-mu0)/x.sd c(t.stat,qt(1-alpha, df)) ## [1] 1.270806 1.664885 We therefore don’t reject $$H_0$$ in our example since $$T_{\rm stat} < qt_{1-\alpha,n-1}$$. We can equivalently compute the significance level for which the test becomes significant. This value is called the p-value: \begin{aligned} p_{\rm value} & = \max{\mathbb P}_{H_0}(T_{\rm stat} > T_{\rm stat}^{\rm obs}) \\ & = {\mathbb P}(T_{\rm stat} > T_{\rm stat}^{\rm obs} \ | \ \mu=\mu_0) \\ &= 1 - \prob{t_{n-1} \leq T_{\rm stat}^{\rm obs}} \end{aligned} Now, $$T_{\rm stat} > qt_{1-\alpha,n-1}$$ under $$H_0$$ if and only if $$\prob{t_{n-1} \leq T_{\rm stat}^{\rm obs}} \geq 1-\alpha$$. Then, the test is significant at the level $$\alpha$$ if and only if $$p_{\rm value}\leq \alpha$$.

( p.value <- 1 - pt(t.stat,df) )
## [1] 0.1038119

Here, we would reject $$H_0$$ for any significance level $$\alpha \geq$$ 0.104.

Important: The fact that the test is not significant at the level $$\alpha$$ does not allow us to conclude that $$H_0$$ is true, i.e. that $$\mu$$ is less than or equal to 500. We can only say that the data does not allow us to conclude that $$\mu>500$$.

Imagine now that we want to test if $$\mu \geq 515$$ for instance. The alternative here is $$H_1: \ \mu < 515$$.

mu0 <- 515
t.test(x, alternative="less", mu=mu0)
##
##  One Sample t-test
##
## data:  x
## t = -1.793, df = 77, p-value = 0.03845
## alternative hypothesis: true mean is less than 515
## 95 percent confidence interval:
##     -Inf 514.373
## sample estimates:
## mean of x
##  506.2218

More generally, we may want to test $H_0: \ \mu \geq \mu_0" \quad \text{versus} \quad H_1: \ \mu < \mu_0"$ We still use the statistic $$T_{\rm stat} = \sqrt{n}(\bar{x}-\mu_0)/s$$ for this test, but the rejection region is now the area that lies to the left of the critical value $$qt_{\alpha,n-1}$$ since

\begin{aligned} \prob{\text{reject } H_0 \ | \ H_0 \ \text{true}} &= {\mathbb P}(T_{\rm stat} < qt_{\alpha,n-1} \ | \ \mu \geq \mu_0) \\ & \leq {\mathbb P}(T_{\rm stat} < qt_{\alpha,n-1} \ | \ \mu = \mu_0) \\ & \leq \alpha \end{aligned}

t.stat <- sqrt(n)*(x.mean-mu0)/x.sd
p.value <- pt(t.stat,df)
c(t.stat, df, p.value)
## [1] -1.79295428 77.00000000  0.03845364

Here, the p-value is less than $$\alpha=0.05$$: we then reject the null hypothesis at the $$5\%$$ level and conclude that $$\mu < 515$$.

### 2.1.2 Two sided test

A two sided test (or two tailed test) can be used to test if $$\mu=500$$ for instance

mu0 = 500
t.test(x, alternative="two.sided", mu=mu0)
##
##  One Sample t-test
##
## data:  x
## t = 1.2708, df = 77, p-value = 0.2076
## alternative hypothesis: true mean is not equal to 500
## 95 percent confidence interval:
##  496.4727 515.9709
## sample estimates:
## mean of x
##  506.2218

More generally, we can test $H_0: \ \mu = \mu_0" \quad \text{versus} \quad H_1: \ \mu \neq \mu_0"$ The test also uses the statistic $$T_{\rm stat} = \sqrt{n}(\bar{x}-\mu_0)/s$$, but the rejection region has now two parts: we reject $$H_0$$ if $$|T_{\rm stat}| > qt_{1-\alpha/2}$$. Indeed,

\begin{aligned} \prob{\text{reject } H_0 \ | \ H_0 \ \text{true}} &= {\mathbb P}(|T_{\rm stat}| > qt_{1 -\frac{\alpha}{2},n-1} \ | \ \mu = \mu_0) \\ & = {\mathbb P}(T_{\rm stat} < qt_{\frac{\alpha}{2},n-1} \ | \ \mu = \mu_0) + {\mathbb P}(T_{\rm stat} > qt_{1-\frac{\alpha}{2},n-1} \ | \ \mu = \mu_0)\\ &= \prob{t_{n-1} \leq qt_{\frac{\alpha}{2},n-1}} + \prob{t_{n-1} \geq qt_{1-\frac{\alpha}{2},n-1}} \\ &= \frac{\alpha}{2} + \frac{\alpha}{2} \\ & = \alpha \end{aligned}

The p-value of the test is now

\begin{aligned} p_{\rm value} & = {\mathbb P}_{H_0}(|T_{\rm stat}| > |T_{\rm stat}^{\rm obs}|) \\ & = {\mathbb P}_{H_0}(T_{\rm stat} < -|T_{\rm stat}^{\rm obs}|) + {\mathbb P}_{H_0}(T_{\rm stat} > |T_{\rm stat}^{\rm obs}|)\\ &= \prob{t_{n-1} \leq -|T_{\rm stat}^{\rm obs}|} + \prob{t_{n-1} \geq |T_{\rm stat}^{\rm obs}|} \\ &= 2 \,\prob{t_{n-1} \leq -|T_{\rm stat}^{\rm obs}|} \end{aligned}

t.stat <- sqrt(n)*(x.mean-mu0)/x.sd
p.value <- 2*pt(-abs(t.stat),df)
c(t.stat, df, p.value)
## [1]  1.2708058 77.0000000  0.2076238

Here, $$p_{\rm value}=$$ 0.208. Then, for any significance level less than 0.208, we cannot reject the hypothesis that $$\mu = 500$$.

