One Sample Tests

One Sample Tests

We focus on one sample tests.

Tests on a Gaussian sample

Consider a random sample (X₁, …, X_n) of a distribution with mean μ, standard deviation σ. Recall that

• The empirical mean is $\bar X = \frac{X_1 + \ldots + X_n}n$
• The empirical variance is $S^2 = \frac{n}{n-1} \left(\frac{X^2_1 +\ldots + X^2_n}n - \bar X^2\right)$.

Test of the mean

A first test is on the mean of the sample.

Example. For an adult, the logarithm of the D-dimer concentration, denoted by X, is modeled by a normal random variable with mean μ and standard deviation σ. The variable X is an indicator for the risk of thrombosis: it is considered that for healthy individuals, μ is −1, whereas for individuals at risk μ is 0. The influence of olive oil on thrombosis risk must be evaluated. A group of 13 patients, previously considered as being at risk, had an olive oil enriched diet. After the diet, their value of X was measured, and this gave an empirical mean of −0.15.

The doctor would like to decide if the olive oil diet has improved the D-dimer concentration.

The decision rule depends on H₁. It consists in computing the bounds at which we reject H₀. The bounds depend also on the risk of the test (the first kind risk).

R> alpha <- 0.05
R> qnorm(alpha/2,0,1)
[1] -1.959964
R> qnorm(1-alpha/2,0,1)
[1] 1.959964

R> alpha <- 0.05
R> qnorm(alpha,0,1, lower.tail=FALSE)
[1] 1.644854

R> alpha <- 0.05
R> qnorm(alpha,0,1)
[1] -1.644854

Back to the example. We assume that the sample of 13 patients is a Gaussian sample. The standard deviation σ is supposed to be known and equal to 0.3. We want to test
H₀ : μ = 0 versus H1 : μ = −1

The test statistic is
$$ T = \sqrt{13} \left(\frac{\bar X - 0}{0.3}\right)$$
According to the null hypothesis H₀, T follows the normal distribution 𝒩(0, 1). The hypothesis H₀ is rejected when T takes low values. At risk 5%, the bound is

R> qnorm(0.05,0,1)
[1] -1.644854

The decision rule is Reject H_0 if T  <  (−1.6449).

For $\bar X= -0.15$, the test statistic takes the value

R> n<-13
R> Xbar<--0.15
R> sig<-0.3
R> mu0<-0
R> t<-sqrt(n)*(Xbar-mu0)/sig
R> t
[1] -1.802776

The previous case assumes that the standard deviation σ is known. This is usually not the case in practice. The adaptation to the test of the mean, with an unknown variance is the following.

The decision rules are before, but the bounds are computing from the Student distribution instead of the normal distribution.

R> alpha <- 0.05
R> n<-13
R> qt(alpha/2,n-1)
[1] -2.178813
R> qt(1-alpha/2,n-1)
[1] 2.178813

R> alpha <- 0.05
R> n<-13
R> qt(alpha,n-1)
[1] -1.782288

R> alpha <- 0.05
R> n<-13
R> qt(alpha,n-1)
[1] -1.782288

Back to the example. We assume that the standard deviation σ is unknown and estimated to 0.3. We want to test
H₀ : μ = 0 versus H1 : μ = −1

The test statistic is
$$ T = \sqrt{13} \left(\frac{\bar X - 0}{0.3}\right)$$
According to the null hypothesis H₀, T follows a Student distribution with 12 degrees of freedom. The hypothesis H₀ is rejected when T takes low values. At risk 5%, the bound is

R> qt(0.05,12)
[1] -1.782288

The decision rule is Reject H_0 if T  <  (−1.6449).

For $\bar X= -0.15$, the test statistic takes the value

R> n<-13
R> Xbar<--0.15
R> s<-0.3
R> mu0<-0
R> t<-sqrt(n)*(Xbar-mu0)/s
R> t
[1] -1.802776

The function computes the test statistic of Student’s T-test comparing mean(X) to mu, and the corresponding p-value according to the alternative. The null hypothesis H₀ is “the mean is equal to mu”.
The alternative is in « two.sided » (default), « less », « greater »; they are understood as:

• two.sided: the mean is not equal mu,

• less: the mean is less than mu,

• greater: the mean is greater than mu.

