Master Data Science and Statistical Learning
Statistical thinking concerns the relation of quantitative data to a real-world problem, often in the presence of variability and uncertainty. It attempts to make precise and explicit what the data has to say about the problem of interest
The starting point in Statistics is a scientific enquiry, an investigation concerning a substantive question.
The first problem (the zeroth problem) to be studied is then to understand what is the relevant population.
In other words one has to decide what the relevant data are, and how these relate to the purpose of the statistical study.
One important issue in demography is fertility (the number of children ever born of a woman) and its relation with other factors. Some of the basic questions are:
The Fiji Fertility Survey (which is part of the World Fertility Survey, see Fiji Fertility survey, 1974) collected data of fertility and other relevant factors.
The data below are a subset of the full data set, regarding 1703 ever-married women Aged 20-49 in 1974 and concerning the variables:
source("fiji_data.R")
set.seed(457)
X <- fiji_data
x10 <- X[sample(1703, 10, replace = TRUE), ]
x10 F A M E I U
56 5 35 14 4 1 1
1544 7 42 20 0 1 0
931 11 47 12 0 1 0
1664 4 35 23 10 1 0
314 4 40 21 4 0 0
1572 7 44 16 0 1 0
1678 7 46 24 8 0 0
480 4 40 22 5 0 0
1630 6 48 17 3 1 1
1412 0 47 20 0 1 1Fertility is a primary outcome depending on age, age at first marriage, education, and residence and ethnicity
The variables M and E are intermediate variables, while A, U and I can be interpreted as background variables that may be considered fixed and on the same standing.
One reasonable ordering of the variables is the following \left\{ \begin{array}{c} I\\ U \\ A \end{array} \right\} \quad \{E\} \quad \{M\} \quad\{F\}
Each variable potentially depends on the variables to the left.
What’s the meaning of statistical inference?
A statistical inference will be defined as a statement about statistical populations made from given observations with measured uncertainty.
An inference in general is an uncertain conclusion.
Two things mark out statistical inferences.
First the information on which they are based is statistical, i.e. consists of observations subject to random fluctuations.
Secondly we explicitly recognize that our conclusion is uncertain and attempt to measure, as objectively as possible, the uncertainty involved.
The survey sampling approach is typically used for finite populations
The focus is on population and sub-population means, or totals, of important variables.
The simplest problem of inference is how to relate the sample mean \bar x of a variable X, from a random sample of size n, to the population mean \mu in the finite population
Inference is this case is based on the repeated sampling principle:
Statistical procedures should be evaluated on the basis of their behaviour in hypothetical repetitions of the experiment that generated the original data.
This principle is the basis of the frequentist approach
The previous sample of size n = 10 is a simple random sample with replacement i.e. we know that the observation are
To estimate the mean of the fertility \mu_F in the population we use the sample mean (called a statistic) \bar x_F = \frac{1}{n} \sum_{i=1}^n x_i
The mean \bar x_F of the sample and \mu_F of the population are in general different
mu <- mean(X[, "F"])
xbar <- mean(x10[,"F"])\bar x_F = 5.5, \quad \mu_F = 5.9
We need a measure of the uncertainty of the estimate \bar x_F !
According to the repeated sampling principle we measure the uncertainty as follows.
We consider the means in repeated random samples, denoted by the random variable \bar X_F.
The random variable \bar X_F has a probability distribution that can be studied. It is called the estimator of \mu_F
We measure the uncertainty of the estimator with the Mean Squared Error (MSE)
MSE = E\{(\bar X_F - \mu_F)^2\}
The square root of MSE, called the standard error, is the measure of uncertainty of the estimator.
There is a formula valid for simple random samples with replacement \begin{align*} E(\bar X_F) &= \mu_F \\ MSE &= \mathrm{var}(X_F) = \sigma^2 / n \\ \end{align*} where \sigma^2 is the variance of the population.
The standard error of the estimator of the mean is therefore SE (\bar X_F) = \sigma/\sqrt{n}
The standard error is however an unknown parameter because the standard deviation \sigma concerns the population that is unkown!