### 2.1.3 Confidence interval for the mean

We have just seen that the data doesn’t allow us to reject the hypothesis that $$\mu = 500$$. But we would come to the same conclusion with other values of $$\mu_0$$. In particular, we will never reject the hypothesis that $$\mu = \bar{x}$$:

t.test(x, mu=x.mean, conf.level=1-alpha)$p.value ## [1] 1 For a given significance level ($$\alpha = 0.05$$ for instance), we will not reject the null hypothesis for values of $$\mu_0$$ close enough to $$\bar{x}$$. pv.510 <- t.test(x, mu=510, conf.level=1-alpha)$p.value
pv.497 <- t.test(x, mu=497, conf.level=1-alpha)$p.value c(pv.510, pv.497) ## [1] 0.44265350 0.06340045 On the other hand, we will reject $$H_0$$ for values of $$\mu_0$$ far enough from $$\bar{x}$$: pv.520 <- t.test(x, mu=520, conf.level=1-alpha)$p.value
pv.490 <- t.test(x, mu=490, conf.level=1-alpha)$p.value c(pv.520, pv.490) ## [1] 0.006204188 0.001406681 There exist two values of $$\mu_0$$ for which the decision is borderline pv1 <- t.test(x, mu=495.892, conf.level=1-alpha)$p.value
pv2 <- t.test(x, mu=515.5443, conf.level=1-alpha)p.value c(pv1,pv2) ## [1] 0.03811820 0.06062926 In fact, for a given $$\alpha$$, these two values $$\mu_{\alpha,{\rm lower}}$$ and $$\mu_{\alpha,{\rm upper}}$$ define a confidence interval for $$\mu$$: We are confident’’ at the level $$1-\alpha$$ that any value between $$\mu_{\alpha,{\rm lower}}$$ and $$\mu_{\alpha,{\rm upper}}$$ is a possible value for $$\mu$$. mu <- seq(490,520,by=0.25) t.stat <- (x.mean-mu)/x.sd*sqrt(n) pval <- pmin(pt(-t.stat,df) + (1- pt(t.stat,df)),pt(t.stat,df) + (1- pt(-t.stat,df))) dd <- data.frame(mu=mu, v.pval=pval) CI <- x.mean + x.sd/sqrt(n)*qt(c(alpha/2,1-alpha/2), df) ggplot(data=dd) + geom_line(aes(x=mu,y=pval)) + geom_vline(xintercept=x.mean,colour="red", linetype=2)+ geom_hline(yintercept=alpha,colour="green", linetype=2)+ geom_vline(xintercept=CI,colour="red") + scale_x_continuous(breaks=round(c(490,500,510,520,CI,x.mean),2))  By construction, \begin{aligned} 1-\alpha &= \prob{ qt_{\frac{\alpha}{2},n-1} < T_{\rm stat} < qt_{1-\frac{\alpha}{2},n-1} } \\ &= \prob{ qt_{\frac{\alpha}{2},n-1} < \frac{\bar{x}-\mu}{s/\sqrt{n}} < qt_{1-\frac{\alpha}{2},n-1} } \\ &= \prob{ \bar{x} +\frac{s}{\sqrt{n}}qt_{\frac{\alpha}{2},n-1} < \mu < \bar{x} +\frac{s}{\sqrt{n}}qt_{1-\frac{\alpha}{2},n-1} } \end{aligned} The confidence interval of level $$1-\alpha$$ for $$\mu$$ is therefore the interval ${\rm CI}_{1-\alpha} = [\bar{x} +\frac{s}{\sqrt{n}}qt_{\frac{\alpha}{2},n-1} \ \ , \ \ \bar{x} +\frac{s}{\sqrt{n}}qt_{1-\frac{\alpha}{2},n-1}]$ (CI <- x.mean + x.sd/sqrt(n)*qt(c(alpha/2,1-alpha/2), df)) ## [1] 496.4727 515.9709 Remark 1: The fact that $$\prob{ \mu \in {\rm CI}_{1-\alpha}} = 1- \alpha$$ does not mean that $$\mu$$ is a random variable! It is the bounds of the confidence interval that are random because they are function of the data. A confidence interval of level $$1-\alpha$$ should be interpreted like this: imagine that we repeat the same experiment many times, with the same experimental conditions, and that we build a confidence interval for $$\mu$$ for each of these replicate. Then, the true mean $$\mu$$ will lie in the confidence interval $$(1-\alpha)100\%$$ of the times. Let us check this property with a Monte Carlo simulation. L <- 100000 n <- 100 mu <- 500 sd <- 40 R <- vector(length=L) for (l in (1:L)) { x <- rnorm(n,mu,sd) ci.l <- mean(x) + sd(x)/sqrt(n)*qt(c(alpha/2, 1-alpha/2),n-1) R[l] <- (mu > ci.l[1] && mu < ci.l[2]) } mean(R) ## [1] 0.94804 Remark 2: The decision rule to reject or not the null hypothesis can be derived from the confidence interval. Indeed, the confidence interval plays the role of an acceptance region: we reject $$H_0$$ if $$\mu_0$$ does not belong to $${\rm CI}_{1-\alpha}$$. In the case of a one sided test, the output of t.test called confidence interval is indeed an acceptance region for $$\mu$$, but not a confidence interval’’ (we cannot seriouly consider that $$\mu$$ can take any value above 500 for instance ) rbind( c(x.mean + x.sd/sqrt(n)*qt(alpha,df) , Inf), c(-Inf, x.mean + x.sd/sqrt(n)*qt(1-alpha,df))) ## [,1] [,2] ## [1,] 499.0229 Inf ## [2,] -Inf 513.4207 ## 2.2 Two samples t-test ### 2.2.1 What should we test? Let us now compare the weights of the male and female rats. y <- data[datagender=="Female" ,"weight"]

dmean <- data.frame(x=c(mean(x),mean(y)),gender=c("Male","Female"))

ggplot(data=subset(data,regime=="Control")) + geom_point(aes(x=weight,y=0,colour=gender)) +
geom_point(data=dmean, aes(x,y=0,colour=gender), size=4) +
ylab(NULL) + scale_y_continuous(breaks=NULL) + xlab("weight (g)")

