Hypothesis testing Null hypothesis - PowerPoint Presentation

Slide1

Hypothesis testingSlide2

Null hypothesis

Ho - this hypothesis holds that if the data deviate from the norm in any way, that deviation is due strictly to chance.

Alternative hypothesis

Ha - the data show something important.

Doing decision = accept/reject Ho (the decision centers around null hypothesis)Slide3

Errors in hypothesis testing

Type I – False Positive

Type II – False Negative

The probability of Type I error: αThe probability of Type II error: βSlide4

Test involving sample from a normally distributed population.

Because it’s a normal distribution, you use z-scores in the hypothesis test.

The z-score here is called

test statistics.The test statistics constructed according the above formula holds only for the mean.Tests for other statistics (e.g. variance) use different formulas. Slide5

Suppose you think that people living in a particular zip code have higher-than-average IQs. Your data are given in sheet ZIP, test this hypothesis.

n

= 16,

μZIP = 107.75, α = 0.05We know about IQ scores: μ = 100, σ = 16

What will be the Ho and Ha?Ha: μ

ZIP > 100Ho: μZIP ≤

100

Can you reject Ho?Slide6

What is the value of z that cuts off 5% of the area in a standard normal distribution?

It’s exactly 1.645.

So what’s the decision?

The calculated value, 1.94, exceeds 1.645, so it’s in the rejection region. The decision is to reject Ho.Slide7

This hypothesis test is called

one tailed

(

one sided).The rejection region is in one tail of the sampling distribution.A hypothesis test can be one tailed in the other direction.Ha: μ

ZIP < 100Ho:

μZIP ≥ 100What is the critical

value?

-1.645

http://www.emathzone.com/tutorials/basic-statisticsSlide8

Test can be also

two-tailed

.

The rejection region is in both tails of the Ho sampling distribution.Ho: μZIP = 100Ha: μZIP ≠ 100

What is the critical value now?

Find z-score that cuts 2.5% from right (=1.96) and from left (=-1.96). 1.94 does not exceed 1.96, we do not reject Ho.

http://www.emathzone.com/tutorials/basic-statisticsSlide9

Using one tailed test we rejected Ho, while using two tailed test we did not!!

A two tailed test indicates that you’re looking for a difference between the sample mean and the null-hypothesis mean, but you don’t know in which direction.

A one tailed test shows that you have a pretty good idea of how the difference should come out.

For practical purposes, this means you should try to have enough knowledge to be able to specify a one tailed test.Slide10

z-test in Excel

ZTEST

Do now:

examples2.xlsx | ZIPprovide sample IQ data, null hypothesis value, σ (if omitted, s is used)p-value is returnedIf p-value <

α, reject Ho.Will you reject Ho or not?Slide11

This is the result of ZTEST Slide12

Critical value for one tailed test

For one tailed test you reject Ho.

What if you do two tailed test?

α

= 0.05

Our actual value (red line, p-value = 0.026) is in the rejection region of one tailed test (0.026 < 0.05).

However, it is outside rejection region for two tailed test. To see this, you must compare 0.026 > 0.025. Or you can 2x multiply this equation → 0.052 > 0.05.

So if you have

α

set to 0.05, and you get p-value for one sided test, you get p-value for two sided test doubling the one sided p-value.Slide13

t

for one

In the real world you typically don’t have the luxury of working with such well-defined populations as results of IQ test.

Real world:small samplesyou often don’t know the population parametersWhen that’s the case, you use the sample data to estimate the population standard deviation

you treat the sampling distribution of the mean as a t-distributionYou use

t as a test statisticSlide14

The formula for the test statistic

with DF =

n – 1. The higher the DF, the more closely the t-distribution resembles the normal distribution.Slide15

Company claims their vacuum cleaner averages four defects per unit. A consumer group believes this average is higher. The consumer group takes a sample of 9 cleaners and finds an average of 7 defects, with a standard deviation of 3.16.

Is companie’s claim correct or not?

Ho, Ha?

Ho: μ ≤ 4Ha: μ > 4And what else is missing in defining the hypothesis?

α = 0.05Slide16

Now calculate t test statistic

Can you reject Ho

?

Get critical value from tables or TINV.Use Excel TDISTreturns p-valuereject

HoSlide17

Testing a variance

The family of distributions for the test is called

chi-square

- χ2The formula

for test statistics

With this test, you have to assume that what you’re measuring has a normal distribution.Slide18

Solve the following example using CHIDIST.

You produce a part of some machine that has to be a certain length with at most a standard deviation of 1.5 cm.

After measuring a sample of 26 parts, you find a standard deviation of 1.8 cm.

Is your process producing these parts OK?Ho: σ2 ≤ 2.25 (remember to square the “at-most” standard deviation of 1.5 cm)Ha: σ

2 > 2.25

α = 0.05

p-value = 0.0716. Do not reject Ho.Slide19

Two sample hypothesis testing

Compare one sample with another.

