statistical tests (hypothesis testing)mwr/lecture 2004-12-09/2004-12-02_g… · 02/12/2004 · for...

Outlines

Statistical Tests (Hypothesis Testing)

Gautam Chandekar

December 2, 2004

Gautam Chandekar Statistical Tests (Hypothesis Testing)

Outlines Statistical Tests (Hypothesis Testing)

Introduction

Steps for Hypothesis TestingDefine a PopulationState Hypotheses and Significance LevelPerform calculations, reach a conclusion.

Statistical Tests (Normal Distribution)IntroductionExamplesHomework

Much of this material is from Allan Bluman’s ElementaryStatistics: A Brief Version, Second Edition.

IntroductionSteps for Hypothesis Testing

Statistical Tests (Normal Distribution)

Part I

Example

A medical researcher is interested in finding out whether anew medication will have any undesirable side effects,particularly with respect to heart rate. Will a patient’s pulserate, increase, decrease, or remain unchanged after taking thismedication?

After conducting controlled tests on a sample of patients, theresearcher finds that the mean pulse rate of the group takingthe new medicine is higher than average. But is this increasesignificant, or simply due to random chance?

Statistical tests (or hypothesis testing) can be used to answerthat question and give the conclusion some certainty.

Example

Define a PopulationState Hypotheses and Significance LevelPerform calculations, reach a conclusion.

Steps for Hypothesis Testing

Define a population under study.

State the particular hypotheses that will be investigated, andgive the significance level.

Select a sample from the population, and collect data.

Perform calculations for the statistical test, and reach aconclusion.

Define a population under study

Recall that a population is the collection of all possible objectsthat have a common measurable or otherwise observablefeature or characteristic.

For the purposes of hypothesis testing, we’re often concernedwith knowing the population’s mean and standard deviation.

In the case of the medical researcher, the population is alladult men, and the mean pulse rate was 70 beats per minute(bpm), with a standard deviation of 8 bpm.

State the hypotheses to be investigated

A statistical hypothesis is a conjecture about a populationparameter. The conjecture may or may not be true.

There are two types of statistical hypotheses in each test: thenull hypothesis and the alternative hypothesis.

The null hypothesis (H0), is a hypothesis that states there isno difference between a parameter and a specific value, orthat there is no difference between two parameters.

The alternative hypothesis (H1) is a hypothesis that states theexistance of difference between a parameter and a specificvalue, or states that there is a difference between twoparameters.

Hypothesis examples

H0: “There is no difference in pulse between these patientsand the nationwide population (µ = 70).” H1: “The pulse ofpatients taking this medication is substantially different thanthe nationwide population (µ 6= 70).”

H0: “This new additive does nothing to increase the life ofautomobile batteries (µ ≤ 36).” H1: “This additivesignificantly increases battery life (µ > 36).”

H0: “A new type of insulation does not help decrease monthlyheating costs (µ ≥ 78).” H1: “This new insulation reducesmonthly heating costs (µ < 78).”

Hypothesis examples

Types of Statistical Tests and Hypotheses

The null and alternative hypotheses are always statedtogether.

Written mathematically, the null hypothesis always containsan equal sign.

The mathematical form of the null and alternative hypothesesfor different types of tests are as follows:

Two-tailed test Right-tailed test Left-tailed testH0: µ = k H0: µ ≤ k H0: µ ≥ kH1: µ 6= k H1: µ > k H1: µ < k

Designing the study, defining the significance level

The simplest design of the medication study would be to givethe drug to a sample of patients, wait long enough for thedrug to be absorbed, and measure the patients’ heart rates.

However, regardless of the medication or the sampling, themean heart rate of the sampled patients will almost never beexactly 70 bpm.

Two possibilities exist:

Either the null hypothesis is true and any differences betweensample mean and population mean are purely due to chance,or the null hypothesis is false and the observed difference inheart rate is due to side effects of the medication.

Either the null hypothesis is true and any differences betweensample mean and population mean are purely due to chance,

or the null hypothesis is false and the observed difference inheart rate is due to side effects of the medication.

Qualitatively, we would expect that if the sample mean pulserate is 71 or 72 bpm, that the difference is due to randomchance.

Similarly, we would expect that if the sample mean pulse rateis 80 or 85 bpm, that the difference is due to the medicine.

But these decisions cannot be made qualitatively, there mustbe a quantifiable standard limits the probability thatdifferences are due to chance.

That is where the level of significance appears.

Types of Errors

There are four possible outcomes from a hypothesis test:

The null hypothesis is actually true, but the sample data doesnot support it and we decide to reject the null hypothesis.

The null hypothesis is actually false, but the sample data leadsus to believe that we should not reject the null hypothesis.

The null hypothesis is actually true, and the sample datasupports this fact.

The null hypothesis is actually false, and the sample datahelps us reject it.

Two of these situations result in a correct decision, and two ofthem result in an incorrect decision.