Looking at the data is more than enough for concluding that the mean weight of the males is (much) larger than the mean weight of the females Computing a $$p$$-value here is of little interest

t.test(x, y)
##
##  Welch Two Sample t-test
##
## data:  x and y
## t = 46.912, df = 166.81, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  207.0330 225.2244
## sample estimates:
## mean of x mean of y
##  498.9312  282.8025

Let us see now what happens if we compare the control and GMO groups for the male rats.

x <- data[data$gender=="Male" & data$regime=="Control","weight"]
y <- data[data$gender=="Male" & data$regime=="GMO","weight"]

dmean <- data.frame(x=c(mean(x),mean(y)),regime=c("Control","GMO"))
ggplot(data=data[datagender=="Male",]) + geom_point(aes(x=weight,y=as.numeric(regime),colour=regime)) + geom_point(data=dmean, aes(x,y=as.numeric(regime),colour=regime), size=4) + ylab(NULL) + scale_y_continuous(breaks=NULL, limits=c(-6,9)) + xlab("weight (g)") We observe a difference between the two empirical means (the mean weight after 14 weeks is greater in the control group), but we cannot say how significant this difference is by simply looking at the data. Performing a statistical test is now necessary. Let $$x_{1}, x_{2}, \ldots, x_{n_x}$$ be the weights of the $$n_x$$ male rats of the control group and $$y_{1}, y_{2}, \ldots, y_{n_x}$$ the weights of the $$n_y$$ male rats of the GMO group. We will assume normal distributions for both $$(x_{i})$$ and $$(y_{i})$$: $x_{i} \iid {\cal N}(\mu_x \ , \ \sigma^2_x) \quad ; \quad y_{i} \iid {\cal N}(\mu_y \ , \ \sigma^2_y)$ We want to test $H_0: \ \mu_x = \mu_y" \quad \text{versus} \quad H_1: \ \mu_x \neq \mu_y"$ ### 2.2.2 Assuming equal variances We can use the function t.test assuming first equal variances ($$\sigma^2_x=\sigma_y^2$$) alpha <- 0.05 t.test(x, y, conf.level=1-alpha, var.equal=TRUE) ## ## Two Sample t-test ## ## data: x and y ## t = 1.5426, df = 76, p-value = 0.1271 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## -4.358031 34.301621 ## sample estimates: ## mean of x mean of y ## 513.7077 498.7359 The test statistic is $T_{\rm stat} = \frac{\bar{x} - \bar{y}}{s_p \sqrt{\frac{1}{n_x}+\frac{1}{n_y}}}$ where $$s_p^2$$ is the pooled variance: $s_p^2 = \frac{1}{n_x+n_y-2} \left(\sum_{i=1}^{n_x} (x_{i}-\bar{x})^2 + \sum_{i=1}^{n_y} (y_{i}-\bar{y})^2 \right)$ Under the null hypothesis, $$T_{\rm stat}$$ follows a $$t$$-distribution with $$n_x+n_y-2$$ degree of freedom. The $$p$$-value is therefore \begin{aligned} p_{\rm value} & = {\mathbb P}_{H_0}(|T_{\rm stat}| > |T_{\rm stat}^{\rm obs}|) \\ &= \prob{t_{n_x+n_y-2} \leq -T_{\rm stat}^{\rm obs}} + 1 - \prob{t_{n_x+n_y-2} \leq T_{\rm stat}^{\rm obs}} \end{aligned} nx <- length(x) ny <- length(y) x.mean <- mean(x) y.mean <- mean(y) x.sc <- sum((x-x.mean)^2) y.sc <- sum((y-y.mean)^2) xy.sd <- sqrt((x.sc+y.sc)/(nx+ny-2)) t.stat <- (x.mean-y.mean)/xy.sd/sqrt(1/nx+1/ny) df <- nx + ny -2 p.value <- pt(-t.stat,df) + (1- pt(t.stat,df)) c(t.stat, df, p.value) ## [1] 1.5426375 76.0000000 0.1270726 The confidence interval for the mean difference $$\mu_x-\mu_y$$ is computed as ${\rm CI}_{1-\alpha} = [\bar{x} - \bar{y} +s_p \sqrt{\frac{1}{n_x}+\frac{1}{n_y}}qt_{\frac{\alpha}{2},n_x+n_y-2} \ \ , \ \ \bar{x} - \bar{y} +s_p \sqrt{\frac{1}{n_x}+\frac{1}{n_y}}qt_{1-\frac{\alpha}{2},n_x+n_y-2} ]$ x.mean-y.mean + xy.sd*sqrt(1/nx+1/ny)*qt(c(alpha/2,1-alpha/2),df) ## [1] -4.358031 34.301621 ### 2.2.3 Assuming different variances Assuming equal variances for the two groups may be disputable. aggregate(dataweight ~ data$regime, FUN= "sd" ) ## data$regime dataweight ## 1 Control 123.0689 ## 2 GMO 111.8559 We can then use the t.test function with different variances (which is the default) t.test(x, y, conf.level=1-alpha) ## ## Welch Two Sample t-test ## ## data: x and y ## t = 1.5426, df = 75.976, p-value = 0.1271 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## -4.358129 34.301719 ## sample estimates: ## mean of x mean of y ## 513.7077 498.7359 The Welch (or Satterthwaite) approximation to the degrees of freedom is used instead of $$n_x+n_y-2=$$ 76: $\df_W = \frac{(c_x + c_y)^2}{{c_x^2}/{(n_x-1)} + {c_y^2}/{(n_y-1)}}$ where $$c_x = \sum (x_{i}-\bar{x})^2/(n_x(n_x-1))$$ and $$c_y = \sum (y_{i}-\bar{y})^2/(n_y(n_y-1))$$. Furthermore, unlike in Student’s t-test with equal variances, the denominator is not based on a pooled variance estimate: $T_{\rm stat} = \frac{\bar{x} - \bar{y}}{ \sqrt{{s_x^2}/{n_x}+{s_y^2}/{n_y}}}$ where $$s_x^2$$ and $$s_y^2$$ are the empirical variances of $$(x_i)$$ and $$(y_i)$$: $s_x^2 = \frac{1}{n_x-1}\sum_{i=1}^{n_x} (x_{i}-\bar{x})^2 \quad ; \quad s_y^2 = \frac{1}{n_y-1}\sum_{i=1}^{n_y} (y_{i}-\bar{y})^2$ sbar.xy <- sqrt(var(x)/nx+var(y)/ny) t.stat <- (x.mean-y.mean)/sbar.xy cx <- x.sc/(nx-1)/nx cy <- y.sc/(ny-1)/ny dfw <- (cx + cy)^2 / (cx^2/(nx-1) + cy^2/(ny-1)) p.value <- pt(-t.stat,dfw) + (1- pt(t.stat,dfw)) c(t.stat, dfw, p.value) ## [1] 1.5426375 75.9760868 0.1270739 The confidence interval for $$\mu_x-\mu_y$$ is now computed as ${\rm CI}_{1-\alpha} = [\bar{x} - \bar{y} +\sqrt{\frac{s_x^2}{n_x}+\frac{s_y^2}{n_y}} \ qt_{\frac{\alpha}{2},\df_W} \ \ , \ \ \bar{x} - \bar{y} +\sqrt{\frac{s_x^2}{n_x}+\frac{s_y^2}{n_y}} \ qt_{1-\frac{\alpha}{2},\df_W} ]$ x.mean-y.mean + sbar.xy*qt(c(alpha/2,1-alpha/2),dfw) ## [1] -4.358129 34.301719 ## 2.3 Power of a t-test Until now, we have demonstrated that the experimental data does not highlight any significant difference in weight between the control group and the GMO group. Of course, that does not mean that there is no difference between the two groups. Indeed, absence of evidence is not evidence of absence. In fact, no experimental study would be able to demonstrate the absence of effect of the diet on the weight. Now, the appropriate question is rather to evaluate what the experimental study can detect. If feeding a population of rats with GMOs has a signicant biological effect on the weight, can we ensure with a reasonable level of confidence that our statistical test will reject the null hypothesis and conclude that there is indeed a difference in weight between the two groups? A power analysis allows us to determine the sample size required to detect an effect of a given size with a given degree of confidence. Conversely, it allows us to determine the probability of detecting an effect of a given size with a given level of confidence, under sample size constraints. For a given $$\delta \in {\mathbb R}$$, let $$\beta(\delta)$$ be the type II error rate, i.e. the probability to fail rejecting $$H_0$$ when $$\mu_x-\mu_y = \delta$$, with $$\delta\neq 0$$. The power of the test is the probability to reject the null hypothesis when it is false. It is also a function of $$\delta =\mu_x-\mu_y$$ defined as \begin{aligned} \eta(\delta) &= 1 - \beta(\delta) \\ &= \prob{\text{reject } H_0 \ | \ \mu_x-\mu_y=\delta } \end{aligned} Remember that, for a two sided test, we reject the null hypothesis when $$|T_{\rm stat}| > qt_{1-\alpha/2, \df}$$, where $$\df$$ is the appropriate degree of freedom. On the other hand, $${(\bar{x} - \bar{y} - \delta)}/{s_{xy}}$$, where $$s_{xy} = \sqrt{{s_x^2}/{n_x}+{s_y^2}/{n_y}}$$, follows a $$t$$-distribution with $$\df$$ degrees of freedom. Thus, \begin{aligned} \eta(\delta) &= 1 - {\mathbb P}(qt_{\frac{\alpha}{2},\df} < T_{\rm stat} < qt_{1 -\frac{\alpha}{2},\df} \ | \ \mu_x-\mu_y=\delta) \\ & = 1- {\mathbb P}(qt_{\frac{\alpha}{2},\df} < \frac{\bar{x} - \bar{y}}{s_{xy}} < qt_{1-\frac{\alpha}{2},\df} \ | \ \mu_x-\mu_y=\delta) \\ &= 1- {\mathbb P}(qt_{\frac{\alpha}{2},\df} - \frac{\delta}{s_{xy}} < \frac{\bar{x} - \bar{y} - \delta}{s_{xy}} < qt_{1-\frac{\alpha}{2},\df} - \frac{\delta}{s_{xy}} \ | \ \mu_x-\mu_y=\delta) \\ &= 1 - Ft_{\df}(qt_{1-\frac{\alpha}{2},\df} - \frac{\delta}{s_{xy}}) + Ft_{\df}(qt_{\frac{\alpha}{2},\df} - \frac{\delta}{s_{xy}}) \end{aligned} As an example, let us compute the probability to detect a difference in weight of 10g with two groups of 80 rats each and assuming that the standard deviation is 30g in each group. alpha=0.05 nx.new <- ny.new <- 80 delta.mu <- 10 x.sd <- 30 df <- nx.new+ny.new-2 dt <- delta.mu/x.sd/sqrt(1/nx.new+1/ny.new) 1-pt(qt(1-alpha/2,df)-dt,df) + pt(qt(alpha/2,df)-dt,df)  ## [1] 0.5528906 The function pwr.t.test allows to compute this power: library(pwr) pwr.t.test(n=nx.new, d=delta.mu/x.sd, type="two.sample", alternative="two.sided", sig.level=alpha) ## ## Two-sample t test power calculation ## ## n = 80 ## d = 0.3333333 ## sig.level = 0.05 ## power = 0.5538758 ## alternative = two.sided ## ## NOTE: n is number in *each* group Let us perform a Monte Carlo simulation, to check this result and better understand what it means. Imagine that the true’’ difference in weight is $$\delta=10$$g. Then, if could repeat the same experiment a (very) large number of times, we would reject the null hypothesis in $$55\%$$ of cases. L <- 100000 mux <- 500 muy <- mux + delta.mu Rt <- vector(length=L) for (l in (1:L)) { x.sim <- rnorm(nx.new,mux,x.sd) y.sim <- rnorm(ny.new,muy,x.sd) Rt[l] <- t.test(x.sim, y.sim, alternative="two.sided")p.value < alpha
}
mean(Rt)
## [1] 0.55311