To get the standard error you need to estimate \sigma from the same sample using the estimator
s_F = \sqrt{\sum_{i=1}^n (X_i - \bar X_F)^2/(n-1) }
The sample standard deviation of fertility in our sample is
sf <- sd(X[,"F"])
se <- sf/sqrt(10)
cat("Estimated SE: ", se, "\n")Estimated SE: 0.9737265 Our inference about the mean fertility is summarized by \bar x_F = 5.5,\quad (\text{se} = 0.91)
meaning that the error is about 1 child.
Consider the true distribution of the population and our estimate.
p <- table(X[,"F"])/1703
plot(0:17, p, type= "h", lwd = 5,
col = "RoyalBlue", axes = FALSE,
xlab = "X", xlim = c(0, 20))
axis(1, pos = 0)
axis(2, pos = 0)
points(5.5, 0, pch =19, cex =1.5, col = "red")
Often we start not from a finite population but from a statistical model i.e., from an assumption concerning the distribution of an infinite population
For example, we assume that the distribution of the population is Gaussian (Normal) and that our sample is a random sample from that population with unknown mean \mu and standard deviation \sigma
In this example, as before we want to estimate these two unknown parameters \mu and \sigma.
The model-based theory of inference are essentially two:
Both theories are based on
In this model the data are binary x = 0,1, where 0 indicates failure and 1 success with probability mass f(x; \theta) = \begin{cases} 1-\theta & \text{ if } x = 0 \\ \theta & \text{ if } x = 1 \end{cases}
Assume that the design is a random sample (X_1, \dots, X_n) where the variables are independent and identically distributed (i.i.d)
The parameter \theta is the probability of success, and assumes values in a parameter space 0 <\theta<1.
The mean of the Bernoulli is also equal to \theta
The variance is \theta(1-\theta)
Suppose that we have data from a random sample in the population of placenta previa births in Germany, and we are interested in the proportion of female births.
Therefore we can specify a Bernoulli model with parameter \theta equal to the probability of a female birth.
If the size of the sample is n = 980 and the number of females in the sample is s = 437 we can estimate \theta as \hat \theta = 437 /980 = 0.446
The first inference is to obtain the standard error of the estimator.
In general, given an i.i.d sample (X_1, \dots, X_n) from the Bernoulli, the proportion of successes is \hat \theta = \frac{1}{n} \sum_{i = 1}^n X_i = \frac{T}{n} where T is the number of females in the sample.
The sampling distribution of T has a Binomial distribution with mean E(T) = n\theta and variance \mathrm{var}(T) = n\theta(1-\theta).
The MSE of \hat \theta is equal to the variance \mathrm{var}(\hat \theta) = \mathrm{var}(T/n) = \frac{1}{n^2}\mathrm{var}(T) = \frac{\theta(1-\theta)}{n}. Therefore the standard error is \text{SE}(\hat \theta) = \sqrt{\frac{\theta(1-\theta)}{n}}
Again the standard error is a function of the unknown parameter \theta
The estimated standard error is obtained by \text{se}(\hat \theta) = \sqrt{\frac{\hat \theta(1-\hat \theta)}{n}}
The inference concerning \hat \theta in the example of placenta previa is \hat \theta = 0.446, \quad (\text{se} = 0.016 )
The conclusion is that the proportion of females in the population of placenta previa births is 44.6\% with a standard error of 1.6%.
The sampling distribution of the proportion \hat \theta is a Binomial T divided by the sample size n
In the placenta previa example, assuming that the true proportion is \theta the distribution of \hat \theta is shown in figure below
We have two fundamental results obtained using the large sample property
The proportion T/n converges in probabilityto the true parameter \theta, for n \to \infty
The proportion T/n converges in law to the Normal, for n \to \infty
When \theta is not too close to 0 or 1, the Binomial distribution \text{Bin}(n, \theta) in large samples approximates the Normal distribution N(n\theta,n\theta(1-\theta)).
Given any population with a continuous distribution with a finite mean \mu and variance \sigma^2, in large samples, the distribution of the sample mean under random sampling approximates the Normal distribution N(\mu, \sigma^2/n)
The previous mathematical results imply a central role of the normal distribution in statistical inference.
The standard normal random variable Z is defined by the density \phi(z) = \frac{1}{\sqrt{2\pi}} e^{-z^2/2}, \quad -\infty < z < \infty
We know that Z is symmetric unimodal with respect to \mu. Moreover E(Z) = 0 and \mathrm{var}(Z) = 1
The family of normal distributions is defined by a linear transformation of the standard normal X = \mu + \sigma Z where \mu \in \mathbb{R} is the mean parameter and \sigma^2>0 is the variance parameter. The family is denoted by X \sim N(\mu, \sigma^2).
