Inferential Statistics Part C Hypothesis tests

Inferential Statistics Part C

Hypothesis tests

Eva Riccomagno, Maria Piera Rogantin

DIMA – Universit`a di Genova

http://www.dima.unige.it/~rogantin/UnigeStat/

Part C

Hypothesis tests on the mean of a population

0. Review

1. Aside of probability. Normal (or Gaussian) random variable 2. Test on the mean of a Normal variable – known variance

(running example: toxic algae)

(a) Composite hypotheses (b) The power function

0. Review

A hypothesis testing is formed by - a null and alternative hypothesis

- a test statistics (function of the sample variables) - a rejection region/decision rule

Significance level/Critical value/Test statistic: all contribute to the definition of the rejection rule

Type I and II errors

Example: if R₀ = {x such that T (x) > s} then:

α = P(T > s | H0) reject H₀ when it’s true β = P(T ≤ s | H1) retain H₀ when it’s false

Formulation of the hypotheses and “form” of R₀

Example. The quantity of tomatoes in a can is nominally set at 100 g.

Formulate a statistical hypothesis test as if you were:

1. the Federal Trade Commission 2. an unscrupulous shareholder

3. the worker in charge of the quality control of the can tomato filling machine.

Indicate to which hypotheses the following R₀ correspond:

R₀ = {x s.t. T (x) > s}

R₀ = {x s.t. T (x) < s₁} ∪ {x s.t. T (x) > s₂} R₀ = {x s.t. T (x) < s}

In case of two-sided test s₁ and s₂ are such that

α₁ = P(T < s1 | H₀) α₂ = P(T > s2 | H₀) with α₁ + α₂ = α If the test statistic has a symmetrical distribution often α₁ and α₂ are set as: α₁ = α₂ = α/2

Consider Ex. 2 of Assignment 5. X ∼ B(20, p)

What happens when the hypotheses change?

1. H0 : p = 0.3 H1 : p = 0.5 α = 0.05 one-sided right 2. H0 : p = 0.5 H1 : p = 0.3 α = 0.05 one-sided left

> p_0=0.3;p_1=0.5;

> s05_right=qbinom(1-0.05,20,p_0);s05_right [1] 9 ### same result as p_1=0.7

> s05_left=qbinom(0.05,20,p_1)-1;s05_left ### note -1 [1] 5

One-sided right R0 = {x > 9}; one-sided left R₀ = {x < 5}.

Notice that for x = 5, 6, 7, 8, 9 the decision of the two tests is different (small size and p0 “close” to p1)

What happens when α change?

1. H0 : p = 0.3 H1 : p = 0.5 α = 0.05 3. H0 : p = 0.3 H1 : p = 0.5 α = 0.01

> s01_right=qbinom(1-0.01,20,p_0);s01_right [1] 11

R0 with α = 0.01 is smaller than R0 with α = 0.05

1. Aside of Probability. Normal (or Gaussian) random variable

Probability density functions and cumulative distribution functions

µ = −1,0, 5 σ = 1

−4 −2 0 2 4 6 8

0.00.20.4

Normal −− sigma = 1

−4 −2 0 2 4 6 8

0.00.20.4

Normal −− sigma = 1

−4 −2 0 2 4 6 8

0.00.20.4

Normal −− sigma = 1

−4 −2 0 2 4 6 8

0.00.40.8

Normal −− sigma = 1

mu= − 1 mu = 0 mu = 5

−4 −2 0 2 4 6 8

0.00.40.8

Normal −− sigma = 1

mu= − 1 mu = 0 mu = 5

−4 −2 0 2 4 6 8

0.00.40.8

Normal −− sigma = 1

mu= − 1 mu = 0 mu = 5

µ = 0

σ = 0.5,1, 3

−5 0 5

0.00.20.40.60.8

Normal −− mu = 0

−5 0 5

0.00.20.40.60.8

Normal −− mu = 0

−5 0 5

0.00.20.40.60.8

Normal −− mu = 0

−5 0 5

0.00.40.8

Normal −− mu = 0

sigma=0.5 sigma=1 sigma=3

−5 0 5

0.00.40.8

Normal −− mu = 0

sigma=0.5 sigma=1 sigma=3

−5 0 5

0.00.40.8

Normal −− mu = 0

sigma=0.5 sigma=1 sigma=3

The density function of X ∼ N (µ, σ²) is: f (x) = ^{√ 1}

2π σ² exp



−^(x−µ)

2

2σ²



Some properties of the Normal random variables

• For X ∼ N (µ, σ²) let Y = aX + b. Then Y ∼ N (aµ + b, a²σ²) In particular

Z = X − µ

σ ∼ N (0, 1) Z is called standard Normal variable

• The sum of Normal variables is a Normal variable.

In particular for X₁, . . . , X_n independent and identically dis- tributed (i.i.d.) with X_i ∼ N (µ, σ²) for all i = 1, . . . , n, then

X ∼ N µ, σ² n

!

2. Test on the mean of a Normal random variable with known variance

2. (a) Composite hypotheses

Running example:

X models the concentration of toxic alga blooms

The statistical model is:

X ∼ N (µ, σ²) with σ² known

and experts set a bathing alert if µ > 10000 cells/liter. Thus H₀ : µ ≥ 10000 H₁ : µ < 10000

The significance level of the test is set at α = 5%

If we reject H₀ we can swim with 5% probability of side effects due to the alga

Test statistics X

Sample size: n = 10 σ = 2100 cells/liter Set α = 5%

The test hypotheses are both composite

First we consider the two cases 1. H₀ : µ=10000 H₁ : µ= 8500 2. H₀ : µ=10000 H₁ : µ<10000 and next

3. H₀ : µ≥10000 H₁ : µ<10000

10000

Case 1.