We may consider this probability as too small. If our objective is a power of 80% at least, with the same significance level, we need to increase the sample size.

pwr.t.test(power=0.8, d=delta.mu/x.sd, sig.level=alpha) 
##
##      Two-sample t test power calculation
##
##               n = 142.2462
##               d = 0.3333333
##       sig.level = 0.05
##           power = 0.8
##     alternative = two.sided
##
## NOTE: n is number in *each* group

Indeed, we see that $$n\geq$$ 143 animals per group are required in order to reach a power of 80%.

nx.new <- ny.new <- ceiling(pwr.t.test(power=0.8, d=delta.mu/x.sd, sig.level=alpha)$n) df <- nx.new+ny.new-2 dt <- delta.mu/x.sd/sqrt(1/nx.new+1/ny.new) 1-pt(qt(1-alpha/2,df)-dt,df) + pt(qt(alpha/2,df)-dt,df) ## [1] 0.8020466 An alternative for increasing the power consists in increasing the type I error rate pwr.t.test(power=0.8, d=delta.mu/x.sd, n=80, sig.level=NULL)  ## ## Two-sample t test power calculation ## ## n = 80 ## d = 0.3333333 ## sig.level = 0.2067337 ## power = 0.8 ## alternative = two.sided ## ## NOTE: n is number in *each* group If we accept a significance level of about 20%, then we will be less demanding for rejecting $$H_0$$: we will reject the null hypothesis when $$|T_{\rm stat}|>qt_{0.9,158}$$= 1.29, instead of $$|T_{\rm stat}|>qt_{0.975,158}$$= 1.98. This strategy will therefore increase the power, but also the type I error rate. # 3 Mann-Whitney-Wilcoxon test The Mann-Whitney-Wilcoxon test, or Wilcoxon rank sum test, can be used to test if the weight in one of the two groups tends to be greater than in the other group. The Mann-Whitney-Wilcoxon test is a non parametric test: we don’t make the assumption that the distribution of the data belongs to a family of parametric ditributions. The logic behind the Wilcoxon test is quite simple. The data are ranked to produce two rank totals, one for each group. If there is a systematic difference between the two groups, then most of the high ranks will belong to one group and most of the low ranks will belong to the other one. As a result, the rank totals will be quite different and one of the rank totals will be quite small. On the other hand, if the two groups are similar, then high and low ranks will be distributed fairly evenly between the two groups and the rank totals will be fairly similar. In our example, we don’t clearly see any of the two groups on the right or on the left of the scatter plot ggplot(data=data[data$gender=="Male",]) +  geom_point(aes(x=weight,y=as.numeric(regime),colour=regime)) +
ylab(NULL) + scale_y_continuous(breaks=NULL, limits=c(-6,9)) + xlab("weight (g)")

We can check that the Mann-Whitney-Wilcoxon test is not significant (at the level 0.05)

wilcox.test(x, y, alternative="two.sided", conf.level=1-alpha)
## Warning in wilcox.test.default(x, y, alternative = "two.sided", conf.level
## = 1 - : cannot compute exact p-value with ties
##
##  Wilcoxon rank sum test with continuity correction
##
## data:  x and y
## W = 904.5, p-value = 0.1516
## alternative hypothesis: true location shift is not equal to 0

The test statistic $$W_x$$ is computed a follows:

• Assign numeric ranks to all the observations, beginning with 1 for the smallest value. Where there are groups of tied values, assign a rank equal to the midpoint of unadjusted rankings
• define $$R_x$$ (resp. $$R_y$$) as the sum of the ranks for the observations which came from sample $$x$$ (resp. $$y$$)
• Let $$W_x = R_x - {n_x(n_x+1)}/{2}$$ and $$W_y = R_y - {n_y(n_y+1)}/{2}$$
nx <- length(x)
ny <- length(y)
Wx=sum(rank(c(x,y))[1:nx]) - nx*(nx+1)/2
Wy=sum(rank(c(y,x))[1:ny]) - ny*(ny+1)/2
c(Wx, Wy)
## [1] 904.5 616.5

For a two sided tests and assuming that $$W_x^{\rm obs}>W_y^{\rm obs}$$, the $$p$$-value is $p_{\rm value} = \prob{W_y \leq W_y^{\rm obs}} + \prob{W_x \geq W_x^{\rm obs}}$ The distribution of $$W_x$$ and $$W_y$$ are tabulated and this $$p$$-value can then be computed

pwilcox(Wy,ny,nx)+ 1 - pwilcox(Wx,nx,ny)
## [1] 0.1508831

We could of course exchange the roles of $$x$$ and $$y$$. In this case the test statistic would be $$W_y$$ but the p-value would be the same.

wilcox.test(y, x, alternative="two.sided", conf.level=1-alpha)
## Warning in wilcox.test.default(y, x, alternative = "two.sided", conf.level
## = 1 - : cannot compute exact p-value with ties
##
##  Wilcoxon rank sum test with continuity correction
##
## data:  y and x
## W = 616.5, p-value = 0.1516
## alternative hypothesis: true location shift is not equal to 0

Remark: It is easy to show that $$W_x+W_y=n_x n_y$$

c(Wx+Wy, nx*ny)
## [1] 1521 1521

Unlike the t-test, the Mann-Whitney-Wilcoxon does not require the assumption of normal distributions. However, it is nearly as efficient as the t-test on normal distributions. That means that both tests have similar power.

This important property can easily be checked by Monte Carlo simulation. Let us simulate $$L$$ replicates of the experiments under $$H_1$$, assuming that $$\mu_y=\mu_x +15$$. We can then compare the power of both tests by comparing the rajection rates of the null hypothesis.