The tails are rather short:
\begin{align*} \Pr\{ -1 < Z < 1 \} &= 0.683\\ \Pr\{ -2 < Z < 2 \} &= 0.954\\ \Pr\{ -3 < Z < 3 \} &= 0.997\\ \end{align*}
Therefore, the probability that X differs by more than 3 \sigma from the mean is only 0.003.
The squared transformation Z^2 belongs to a different family of distributions called chi-squared and denoted by \chi^2_m. Specifically, Z^2 \sim \chi^2_1.
This model assumes that we get a random sample from a population that has a normal distribution N(\mu, \sigma^2) where both \mu and \sigma^2 are unknown.
In formulae we assume that
The one sample model can be written as X_i = \mu + \varepsilon_i where \varepsilon_i are mutually independent errors with normal distributions N(0, \sigma^2) with the same variance.
The estimators of \mu (parameter of interest) and \sigma^2 (nuisance parameter) are respectively \bar X = \sum_{i=1}^n X_i, \quad S^2 = \sum_{i=1}^n(X_i - \bar X)^2/(n-1)
Standard error of the sample mean:
\text{se}(\bar X) = S /\sqrt{n} - The sampling distributions with the normal model and sample size n are exactly
\bar X \sim N(\mu, \sigma^2/n)
S^2 \sim \sigma^2 \chi^2_{n-1}/(n-1)
The variance estimate has a chi-squared distribution with mean n-1, in statistical jargon called degrees of freedom (details postponed).
sample mean = $ 5138.852 se = $ 69.12335 Let’s discuss the previous example.
The full set of data is in R package Sleuth3
library(Sleuth3)
ex <- case1202
df <- ex[, c("Bsal", "Sex")]
boxplot(Bsal ~ Sex, data = df, horizontal = TRUE,
col = c("Orange", "LightGreen"))
The two samples have sizes n_0 = 61 (females) and n_1= 32 (males).
The two-sample model can be denoted by Y_{i} = \mu + \theta \cdot X_i + \varepsilon_i with errors \varepsilon_i \sim N(0, \sigma^2) mutually independent
The parameters in the two-sample models are three: \mu, \theta and \sigma^2
We can change the parameters defining a 1-1 correspondence. For example, E(Y_i) = \begin{cases} \mu & \text{if } X_i = 0\\ \mu + \theta & \text{if } X_i = 1 \end{cases}
The means of the two normal distributions are \mu_0 = \mu and \mu_1 = \mu+ \theta, so that \begin{align*} \mu &= \mu_0\\ \theta &= \mu_1 - \mu_0 \end{align*} Therefore \theta measures the discrepancy between the two populations.
In the example \theta is the parameter of interest in evaluating the amount of sex discrimination.
Intuitively, we can estimate \mu with the mean of the first sample, and \theta with the difference between the two means. Denoting by \bar Y_0 and \bar Y_1 the sample means, we get the estimators \hat \mu = \bar Y_0, \quad \hat \theta = \bar Y_1 - \bar Y_0 and we can prove that both are normally distributed.
The standard error of \hat \mu is given by \text{SE}(\hat \alpha) = \frac{\sigma}{\sqrt{n_0}} where n_0 is the size of the first sample.
The standard error of \hat \theta is the square root of \begin{align*} \mathrm{var}(\hat \theta) &= \mathrm{var}(\bar Y_1 - \bar Y_0) \\ &= \mathrm{var}(\bar Y_1) + \mathrm{var}(\bar Y_0)\\ &= \frac{\sigma^2}{n_0} + \frac{\sigma^2}{n_1} = \sigma^2 \left(\frac{1}{n_0} + \frac{1}{n_1}\right) \end{align*}
The estimated standard errors depend upon an estimate of the common variance \sigma^2
The estimator of \sigma^2 is the pooled variance S^2_{p} = \frac{1}{n-2}\left[\textstyle \sum_{i \in G_0}(Y_i - \bar Y_0)^2 + \sum_{i \in G_1}(Y_i - \bar Y_1)^2 \right] that is a sum of the deviances within the groups divided by the number of observations minus 2. The explanation is postponed.
