Statistical Models

Lecture 5

Dr. Silvio Fanzon

S.Fanzon@hull.ac.uk

University of Hull

Dr. John Fry

J.M.Fry@hull.ac.uk

University of Hull

Part 1:
One-sample variance test

Estimating mean and variance

Suppose to have a population with normal distribution $N(\mu,\sigma^2)$
- Mean $\mu$ and variance $\sigma^2$ are unknown
Questions about $\mu$ and $\sigma^2$
1. What is my best guess of the value?
2. How far away from the true value am I likely to be?
Answers:
- The one-sample t-test answers questions about $\mu$
- The one-sample variance ratio test answers questions about $\sigma^2$

One-sample one-sided variance ratio test

Assumption: Suppose given sample $X_1,\ldots, X_n$ iid from $N(\mu,\sigma^2)$

Goal: Estimate variance $\sigma^2$ of population

Test:

Suppose $\sigma_0$ is guess for $\sigma$
The one-sided hypothesis test for $\sigma$ is $H_0 \colon \sigma = \sigma_0 \qquad H_1 \colon \sigma > \sigma_0$

One-sample one-sided variance ratio test

What to do?

Consider the sample variance $S^2 = \frac{ \sum_{i=1}^n X_i^2 - n \overline{X}^2 }{n-1}$
Since we believe $H_0$ , the variance is $\sigma = \sigma_0$
$S^2$ cannot be too far from the true variance $\sigma$
Therefore we cannot have that $S^2 \gg \sigma^2 = \sigma_0^2$

One-sample one-sided variance ratio test

What to do?

If we observe $S^2 \gg \sigma_0^2$ then our guess $\sigma_0$ is probably wrong
Therefore we reject $H_0$ if $S^2 \gg \sigma_0^2$
The rejection condition $S^2 \gg \sigma_0^2$ is equivalent to $\frac{(n-1)S^2}{\sigma_0^2} \gg 1$ where $n$ is the sample size

One-sample one-sided variance ratio test

What to do?

Recall that $\frac{(n-1)S^2}{\sigma^2} \sim \chi_{n-1}^2$
Assuming $\sigma=\sigma_0$ , we therefore have $\chi^2 = \frac{(n-1)S^2}{\sigma_0^2} = \frac{(n-1)S^2}{\sigma^2} \sim \chi_{n-1}^2$

One-sample one-sided variance ratio test

Summary

We reject $H_0$ if $\chi^2 = \frac{(n-1)S^2}{\sigma_0^2} \gg 1$
As $\chi^2 \sim \chi_{n-1}^2$ , we decide to rejct $H_0$ if $\chi^2 > \chi_{n-1}^2(0.05)$
By definition the critical value $\chi_{n-1}^2(0.05)$ is such that $P(\chi_{n-1}^2 > \chi_{n-1}^2(0.05) ) = 0.05$

One-sample one-sided variance ratio test

p-value

Given the test statistic $\chi^2$ the p-value is defined as $p := P( \chi_{n-1}^2 > \chi^2 )$
Notice that $p < 0.05 \qquad \iff \qquad \chi^2 > \chi_{n-1}^2(0.05)$

This is because $\chi^2 > \chi_{n-1}^2(0.05)$ iff $p = P(\chi_{n-1}^2 > \chi^2) < P(\chi_{n-1}^2 > \chi_{n-1}^2(0.05) ) = 0.05$

One-sample one-sided variance ratio test

Procedure

Suppose given

Sample $x_1, \ldots, x_n$ of size $n$ from $N(\mu,\sigma^2)$
Guess $\sigma_0$ for $\sigma$

The one-sided hypothesis test is $H_0 \colon \sigma = \sigma_0 \qquad H_1 \colon \sigma > \sigma_0$ The variance ratio test consists of 3 steps

One-sample one-sided variance ratio test

Procedure

Calculation: Compute the chi-squared statistic $\chi^2 = \frac{(n-1) s^2}{\sigma_0^2}$ where sample mean and variance are $\overline{x} = \frac{1}{n} \sum_{i=1}^n x_i \,, \qquad s^2 = \frac{\sum_{i=1}^n x_i^2 - n \overline{x}^2}{n-1}$

One-sample variance ratio test: Example

Month	J	F	M	A	M	J	J	A	S	O	N	D
Cons. Expectation	66	53	62	61	78	72	65	64	61	50	55	51
Cons. Spending	72	55	69	65	82	77	72	78	77	75	77	77
Difference	-6	-2	-7	-4	-4	-5	-7	-14	-16	-25	-22	-26

Remark: Monthly data on CE and CS can be matched
- Hence consider: $\quad$ Difference $=$ CE $-$ CS
- CE and CS normal $\quad \implies \quad$ Difference $\sim N(\mu,\sigma^2)$
Question: Test the following hypothesis: $H_0 \colon \sigma = 1 \qquad H_1 \colon \sigma > 1$