L <- 10000
alpha <- 0.05
mux <- 500
muy <- 520
sdx <- sdy <- 30
nx <- ny <- 40
Rt <- vector(length=L)
Rw <- vector(length=L)
for (l in (1:L)) {
x.sim <- rnorm(nx,mux,sdx)
y.sim <- rnorm(ny,muy,sdy)
Rt[l] <- t.test(x.sim, y.sim)$p.value < alpha Rw[l] <- wilcox.test(x.sim, y.sim)$p.value < alpha
}
c(mean(Rt), mean(Rw))
## [1] 0.8392 0.8230

On the other hand, the Wilcoxon test may be much more powerful than the t-test for non normal distributions when the empirical mean converges lowly in distribution to the normal distribution. Such is the case, for instance, of the log-normal ditribution which is strongly skewed for large variances.

mux <- 5
muy <- 6
sdx <- sdy <- 1
nx <- ny <- 20
Rt <- vector(length=L)
Rw <- vector(length=L)
for (l in (1:L)) {
x.sim <- exp(rnorm(nx,mux,sdx))
y.sim <- exp(rnorm(ny,muy,sdy))
Rt[l] <- t.test(x.sim, y.sim, alternative="two.sided")$p.value < alpha Rw[l] <- wilcox.test(x.sim, y.sim, alternative="two.sided")$p.value < alpha
}
c(mean(Rt), mean(Rw))
## [1] 0.6679 0.8487

# 4 The limited role of the p-value

First of all it is important to emphasis that statistics is a tool for supporting decision-making. It is not a decision tool that can be used blindly.

A $$p$$-value below the sacrosanct 0.05 threshold does not mean that GMOs have some negative impacts on human health, or that a drug is better than another one. On the other hand, a $$p$$-value above 0.05 does not mean that GMOs are safe or that a drug has no effect.

The American Statistical Association (ASA) has released a “Statement on Statistical Significance and P-Values” with six principles underlying the proper use and interpretation of the p-value.

“The p-value was never intended to be a substitute for scientific reasoning,” said Ron Wasserstein, the ASA’s executive director. “Well-reasoned statistical arguments contain much more than the value of a single number and whether that number exceeds an arbitrary threshold. The ASA statement is intended to steer research into a ‘post p<0.05 era.’” “Over time it appears the p-value has become a gatekeeper for whether work is publishable, at least in some fields,” said Jessica Utts, ASA president. “This apparent editorial bias leads to the ‘file-drawer effect,’ in which research with statistically significant outcomes are much more likely to get published, while other work that might well be just as important scientifically is never seen in print. It also leads to practices called by such names as ‘p-hacking’ and ‘data dredging’ that emphasize the search for small p-values over other statistical and scientific reasoning.”

The statement’s six principles, many of which address misconceptions and misuse of the pvalue, are the following:

1. P-values can indicate how incompatible the data are with a specified statistical model.
2. P-values do not measure the probability that the studied hypothesis is true, or the probability that the data were produced by random chance alone.
3. Scientific conclusions and business or policy decisions should not be based only on whether a p-value passes a specific threshold.
4. Proper inference requires full reporting and transparency.
5. A p-value, or statistical significance, does not measure the size of an effect or the importance of a result.
6. By itself, a p-value does not provide a good measure of evidence regarding a model or hypothesis.

As an illustration, assume that the diet has a real impact on the weight: after 14 weeks, rats fed with GMOs weigh in average 15g less than control rats.

It will be extremly unlikely to conclude that there is an effect with only 10 rats per group, even if we observe a difference of 15g in the two samples.

n <- 10
mu <- 500
delta <- 15
sd <- 30
x.sim <- rnorm(n,mu,sd)
y.sim <- x.sim + delta
t.test(x.sim, y.sim)
##
##  Welch Two Sample t-test
##
## data:  x.sim and y.sim
## t = -0.77216, df = 18, p-value = 0.45
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -55.8125  25.8125
## sample estimates:
## mean of x mean of y
##  504.9729  519.9729

This basic example shows that a difference considered as biologically significant may not be statistically significant.

On the other hand, a small difference considered as not biologically significant (1g for instance) may be considered as statistically significant if the group sizes are large enough.

n <- 10000
delta <- 1
x.sim <- rnorm(n,mu,sd)
y.sim <- x.sim + delta
t.test(x.sim, y.sim)
##
##  Welch Two Sample t-test
##
## data:  x.sim and y.sim
## t = -2.3407, df = 19998, p-value = 0.01926
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -1.8373819 -0.1626181
## sample estimates:
## mean of x mean of y
##  499.9298  500.9298

This example confirms the need for a power analysis as a complement of the comparison test. We will see that equivalence testing may also be relevant for evaluating what the data allows to say.

# 5 Equivalence tests

## 5.1 Introduction

Traditional hypothesis testing seeks to determine if means are the same or different. Such approach has several drawbacks:

1. Testing that two means are exactly the same is usually of little interest. A very small difference may exist, even if it is not biologically, or physically significant. It is much more meaningful in such situation to test if some significant difference exists, i.e. if this difference may have some concrete impact.

2. When testing difference between means, the null hypothesis considers that there is no differences: it thus falls to the data to demonstrate the converse. In the absence of enough data, we may fail to detect a significant difference, because of the lack of power. An opposite point of view consists of applying the basic hypothesis that a significant difference exists: it now falls to the data to demonstrate the converse.

On the other hand, equivalence testing determines an interval where the means can be considered equivalent. In other words, equivalence does not mean that two means $$\mu_x$$ and $$\mu_y$$ are equal, but rather that they are close enough“, i.e. $$|\mu_x - \mu_y| < \delta$$ for some equivalence limit $$\delta$$ that should be chosen according to the problem under study.

Thus, in terms of hypothesis testing, we want to test $H_0: \ |\mu_x - \mu_y| \geq \delta" \quad \text{versus} \quad H_1: \ |\mu_x - \mu_y| < \delta"$

These tests are currently used in the field of medication, to put a generic drug on the market. These are bioequivalence tests. Equivalence testing may also be used to determine if new therapies have equivalent or noninferior efficacies to the ones currently in use. These studies are called equivalence/noninferiority studies.