This estimator can be interpreted as a weighted average of the variances of the groups:
S^2_{p} = \frac{(n_0-1) S^2_0 + (n_1-1) S^2_{1}} {(n_0-1) + (n_1-1)}
\begin{align*} \hat \mu &= \$\,5138.8 \quad (\text{se} = 75.2) \\ \hat \theta &= \$\,818.0 \quad (\text{se} = 130.0) \end{align*}
= The salary difference has an exactly normal sample distribution. Intuitively, it seems unlikely that a difference of $818 can happen if the true \beta = 0 given that is 6 times the standard error $130.
Many inferential problems can be classified into three categories:
Her we give a short introduction.
Point estimation tries to find an “optimal guess” of some quantity of interest like
Let’s call \theta a generic parameter of a model and \hat\theta its estimator. Remember that the estimator is a random variable in repeated sampling.
The bias of an estimator is defined by the difference between the true \theta and the expected value of the estimator \text{bias}(\hat \theta) = E(\hat \theta) - \theta If E(\hat \theta) = \theta the bias is 0 and the estimator is said unbiased.
Given an i.i.d. sample the sample mean \bar X is always unbiased for the mean of the population \mu.
Given an i.i.d sample from a Bernoulli, the proportion of successes \hat \theta = T/n is unbiased.
However the estimator (T/n)^2 of \theta^2 of the Bernoulli is biased because the expectation is E(T^2/n^2) = E(T^2)/n^2 = \theta/n^2 and \text{bias}[(T/n)^2] = \theta/n^2 - \theta^2
The sample variance S^2 (with denominator n-1) is unbiased for the variance \sigma^2 of the population.
However the square root of the sample variance, i.e. \sqrt{S^2} is biased for the standard deviation \sigma.
In general, unbiasedness is a non persistent property, and in fact many good estimators are biased provided that the bias tends to zero when the sample size increases.
The important thing for an estimator is not the zero bias, but a property called consistence meaning that the estimator converges to the true parameter value as we collect more and more data.
A point estimator \hat\theta_n of \theta is consistent if \hat\theta_n \to \theta for n \to \infty in probability.
As we have seen before the quality of a point estimate is assessed by the mean squared error \text{MSE} = E_\theta\{ \hat \theta - \theta)^2\}.
Interestingly the MSE can be decomposed as follows
\text{MSE} = \text{bias}^2(\hat \theta) + \mathrm{var}(\hat \theta) And this helps in checking if an estimator is consistent
If the \text{bias} \to 0 and the \to 0 as n \to \infty then the estimator \hat \theta_n is consistent.
A confidence interval of level 1-\alpha is an interval with extremes L (left) and R (right) that are functions of the random sample (therefore random variables) such that \textstyle \Pr_\theta(L < \theta < R) \ge (1-\alpha) \text{ for all } \theta.
The idea is that we have random interval that traps the true parameter \theta with probability 1-\alpha called the coverage.
Typically we use 95 percent confidence levels, which correspond to choosing \alpha = 0.05.
The probability concerns the random interval not the parameter. The parameter is fixed !
(Wasserman) Every day, newspapers report opinion polls. For example, they might say that “83 percent of the population favor arming pilots with guns.” Usually, you will see a statement like “this poll is accurate to within 4 points 95 percent of the time.”
Technically, this means that a study based on a random sample gives a confidence interval with level 0.95 0.83 \pm 0.04 for the true proportion \theta of people wo favor arming pilots.
A typical error is interpreting the observed confidence interval as: \small \Pr( 0.79 < \theta < 0.87) = 0.95 because is conceptually wrong!
If you continue constructing confidence intervals (not only for \theta but also for a sequence of other unrelated parameters, then 95 percent of your intervals will trap the true parameter value.
If we have an asymptotically normal estimator \hat \theta_n \approx N(\theta, \text{se}^2) of \theta, then we can construct an approximate CI for \theta
C_n = (\hat \theta_n - 1.96 \, \text{se} < \mu < \hat \theta_n + 1.96\, \text{se})
with confidence level \to 0.95 for n \to \infty.