H0 True H0 False

Reject H0 Type I Error Correct decision

Do not reject H0 Correct decision Type II Error

Types of Errors

H0 True H0 False

Types of Errors

H0 True H0 False

Types of Errors

H0 True H0 False

Types of Errors

H0 True H0 False

Level of Significance (finally)

The level of significance is defined as the maximum probability ofcommitting a type I error, and is normally shown as α. Forexample, if α = 0.05, there is a 5% chance that we will reject atrue null hypothesis. If α = 0.01, there is a 1% chance that we willreject a true null hypothesis.(The probability of committing a type II error is normally shown asβ, and is normally not easily computed. However, for a giventesting situation, lowering α increases β, and raising α lowers β.)

Perform calculations for a statistical test

The significance level α and the type of test (two-tailed, left-tailed,right-tailed) lead to the selection of a critical value from statisticaltables.

The critical value(s) separates the critical region of theprobability distribution curve from the noncritical region.

The critical region is the range of values of the test valuethat indicate a significant difference from the population andthat the null hypothesis should be rejected.

The noncritical region is the range of values of the test valuethat indicate any differences from the population are probablydue to chance, and that the null hypothesis should not berejected.

Example Critical Values: Right-Tailed Test

For a right-tailed test on a normal distribution, the critical value isthe z value that gives Φ(z > zcrit) = α.

−3 −2 −1 0 1 2 30

z*=2.33α=Φ(z>z*)=0.01

Example Critical Values: Left-Tailed Test

For a left-tailed test on a normal distribution, the critical value isthe z value that gives Φ(z < zcrit) = α.

−3 −2 −1 0 1 2 30

z*=−2.33α=Φ(z<z*)=0.01

Example Critical Values: Two-Tailed Test

For a two-tailed test on a normal distribution, the critical value isthe z value that gives Φ(z > zcrit) + Φ(z < −zcrit) = α.

−3 −2 −1 0 1 2 30

z*=±2.58α=Φ(z<z*)=0.005α=Φ(z>z*)=0.005

IntroductionExamplesHomework

The z test is a statistical test for the mean of a population. It canbe used whenever:

the sample size n ≥ 30, or

if the population is known to be normally distributed and thepopulation standard deviation σ is known (regardless of n).

The formula for the z test is

z =X̄ − µ

σ/√

If this z value falls into the critical region, we reject the nullhypothesis, and conclude that there is not enough evidence tosupport the idea that there is a significant difference between thesample and the overall population.

z =X̄ − µ

σ/√

z =X̄ − µ

σ/√

z =X̄ − µ

σ/√

Example 1

Let’s revisit our medical researcher. The mean pulse rate of theentire population of adult men is 70 beats per minute (bpm), witha standard deviation of 8 bpm. A sample of 10 adult male patientsis given a new drug, and after several minutes, their mean pulserate is measured at 75 bpm. With a level of significance ofα = 0.01, is this change in pulse rate due to random chance?

Example 1 Solution (1)

State the null hypothesis: “There is no significant differencein pulse between these patients and the population as awhole.” (H0 : µ = 70)

State the alternative hypothesis: “There is a significantdifference in pulse between these patients and the populationas a whole.” (H1 : µ 6= 70).

The null hypothesis is given in the form corresponding to atwo-tailed test (H0 : µ = k, H1 : µ 6= k), and for α = 0.01,the critical z values are z = ±2.58.

Calculate the test z value:

z =X̄ − µ

σ/√

with X̄ = 75, µ = 70, σ = 8, n = 10.

In this case, z = 1.98. Since z falls within the noncriticalregion, our conclusion is to not reject the null hypothesis.

We conclude that there is not sufficient evidence to indicatethat the change in pulse rate is due to anything other thanrandom chance.

z =X̄ − µ

σ/√

with X̄ = 75, µ = 70, σ = 8, n = 10.

z =X̄ − µ

σ/√

with X̄ = 75, µ = 70, σ = 8, n = 10.

Example 2

Our medical researcher continues with the pulse tests. A largersample of 50 adult male patients has been tested identically to thefirst group, and their mean pulse rate is also measured at 75 bpm.With a level of significance of α = 0.01, is this change in pulse ratedue to random chance?

z =X̄ − µ

σ/√

with X̄ = 75, µ = 70, σ = 8, n = 50.

In this case, z = 4.42. Since z falls within the critical region,our conclusion is to reject the null hypothesis.

We conclude that there is sufficient evidence to indicate thatthe change in pulse rate is due to something other thanrandom chance.

z =X̄ − µ

σ/√

with X̄ = 75, µ = 70, σ = 8, n = 50.

z =X̄ − µ

σ/√

with X̄ = 75, µ = 70, σ = 8, n = 50.

Homework (hand in at final exam time)

A manufacturer states that the average lifetime of its lightbulbs is36 months, with a standard deviation of 10 months. Thirty bulbsare selected, and their average lifetime is found to be 32 months.Should the manufacturer’s statement be rejected at α = 0.01?Should it be rejected if you tested 50 bulbs and found an averagelifetime of 32 months?

statistical tests (hypothesis testing)mwr/lecture 2004-12-09/2004-12-02_g… · 02/12/2004 · for...

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