In the field of GMO risk assessment, equivalence testing is required to demonstrate that a GMO crop is compositionally equivalent and as safe as a conventional crop.

## 5.2 Two samples test

### 5.2.1 The TOST procedure

The simplest and most widely used approach to test equivalence is the two one-sided test (TOST) procedure. Let $$\mu_d = \mu_x - \mu_y$$, then TOST consists in performing the two tests:

$\begin{eqnarray} &H_0^{(+)}: \ \mu_d \geq \delta" &\text{versus} \ &H_1^{(+)}: \ \mu_d < \delta" \\ &H_0^{(-)}: \ \mu_d \leq -\delta" &\text{versus} \ &H_1^{(-)}: \ \mu_d > -\delta" \end{eqnarray}$

which is equivalent to test

$\begin{eqnarray} &H_0^{(+)}: \ \mu_d = \delta" &\text{versus} \ &H_1^{(+)}: \ \mu_d < \delta" \\ &H_0^{(-)}: \ \mu_d = -\delta" &\text{versus} \ &H_1^{(-)}: \ \mu_d > -\delta" \end{eqnarray}$

Proposition: Using TOST, equivalence is established at the $$\alpha$$ significance level if a $$(1-2\alpha) × 100\%$$ confidence interval for the difference $$\mu_x-\mu_y$$ is contained within the interval $$(-\delta,\delta)$$.

Proof: Let $$d_i = x_i-y_i$$ and $$\bar{d}=\bar{x}-\bar{y}$$. Let $$s_{\bar{d}}$$ be the estimated standard deviation of $$\bar{d}$$. $s_{\bar{d}} = \left\{ \begin{array}{ll} s_p \sqrt{\frac{1}{n_x}+\frac{1}{n_y}} \quad \text{if} \ \sigma^2_x=\sigma^2_y \\ \sqrt{{s_x^2}/{n_x}+{s_y^2}/{n_y}} \quad \text{otherwise} \end{array} \right.$

Then, we

• reject $$H_0^{(+)}$$ if $$T_{\rm stat}^{(+)} = (\bar{d}-\delta)/s_{\bar{d}}<qt_{\alpha,\nu}$$
• reject $$H_0^{(-)}$$ if $$T_{\rm stat}^{(-)} = (\bar{d}+\delta)/s_{\bar{d}}>qt_{1-\alpha,\nu}$$

where $$\nu$$ is the appropriate degree of freedom. Using the fact that $$qt_{\alpha,\nu} = -qt_{1-\alpha,\nu}$$, these decision rules are equivalent to:

• reject $$H_0^{(+)}$$ if $$\bar{d} + qt_{1-\alpha,\nu}s_{\bar{d}} < \delta$$
• reject $$H_0^{(-)}$$ if $$\bar{d} + qt_{\alpha,\nu}s_{\bar{d}} > -\delta$$

By definition, $$\bar{d} + qt_{\alpha,\nu}s_{\bar{d}}$$ and $$\bar{d} + qt_{1-\alpha,\nu}s_{\bar{d}}$$ are the bounds of a $$(1-2\alpha) × 100\%$$ confidence interval for the difference mean $$\mu_d$$.

We therefore reject the null hypothesis $$H_0$$ when both $$H_0^{(+)}$$ and $$H_0^{(-)}$$ are rejected, i.e. when a $$(1-2\alpha) × 100\%$$ confidence interval for $$\mu_d$$ is contained within the equivalence limits $$(-\delta, \delta)$$. $$\square$$

In our example, assuming unequal variance, the 90% confidence interval for $$\mu_d=\mu_x-\mu_y$$ is now

d.mean <- x.mean-y.mean
s.d.mean <- sqrt(var(x)/length(x) + var(y)/length(y))
d.mean + s.d.mean*qt(c(alpha,1-alpha),dfw)
## [1] -1.189099 31.132689

The calculated confidence interval can be plotted together with the value 0 (for difference testing) and the equivalence limits $$-\delta$$, $$\delta$$. Such a plot will immediately reveal whether the GMO is significantly different from the conventional counterpart (at the $$2(1-\alpha)$$ confidence level), and/or equivalence can be claimed or denied (at the $$1-\alpha$$ confidence level).

Considering for instance that a difference of weight larger than 20g is biologically significant leads to choose $$\delta=20$$. Since the 90% confidence interval is not contained in the equivalence interval $$(-\delta, \delta)$$, we don’t reject the null hypothesis for a significance level of 0.05. We therefore cannot conclude to equivalence, even if the difference in mean is not statistically significant.

On the other hand, we will conclude that the two regimens are nutritionally equivalent if we choose 40g as a limit since the new equivalence interval $$(-40, 40)$$ contains the CI.

When it is considered useful to have results also in the form of a $$p$$-value from a statistical significance test, then it can be easily calculated as follows:

\begin{aligned} p_{\rm value} &= p_{\rm value}^{(+)} + p_{\rm value}^{(-)} \\ &= \prob{ \frac{\bar{d}-\delta}{s_\bar{d}} < \frac{\bar{d}^{\rm obs}-\delta}{s_\bar{d}} \ | \ \mu_d=\delta} + \prob{ \frac{\bar{d}+\delta}{s_\bar{d}} > \frac{\bar{d}^{\rm obs}+\delta}{s_\bar{d}}\ | \ \mu_d=-\delta} \\ &= Ft_{\df_W}\left(\frac{\bar{d}^{\rm obs}-\delta}{s_\bar{d}}\right) + 1 - Ft_{\df_W}\left(\frac{\bar{d}^{\rm obs}+\delta}{s_\bar{d}}\right) \end{aligned}

Using $$\delta=20$$, the $$p$$-value is greater than the significance level $$\alpha=0.05$$:

delta = 20
pt((d.mean-delta)/s.d.mean, dfw) + 1 - pt((d.mean+delta)/s.d.mean, dfw)
## [1] 0.3032307