The approximate confidence interval for the mean fertility considering a random sample with size n = 200 is calculated as follows
X <- fiji_data$F
x200 <- X[sample(1703, 200, replace = TRUE)]
M <- mean(x200)
se <- sd(x200)/sqrt(200)
L <- M - 1.96*se
U <- M + 1.96*se
cat("mean: ", M, ", se: ", se, "n: ", 200 , "\n" )mean: 6.18 , se: 0.236469 n: 200 95 percent Confidence Interval
( 5.716521 6.643479 )
In hypothesis testing we start from a study concerning a substantive research, like for example the study concerning the starting salaries of men and women.
Also we have a sort of default theory like that the mean salaries for men and women are the same.
The deafult theory must be converted into a statistical hypothesis that makes statements about population parameters, called the null hypothesis. In the example
H_0: \theta = \mu_{\text{M}} - \mu_{\text{F}} = 0
We can also identify an alternative hypothesis H_1 that specifies the directions of the departures from H_0. For example H_0: \theta > 0
Then we ask if the data provide sufficient evidence to reject the theory. If not we retain the null hypothesis.
Calculate some function t({\boldsymbol{Y}}) of the difference between the means of males and females. Then:
\small \begin{cases} \text{if $T$ is large} & \text{ reject } H_0 \\ \text{if $T$ is small} & \text{ mantain } H_0 \end{cases}
In the study concerning the gender of the births with placenta previa, we could ask if the proportion of males is 1/2. Then we can compare a null hypothesis H_0 : \theta = \textstyle\frac{1}{2} with an alternative hypothesis H_1: \theta \ne \textstyle\frac{1}{2}. It seems reasonable to reject H_0 if T = |\hat \theta - \textstyle\frac{1}{2}|. is large.
We can distinguish two different approaches to testing:
Fisher’s approach is directed to evaluate the evidence against a single hypothesis H_0 measured by a normalized value of the distance of the data and the hypothesis, called p-value
Neyman and Pearson’s approach considers two hypotheses H_0 and an alternative H_1 and use a decision procedure on the basis of data, either to accept or reject H_0.
The procedure consist in defining a test statistic, that is a function of the data measuring the difference t({\boldsymbol{Y}}) between the data and what is expected on the null hypothesis. Typically,
t({\boldsymbol{Y}}) = \frac{\text{observed} - \text{expected}}{\text{se}}
The observed value of the test statistic is denoted by t_{\text{obs}}.
Then we calculate the chance of getting a test statistic es extreme as or more extreme than t_{\text{obs}}, computed on the basis that H_0 is right:
p = \Pr(t({\boldsymbol{Y}}) \ge t_{\text{obs}}; H_0)
This probability is called the observed significance level or briefly the p-value.
Start from a general model assumed to be the generating process of the observations Y_i = \mu + \theta X_i + \varepsilon_i, \quad \epsilon_i \sim N(0, \sigma^2), \text{ for } i = 1, \dots, n where X_i is a binary variable defining two samples.
We define a null hypothesis H_0: \theta = 0 asserting that the two populations are identical: Y_i = \mu + \epsilon_i, \quad \varepsilon_i \sim N(0, \sigma^2) meaning that the mean salaries are equal.
An alternative hypothesis can be H_1: \theta > 0 so that \mu_{\text{M}} > \mu_{\text{F}}.
We define a test statistic
t({\boldsymbol{Y}}) = W = \frac{\bar Y_1 - \bar Y_0}{\text{se}(\hat\beta)}
that is the difference between the means divided by the estimated standard error, calculated with the pooled variance.
In our data about salaries the value of the test statistic is t_{\text{obs}} = \frac{818}{130}= 6.3
To evaluate it this value is small or large we calculate the p-value: p = \Pr(W \ge 6.3; H_0).
To calculate this probability we should find the null distribution of T that fortunately is known as a t- distribution (discovered by William Gosset said “Student”) with parameter n-2.
We will explain its genesis later, but for the moment we just plot it. The following figure shows the density of a T_{n-2}, with n = n_0 + n_1 = 93
As you see the curve is quite similar to the standard normal.
The red dot, representing the value d_{\text{oss}} = 6.3, is quite far in the right tail so that the p-value is extremely
small.