Example To test if the mean of the age in LenzIsample is equal to 60, we run the following code

R> LenzI <- readRDS("data/LenzI.rds")
R> A<-LenzI$age
R> t.test(A, mu=60)

	One Sample t-test

data:  A
t = 1.5019, df = 413, p-value = 0.1339
alternative hypothesis: true mean is not equal to 60
95 percent confidence interval:
 59.6479 62.6323
sample estimates:
mean of x 
  61.1401

The two hypotheses of this t-test are H₀ : μ = 60 and H₁ : μ ≠ 60*. The output reads as follows:

• t=1.5019 is the value of the test statistic

• df=413is the degree of freedom of the Student distribution (equal to n-1)

• p-value = 0.1339is the p-value of the t-test, corresponding to the two-sided test (by default, the alternative is two-sided)

• alternative… recalls the choice of the alternative

• 95 percent confidence interval computes the 95% confidence interval of the mean, assuming the standard deviation unknown

• the last value 61.1401 is the estimation of the mean of the sample

We can also apply a one-sided test by changing the alternative. To test the alternative H₁ : μ ≥ 60, run

R> A<-LenzI$age
R> t.test(A, mu=60, alternative= "greater")

	One Sample t-test

data:  A
t = 1.5019, df = 413, p-value = 0.06695
alternative hypothesis: true mean is greater than 60
95 percent confidence interval:
 59.88867      Inf
sample estimates:
mean of x 
  61.1401

Remark that even if the empirical mean (61.1401) is greater than 60 years, the difference is not significant, and we can not conclude, at a risk of 5%, that the mean is larger than 60. Several reasons could be involved: the variability in the sample is too large (the standard error of the empirical mean is large) or the size of the sample is not large enough (the empirical mean is not estimated with enough precision).

Test of the standard deviation or the variance

Test of the mean for large sample

Finally, a test of the mean exists for large sample, and we don't need the assumption that the sample is Gaussian thanks to the Central Limit Theorem.

With R, the test of the mean for large sample can be applied with the function t.test (as above), assuming that the normal distribution is very closed to a Student distribution with a large degree of freedom.

Test of a proportion

The previous tests are applied to gaussian samples. When the variable of interest is binary (two possible modalities), we are interested in comparing the proportion of the first modality with a theoretical proportion p₀.

Example For a certain disease, there exists a treatment that cures 70% of the cases. A laboratory proposes a new treatment claiming that it is better than the previous one. Out of 100 patients having received the new treatment, 74 of them have been cured. The expert would like to decide whether the new treatment should be authorized.


Back to the example The hypotheses we want to test are H₀ : p = 0.7 versus H1 : p ≥ 0.7.


The null hypothesis H₀ is: “the proportion of x out of n is equal to p”. The alternative is in « two.sided » (default), « less », « greater »; they are understood as:

• two.sided: the proportion x/n is not equal p,

• less: the proportion x/n is less than p,

• greater: the proportion x/n is greater than p.

Back to the example The one-sided test is applied running

R> prop.test(x=74, n= 100, p=0.7, alternative=« greater »)

Note that when the whole binary sample X is available (and not only the count of « successes »), the instruction is prop.test(sum(X), length(X), p, alternative).


Goodness-of-fit tests

A goodness-of-fit test answers the question: could the sample have been drawn at random from a particular distribution?

Example Consider a diploid population with allele frequencies 0.4 for A, 0.6 for a. At the Hardy-Weinberg equilibrium, the probabilities of the three genotypes AA, Aa, aa, are (0.16, 0.48, 0.36). Frequency table of the three genotypes is (1600, 4900, 3500). Is the theoretical model plausible ?


Chi-squared test

For a discrete variable, the goodness-of-fit is measured by a distance between the relative frequencies of the variable, and the probabilities of the target distribution.

Under the alternative, the distance should be large, so that the p-value is computed as the right-tail probability of the chi-squared distribution at the distance.