Then, we don’t reject the null hypothesis. On the contrary, the $$p$$-value obtained with $$\delta=40$$ is smaller than the significance level and the null hypothesis of non equivalence can be rejected:

delta = 40
pt((d.mean-delta)/s.d.mean, dfw) + 1 - pt((d.mean+delta)/s.d.mean, dfw)
## [1] 0.005923576

Function tost {equivanece} computes a TOST for equivalence from paired or unpaired data

library(equivalence)
tost(x, y, alpha=0.05, epsilon=20)
##
##  Welch Two Sample TOST
##
## data:  x and y
## df = 75.976
## sample estimates:
## mean of x mean of y
##  513.7077  498.7359
##
## Epsilon: 20
## 95 percent two one-sided confidence interval (TOST interval):
##  -1.189099 31.132689
## Null hypothesis of statistical difference is: not rejected
## TOST p-value: 0.3029514
tost(x, y, alpha=0.05, epsilon=40)
##
##  Welch Two Sample TOST
##
## data:  x and y
## df = 75.976
## sample estimates:
## mean of x mean of y
##  513.7077  498.7359
##
## Epsilon: 40
## 95 percent two one-sided confidence interval (TOST interval):
##  -1.189099 31.132689
## Null hypothesis of statistical difference is: rejected
## TOST p-value: 0.005923451

### 5.2.2 Difference testing versus equivalence testing

The choice of the test mainly depends on how the conclusion is formulated.

As an example, AFSSA (currently ANSES) was committing a fairly serious error in this area by concluding in relation to the MON863 GMO maize: Considering that no significant difference has been observed between the results obtained for MON863 maize and for the other varieties of maize, one might, therefore, conclude that the new plant is nutritionally equivalent (AFSSA, Saisine 2003-0215, p. 6).

The absence of statistically significant difference does not allow to conclude on equivalence. The European Food Safety Authority (EFSA) published a “Scientific Opinion” on this topic.

In particular, EFSA considers that statistical methodology should not be focussed exclusively on either differences or equivalences, but should provide a richer framework within which the conclusions of both types of assessment are allowed. Both approaches are complementary: statistically significant differences may point at biological changes caused by the genetic modification, but may not be relevant from the viewpoint of food safety. On the other hand, equivalence assessments may identify differences that are potentially larger than normal natural variation, but such cases may or may not be cases where there is an indication for true biological change caused by the genetic modification. A procedure combining both approaches can only aid the subsequent toxicological assessment following risk characterization of the statistical results.

EFSA also propose the following classification of the possible outcomes:

After adjustment of the equivalence limits, a single confidence limit (for the difference) serves visually for assessing the outcome of both tests (difference and equivalence). Here, only the upper adjusted equivalence limit is considered. Shown are: the mean of the GM crop on an appropriate scale (square), the confidence limits (whiskers) for the difference between the GM crop and its conventional couterpart (bar shows confidence interval), a vertical line indicating zero difference (for proof of difference), and vertical lines indicating adjusted equivalence limits (for proof of equivalence).

For outcome types 1, 3 and 5 the null hypothesis of no difference cannot be rejected: for outcomes 2, 4, 6 and 7 the GM crop is different from its conventional counterpart. Regarding interpretation of equivalence, four categories (i) - (iv) are identified: in category (i) the null hypothesis of non-equivalence is rejected in favour of equivalence; in categories (ii) equivalence is more likely than not (further evaluation may be required), (iii) non-equivalence is more likely than not (further evaluation required) and (iv) non-equivalence is established (further evaluation required).

## 5.3 One sample test

Even if equivalence testing is mainly used for testing the equivalence between two populations, a one sample equivalence test is also possible.

Assume that, for some $$\delta>0$$, we want to test $H_0: \ |\mu_x - \mu_0| \geq \delta" \quad \text{versus} \quad H_1: \ |\mu_x - \mu_0| < \delta"$ Let $$z_i = x_i - \mu_0$$, $$i=1,2,\ldots,n_x$$, and let $$\mu_z = \mu_x - \mu_0$$. Then, the test reduces to use the $$(z_i)$$ for testing

$H_0: \ |\mu_z| \geq \delta" \quad \text{versus} \quad H_1: \ |\mu_z| < \delta"$

mu0 <- 500
delta <- 10
alpha <- 0.05
z <- x - mu0
tost(z, alpha=alpha, epsilon=delta)
##
##  One Sample TOST
##
## data:  z
## df = 38
## sample estimates:
## mean of x
##  13.70769
##
## Epsilon: 10
## 95 percent two one-sided confidence interval (TOST interval):
##   2.035311 25.380074
## Null hypothesis of statistical difference is: not rejected
## TOST p-value: 0.7023006

Here, the two-sided null hypothesis is the union of the one sided hypotheses $$\mu_z \geq \delta"$$ and $$\mu_z \leq -\delta"$$. Thus,

\begin{aligned} p_{\rm value} &= \prob{\bar{z} < \bar{z}^{\rm obs} \ | \ \mu_z=\delta} + \prob{\bar{z} > \bar{z}^{\rm obs} \ | \ \mu_z=-\delta} \\ &= \prob{t_{n_z-1} < \frac{\bar{z}^{\rm obs}-\delta}{s_{\bar{z}}}} + \prob{t_{n_z-1} > \frac{\bar{z}^{\rm obs}+\delta}{s_{\bar{z}}}} \\ &= Ft_{n_x-1}\left(\frac{\bar{x}^{\rm obs}-\mu_0-\delta}{s_{\bar{x}}}\right) + 1 - Ft_{n_x-1}\left(\frac{\bar{x}^{\rm obs}-\mu_0+\delta}{s_{\bar{x}}}\right) \end{aligned}

z.mean <- mean(z)
zbar.sd <- sd(z)/sqrt(nx)
p.value <- pt((z.mean-delta)/zbar.sd, nx-1) + 1 - pt((z.mean+delta)/zbar.sd, nx-1)
p.value
## [1] 0.6592176