To find the p-value for the test of \beta=0 we calculate the left-tail area with R using the function pt():
pt(6.3, df = 91, lower.tail = FALSE)[1] 5.202414e-09Traditionally, the p-value is reported on a rough scale: \small \begin{array}{cll} \text{P-value} & \text{Interpretation} & \text{Jargon}\\ \hline \le 0.01 & \text{strong evidence against } H_0 & \text{highly significant} \\ \bumpeq 0.05 & \text{moderate evidence against } H_0 & \text{significant}\\ > 0.1 & \text{reasonable consistency with } H_0 & \text{non-significant}\\ \hline \end{array}
A manufacturing plant produces batches of a chemical using a standard production method (A). Now they try a modified method (B) that is supposed to give a higher yield
The experimenters verify that the batches could be supposed independent and that the order of the runs has no influence on the results.
Problem: decide using an experiment whether method (B) gives significantly higher yields than method (A).
Define the experiment: use a completely randomized experiment by making in sequence 10 batches of a chemical using the standard production method (A) followed by 10 batches using the modified method (B).
Define the model: the data are generated by a two-sample model Y_i = \mu_A + \theta X_i + \varepsilon_i, \quad \varepsilon_i \sim_{\text{ ind}} N(0, \sigma^2) where X_i=0 if chemical i is produced by method (A) and X_i = 1 if is produced by method (B)
Define the null and alternative hypotheses. In general they define a partition of the parameter space \Theta: H_0 : \theta \in \Theta_0 \text{ versus } H_1: \theta \in \Theta_1 For instance : H_0: \mu_B - \mu_A = \theta = 0 \text{ vs } H_1: \theta > 0
A hypothesis is said simple if it defines a single distribution. Otherwise it is said composite
In the previous example both hypotheses are composite!
Usually the rejection region is defined by \small R = \{ {\boldsymbol{Y}} : t({\boldsymbol{Y}}) > c\} where t({\boldsymbol{Y}}) is a test statistic and c is called the critical value.
If we can reduce the type I error the type II error increases and viceversa.
A specific distinction from the Fisher’s approach is the definitions of power function and size of the test.
The power function is the probability of rejecting H_0 for all values of the parameter \theta
The size of the test called \alpha is the maximum probability of type I error when the value of the parameter \theta satisfies the null hypothesis.
Unfortunately, finding the optimal test is rather hard !
However there are several tests that come close to this ideal.
We met before the two-sample model with normal errors. The t-test concerns the test about the difference of the two means \theta = \mu_1 - \mu_0.
Two-sided test H_0: \theta = \theta_0 \text{ versus } H_1: \theta \ne \theta_0
One-sided test (right) H_0: \theta \le \theta_0 \text{ versus } H_1: \theta \ge \theta_0
Test statistic
W = \frac{\hat \theta - \theta_0}{\text{se}} \quad \text{where }\quad \hat \theta = \bar Y_1 - \bar Y_0
W \sim T_{n -2}.
The two samples of chemical batches are
YA <-c(89.7,81.4,84.5,84.8,87.3,79.7,85.1,81.7,83.7,84.5)
YB <-c(84.7,86.1,83.2,91.9,86.3,79.3,82.6,89.1,83.7,88.5) and a boxplot is
group = factor(c(rep("A", 10), rep("B", 10)))
Y = c(YA, YB)
boxplot(Y ~ group, horizontal = TRUE)
The test statistic is calculated as follows. The estimate of \theta is \hat \theta = \bar y_B - \bar y_A = 85.54 - 84.24 = 1.3 the estimate of the variance \sigma^2 is the pooled variance s^2_p = \frac{\Sigma_{i\in A} (y_i - \bar y_A)^2 + \Sigma_{i \in B} (y_i - \bar y_B)^2}{n_A + n_B -2} = \frac{75.784+119.924}{18} = 10.87 and the estimate of the standard error is \text{se} = \sqrt{s^2_p \left(\frac{1}{n_A} + \frac{1}{n_B}\right)} = 1.47 So that w_{obs} = \frac{1.30 - 0}{1.47} = 0.88
The rejection region for the one-sided test of size \alpha = 0.05 is |W| > t_{18, 0.05}. The region can be found with
cat("0.05 upper quantile:", qt(0.05, df = 18, lower.tail = FALSE), "\n")0.05 upper quantile: 1.734064 that gives R = \{ {\boldsymbol{Y}} : W > 1.734\}. Therefore the null hypothesis is not rejected at the level \alpha = 0.05.
cat("p-value: ", pt(0.88, df = 18, lower.tail = FALSE), "\n")p-value: 0.1952286 so that the test is not significant at the 5 % level.