Motivation of test

If $X \sim N(\mu,\sigma^2)$ then $P( \mu - 2 \sigma \leq X \leq \mu + 2\sigma ) \approx 0.95$
Recall: $\quad$ Difference $=$ (CE $-$ CS) $\sim N(\mu,\sigma^2)$
Hence if $\sigma = 1$ $P( \mu - 2 \leq {\rm CE} - {\rm CS} \leq \mu + 2 ) \approx 0.95$
Meaning of variance ratio test: $\sigma=1 \quad \implies \quad \text{CS index is within } \pm{2\sigma} \text{ of CE index with probability } 0.95$

The variance ratio test

Month	J	F	M	A	M	J	J	A	S	O	N	D
Difference	-6	-2	-7	-4	-4	-5	-7	-14	-16	-25	-22	-26

Calculations: With data on Difference compute

Sample mean: $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} = \frac{-6-2-7-4- \ldots -22-26}{12} = -\frac{138}{12} = -11.5$

The variance ratio test

Month	J	F	M	A	M	J	J	A	S	O	N	D
Difference	-6	-2	-7	-4	-4	-5	-7	-14	-16	-25	-22	-26

Calculations: With data on Difference compute

Sample variance: $\begin{align*} \sum_{i=1}^{n} x_i^2 & = (-6)^2 + (-2)^2 + (-7)^2 + \ldots + (-22)^2 + (-26)^2 = 2432 \\ s^2 & = \frac{\sum_{i=1}^n x^2_i- n \bar{x}^2}{n-1} = \frac{2432-12(-11.5)^2}{11} = \frac{845}{11} = 76.8182 \end{align*}$

The variance ratio test

Conclusion:
- We accept $H_1$ : The standard deviation satisfies $\sigma > 1$
- A better estimate for $\sigma$ could be sample standard deviation $s=\sqrt{\frac{845}{11}}=8.765$
- This suggests: With probability $0.95$ $\text{CS index is within } \pm{2 \times 8.765 = \pm 17.53 } \text{ of CE index}$

Probability Distributions in R

R can natively do calculations with known probability distrubutions
Example: Let $X$ be r.v. with $N(\mu,\sigma^2)$ distribution

R command	Meaning
`pnorm(x, mean = mu, sd = sig)`	$P(X \leq x)$
`qnorm(p, mean = mu, sd = sig)`	$P(X \leq q) = p$
`dnorm(x, mean = mu, sd = sig)`	$f(x)$ where $f$ is distr. of $X$
`rnorm(n, mean = mu, sd = sig)`	$n$ random samples from distr. of $X$

Note: Syntax of commands

norm $=$ normal $\qquad$ p $=$ probability $\qquad$ q $=$ quantile

d $=$ distribution $\qquad$ r $=$ random

R command	Meaning
`pchisq(x, df = n)`	$P(X \leq x)$
`qchisq(p, df = n)`	$P(X \leq q) = p$
`dchisq(x, df = n)`	$f(x)$ where $f$ is distr. of $X$
`rchisq(m, df = n)`	$m$ random samples from distr. of $X$

Probability Distributions in R

Example 1 – Solution

From Tables we found quantile $\chi_{11}^2 (0.05) = 19.68$
Question: Compute such quantile in R

# Compute 0.95 quantile for chi-squared with 11 degrees of freedom

quantile <- qchisq(0.95, df = 11)

cat("The 0.95 quantile for chi-squared with df = 11 is", quantile)

The 0.95 quantile for chi-squared with df = 11 is 19.67514

Probability Distributions in R

Example 2 – Solution

The $\chi^2$ statistic for variance ratio test has distribution $\chi_{n-1}^2$
Question: Compute the p-value $p := P(\chi_{n-1}^2 > \chi^2)$
Observe that $p := P(\chi_{n-1}^2 > \chi^2) = 1 - P(\chi_{n-1}^2 \leq \chi^2)$
The code is therefore

# Compute p-value for chi^2 = chi_squared and df = n

p_value <- 1 - pchisq(chi_squared, df = n)

The variance ratio test in R

Month	J	F	M	A	M	J	J	A	S	O	N	D
Cons. Expectation	66	53	62	61	78	72	65	64	61	50	55	51
Cons. Spending	72	55	69	65	82	77	72	78	77	75	77	77
Difference	-6	-2	-7	-4	-4	-5	-7	-14	-16	-25	-22	-26

Back to Example: Monthly data on CE and CS
Question: Test the following hypothesis: $H_0 \colon \sigma = 1 \qquad H_1 \colon \sigma > 1$

The variance ratio test in R

Compute the p-value $p = P(\chi_{n-1}^2 > \chi^2) = 1 - P(\chi_{n-1}^2 \leq \chi^2)$

# Compute p-value
p_value <- 1 - pchisq(chi_squared, df = n - 1)

# Print p-value
cat("The p-value for one-sided variance test is", p_value)

The full code can be downloaded here variance_ratio_test.R

F-distribution

Definition

The r.v.