Assume H₀: X ∼ N (10000, 2100²/10)

R₀ = {x < s} where s is such that P^X < s|µ = 10000^ = α Get s = 8908 with R

mu0=10000;std=2100/sqrt(10) c1= qnorm(.05,mu0,std);c1

If the test statistic value on the sample is less than 8908, then we reject H₀ Spot the probability of type I error in the plot.

If x is larger than 8908, then the prob- ability of type II error is

β = P ^X > s|µ = 8500^ Get β = 27% with R

mu1=8500;1-pnorm(c1,mu1,std)

We retain H₀ with a large probability of type II error

10000

8500 retain H

reject H_{0 0}

10000

8500 retain H

reject H_{0 0}

Case 2.

H₀ : µ=10000 H₁ : µ<10000 R₀ does not change

as it is computed under H₀

The probability β of II type error be- comes a function of µ as it is com-

puted under H₁ namely µ < 10000 _{reject H} ¹⁰⁰⁰⁰_{retain H}

0 0

Case 3.

H₀ : µ≥10000 H₁ : µ<10000

Keeping the same R₀, the prob- ability of type I error, denoted by α(µ), is smaller than α

α(µ) < α

The probability β of the type II error is a function of µ under H₁, as in

10000 retain H

reject H_{0 0}

2. (b) The power function P⁽θ) of a test

Consider a generic test on a parameter θ:

H₀ : θ ∈ Θ₀ H₁ : θ ∈ Θ₁

Running example. H₀ : µ ≥ 10000 H₁ : µ < 10000 Then Θ₀ = [10000, +∞), Θ₁ = (−∞, 10000)

The power function of a test is the probability to reject H₀ as a function of the parameter θ

P (θ) = P(T ∈ R0|θ) =

( 1 − β(θ) if θ ∈ Θ₁ correct decision α(θ) if θ ∈ Θ₀ type I error

Note that α(θ) ≤ α

1. the critical value is the smallest s s.t. P (T > s|H₀) < α 2. the critical region is the largest R₀ s.t. P (T ∈ R₀|H₀) < α

Power function:

P (θ) = P (T ∈ R₀|θ) =

( 1 − β(θ) if θ ∈ Θ₁ correct decision α(θ) if θ ∈ Θ₀ type I error

Running example:

R₀ = {X < 8908}

P (µ) = P^X < 8908 | µ ∈ R^ P (10000) = 0.05 = α

0 1

α

10000 8500

1-β(8500)

std=2100/sqrt(10);c1=qnorm(.05,10000,std) mu=seq(7000,11000)

p=pnorm(c1,mu,std)

plot(mu,p,type="l",lwd=3, col="red")

Power and sample size

The probability to reject H₀, when it is false, grows as the sam- ple size grows

The probabilities of the two type of errors decrease as the sample size grows

Running example

H₀ : µ ≥ 10000 and H₁ : µ < 10000 R₀ = {X_n < 8908}

P (µ) = P ^X_n < 8908 | µ ∈ R^

n = 10 red n = 20 blue ⁰

1

α

10000 0

1

α

10000

9500 Θ

Θ1 0

If the values of the parameter under H₀ and under H₁ are “close”

(e.g. 10000 and 9500 respectively), only with large sample the probability of correct decision is large

The power for one-sided and two-sided tests

One-sided: H₀ : µ ≥ 10000 and H₁ : µ < 10000

R₀ = {X < 8908} P (µ) = P^X < 8908| µ ∈ R^

Two-sided: H₀ : µ = 10000 and H₁ : µ 6= 10000 R₀ = {X < 8700} ∪ {X > 11300}

P (µ) = P^X < 8700 | µ ∈ R^ + P^X > 11300) | µ ∈ R^

One-sided red

H₀ : µ ≥ 10000 H₁ : µ < 10000

Two-sided blue

H₀ : µ = 10000 H₁ : µ 6= 10000

0 1

α

10000

mu=seq(7000,13000); mu0=10000; std=2100/sqrt(10) c1_u=qnorm(.05,mu0,std); p=pnorm(c1_u,mu,std)

c1_b=qnorm(.025,mu0,std);c2_b=qnorm(.975,mu0,std);p_b=pnorm(c1_b,mu,std)+1-pnorm(c2_b,mu,std) limx=c(7000,13000);limy=c(0,1)

plot(mu,p,type="l",lwd=3,xaxt="n",yaxt="n",xlab=" ",ylab=" ",col="red",xlim=limx,ylim=limy) par(new=T)

plot(mu,p_b,type="l",lwd=3,xaxt="n",yaxt="n",xlab=" ",ylab=" ",col="blue",xlim=limx,ylim=limy) axis(2, at = 0,0, las=1,cex=1.2,col="blue");axis(2, at = 1,1, las=1,cex=1.2,col="blue")

axis(2, at =0.05,expression(alpha), las=1,cex=1.2,col="blue") axis(1, at = mu0,mu0, las=1,cex=1.2,col="blue")

abline(h=c(0,1));abline(h=0.05,v=c(mu0),lty=3,lwd=2);abline(h=0.05,lty=3,lwd=2)