In that command,

• table(X) is a table of absolute frequencies (any vector of integers can be tested)

• The probability distribution p is a vector of probabilities, with same length as table(X)

The answer is “the fit is good”, if the p-value is large (above the risk).

If some frequencies are too small, a warning message may be issued. If one parameter or more have been estimated, the test statistic should be extracted, and the p-value computed as its right-tail probability for the chi-squared distribution with a smaller parameter.

Back to the example The frequency table of the three genotypes AA, Aa, aa is (1600, 4900, 3500). The theoretical probabilities are (0.16, 0.48, 0.36). The chi-squared test is applied running

R> chisq.test(c(1600, 4900, 3500),p=c(0.16, 0.48, 0.36))

	Chi-squared test for given probabilities

data:  c(1600, 4900, 3500)
X-squared = 4.8611, df = 2, p-value = 0.08799

The outputs are

• data the observed data

• X-squared the value of the chi-squared distance

• dfthe degree of freedom of the chi-squared distribution

• p-value the p-value

Kolmogorov-Smirnov test

For a continuous variable, the goodness-of-fit is measured by a distance between the empirical cumulative distribution function (ecdf) of the variable, and the cdf of the target distribution.

The answer is “the fit is good”, if the p-value is large (above the risk). The variable X should not have ties (equal values). If some values are equal, a warning message is issued, indicating that the p-value is not quite as precise. This does not affect the validity of the result.

The null hypothesis H₀ is: “the distribution of the sample is the theoretical cdf”. The alternative is in « two.sided » (default), « less », « greater »; they are understood as:

• two.sided: the ecdf of the sample is different from the theoretical cdf,

• less: the ecdf of the sample is under the theoretical cdf (the values of the sample are larger than those of the theoretical distribution),

• greater: the ecdf of the sample is above the theoretical cdf (the values of the sample are smaller than those of the theoretical distribution).

Example The hypoxy level is given in data set HY. Let us plot the ecdf of Level and those of a normal distribution.

R> HY <- read.table("data/hypoxy.csv", header=TRUE, dec=",")
R> L<-HY$Level
R> plot(ecdf(L))
R> curve(pnorm(x,mean(L), sd(L)), col="red", add=TRUE)

Résultat : #r{codeimg}

The red curve is the one of a normal distribution with parameters μ = 1.2 and σ = 1. The ecdf (black curve) is quite far from the theoretical cdf. The Level variable is probably not normally distributed.

To test if the ecdf of Level is a normal distribution with parameters μ = 1.2 and σ = 1 or not, run

R> ks.test(L, "pnorm", c(1.2,1))

	One-sample Kolmogorov-Smirnov test

data:  L
D = 0.1782, p-value = 2.501e-06
alternative hypothesis: two-sided

The outputs are

• data the observed data

• D the value of the Kolmogorov-Smirnov distance

• p-value the p-value

The histogram of the variable Level reveals a right-skewed distribution, closed to a log-normal distribution. Let us log-transform the data

R> LL<-log(L)

and plot the ecdf (black curve) of the log-Level and the cdf (red curve) of a normal distribution with parameters log(1.2)≈0.2 and 1. #r{codeimg}

The two curves are quite closed. Let us test if the empirical cdf of LL is a normal distribution with parameters log(1.2)≈0.2 and 1

R> ks.test(LL, "pnorm", c(0.2,1))

	One-sample Kolmogorov-Smirnov test

data:  LL
D = 0.51048, p-value < 2.2e-16
alternative hypothesis: two-sided

Remark: the difference on the left of the two curves (ecdf and cdf) is large enough to reject the null hypothesis.

Normality test

Testing whether a variable is normally distributed, is different from testing whether a particular normal distribution with given parameters fits the variable.

Back to the example Test of the normality of the log-level of hypoxy:

R> shapiro.test(LL)

	Shapiro-Wilk normality test

data:  LL
W = 0.95329, p-value = 1.936e-06

The outputs are

• data the observed data

• W the value of the Shapiro-Wilk distance

• p-value the p-value