F

has F-distribution with

p

and

q

degrees of freedom if the pdf is

f_F(x) = \frac{ \Gamma \left(\frac{p+q}{2} \right) }{ \Gamma \left( \frac{p}{2} \right) \Gamma \left( \frac{q}{2} \right) } \left( \frac{p}{q} \right)^{p/2} \, \frac{ x^{ (p/2) - 1 } }{ [ 1 + (p/q) x ]^{(p+q)/2} } \,, \quad x > 0

Notation: F-distribution with $p$ and $q$ degrees of freedom is denoted by $F_{p,q}$

F-distribution

Idea of Proof

Similar to the proof (seen in Homework 3) that $\frac{U}{\sqrt{V/p}} \sim t_p$ where $U \sim N(0,1)$ and $V \sim \chi_p^2$ are independent
In our case we need to prove $X := \frac{U/p}{V/q} \sim F_{p,q}$ where $U \sim \chi_p^2$ and $V \sim \chi_q^2$ are independent

F-distribution

Idea of Proof

$U \sim \chi_{p}^2$ and $V \sim \chi_q^2$ are independent. Therefore $\begin{align*} f_{U,V} (u,v) & = f_U(u) f_V(v) \\ & = \frac{ 1 }{ \Gamma \left( \frac{p}{2} \right) \Gamma \left( \frac{q}{2} \right) 2^{(p+q)/2} } u^{\frac{p}{2} - 1} v^{\frac{q}{2} - 1} e^{-(u+v)/2} \end{align*}$
Consider the change of variables $x(u,v) := \frac{u/p}{v/q} \,, \quad y(u,v) := u + v$

F-distribution

Idea of Proof

The joint pdf $f_{X,Y}$ can be computed by inverting the change of variables $x(u,v) := \frac{u/p}{v/q} \,, \quad y(u,v) := u + v$ and using the formula $f_{X,Y}(x,y) = f_{U,V}(u(x,y),v(x,y)) \, |\det J|$ where $J$ is the Jacobian of the inverse transformation $(x,y) \mapsto (u(x,y),v(x,y))$

F-distribution

Idea of Proof

Since $f_{U,V}$ is known, then also $f_{X,Y}$ is known
Moreover the integral $f_{X}(x) = \int_{0}^\infty f_{X,Y}(x,y) \, dy$ can be explicitly computed, yielding the thesis $f_{X}(x) = \frac{ \Gamma \left(\frac{p+q}{2} \right) }{ \Gamma \left( \frac{p}{2} \right) \Gamma \left( \frac{q}{2} \right) } \left( \frac{p}{q} \right)^{p/2} \, \frac{ x^{ (p/2) - 1 } }{ [ 1 + (p/q) x ]^{(p+q)/2} }$

Properties of F-distribution

Proof of Theorem

Requires a bit of work. It will be left as Homework assignment
By the Theorem in Slide 43 we have $X \sim F_{p,q} \quad \implies \quad X = \frac{U/p}{V/q}$ with $U \sim \chi_p^2$ and $V \sim \chi_q^2$ independent. Therefore $\frac{1}{X} = \frac{V/q}{U/p} \sim F_{q,p}$

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Statistical Models Lecture 5 Dr. Silvio Fanzon S.Fanzon@hull.ac.uk University of Hull Dr. John Fry J.M.Fry@hull.ac.uk University of Hull

Statistical Models
Lecture 5: Hypothesis tests in R Part 2
Outline of Lecture 5
Part 1: One-sample variance test
Estimating mean and variance
Chi-squared distribution
One-sample one-sided variance ratio test
One-sample one-sided variance ratio test
One-sample one-sided variance ratio test
One-sample one-sided variance ratio test
One-sample one-sided variance ratio test
One-sample one-sided variance ratio test
Critical values of chi-squared
Critical values of chi-squared – Tables
One-sample one-sided variance ratio test
One-sample one-sided variance ratio test
One-sample one-sided variance ratio test
One-sample one-sided variance ratio test
One-sample one-sided variance ratio test
Part 2: One-sample variance test: Example
One-sample variance ratio test: Example
One-sample variance ratio test: Example
Motivation of test
The variance ratio test
The variance ratio test
The variance ratio test
The variance ratio test
The variance ratio test
The variance ratio test in R
Probability Distributions in R
Probability Distributions in R
Probability Distributions in R
Probability Distributions in R
Probability Distributions in R
Probability Distributions in R
Probability Distributions in R
Probability Distributions in R
Probability Distributions in R
Probability Distributions in R
The variance ratio test in R
The variance ratio test in R
The variance ratio test in R
The variance ratio test in R
Running the code
Part 3: F-distribution
Overview
Overview
F-distribution
F-distribution
F-distribution
F-distribution
F-distribution
F-distribution
F-distribution
F-distribution
Properties of F-distribution
Properties of F-distribution
Properties of F-distribution
Properties of F-distribution