Usually, this involves tests of hypotheses about population means. You can also test hypotheses about population variances.

Here’s an example. Imagine a new training technique designed to increase IQ. Take a sample of 25 people and train them under the new technique. Take another sample of 25 people and give them no special training. Suppose that the sample mean for the new technique is 107, and for the no-training sample it’s 101.2.Did the technique really increased IQ?Slide20

Same principles: Ho (no difference between means), Ha,

α

one-tailed test

Ho: μ1 – μ2 = 0, Ha:

μ1 –

μ2 > 0Ho: μ

1

–

μ

2

= 0, Ha:

μ1 – μ2 < 0

two-tailed test

Ho:

μ

1

–

μ

2

= 0, Ha:

μ

1

–

μ

2

≠

0

The zero is typical case, but it’s possible to test for any value.Slide21

The first sample in the pair always has the same size, and the second sample in the pair always has the same size. The two sample sizes are not necessarily equal.Slide22

CLT strikes again

If the samples are large, the sampling distribution of the difference between means is approximately a normal distribution.

If the populations are normally distributed, the sampling distribution is a normal distribution even if the samples are small.

The mean of the sampling distributionThe standard deviation of the sampling distribution (

standard error of the difference between means)Slide23
Slide24

Because CLT says that the sampling distribution is approximately normal for large samples (or for small samples from normally distributed populations), you use the z-score as your test statistic.

i.e. you perform a z-test.

The

z test statistics:Slide25

Solve the following

.

Imagine a new training technique designed to increase IQ. Take a sample of 25 people and train them under the new technique. Take another sample of 25 people and give them no special training. Suppose that the sample mean for the new technique sample is 107, and for the no-training sample it’s 101.2.

Did the technique really increased IQ?Slide26

Ho:

μ

1

– μ2 = 0, Ha: μ1

– μ2

> 0, α = 0.05The IQ is known to have a standard deviation of 16, and I assume that standard deviation would be the same in the population of people trained on the new technique.

Use either NORMSDIST (supply 1.28, you get p-value = 1-0.899=0.101) or NORMSINV (probability = 0.95, you get critical value equaling to 1.645).

Accept Ho.Slide27
Slide28

Excel provides a tool z-Test: Two Sample for Means (Data | Data Analysis)

Do now

IQ_Test sheetSlide29

Variable variance is 16

2

= 256 (16 is population standard deviation of IQ test distribution)Slide30

t

for Two

The previous example involves a situation you rarely encounter - known population variances.

Not knowing the variances takes the CLT out of play. This means that you can’t use the normal distribution as an approximation of the sampling distribution of the difference between means. Instead, you use the t-distribution. You perform a t-test.Slide31

Unknown variances lead to two possibilities for hypothesis testing:

although the variances are unknown, you have reason to assume they’re equal

you cannot assume they’re equalSlide32

t for Two – equal variances

Put sample variances together to estimate a population variance –

pooling

DFSlide33

FarKlempt Robotics is trying to choose between two machines to produce a component for its new microrobot. Speed is of the essence, so they have each machine produce ten copies of the component, and time each production run.

Which machine should they choose?

Do

now using Data Analysis, Mechine_speed sheet.Slide34

Ho:

μ

1

- μ2 = 0, Ha: μ1 - μ

2 ≠ 0, α

= 0.05This is a two-tailed test, because we don’t know in advance which machine might be faster.

Get critical value using TINV (+-2.10)

or

p-value using TDIST (0.0252).

Result: reject Ho. Slide35

The worksheet function TTEST eliminates the muss, fuss, and bother of working through the formulas for the t-test.

Do now

– Machines example in examples2.xlsx | Machine_speed

It’s more desirable to use the equal

variances t-test

, which typically provides more degrees of freedom than the unequal variances t-test.Slide36

Do now

– Data|Data Analysis, use t-Test: Two-Sample Assuming Equal VariancesSlide37

t for Two – unequal variances

In the case of unequal variances, the t distribution with (N

1

-1) + (N2-1) DF is not as close an approximation to the sampling distribution.DF must be reduced, fairly involved formulas are used to do this.A pooled estimate is not appropriate. t-test is calculated as Slide38

Testing two variances

classic: Ho:

σ

12 = σ22, Ha: σ12

≠≥≤ σ

22, α=0.05When you test two variances, you don’t subtract one from the other.

Instead, you

divide one by the other to calculate the test statistic

.

This statistics is called

F-ratio

, and you’re doing F-test.Slide39

The family of distributions for the test is called the

F-distribution

.

Each member of the family is associated with two values of DF (each DF is n - 1)!And it makes a difference which DF is in the numerator and which

DF is in the denominator.Slide40

One use of the

F

-distribution is in conjunction with the t-test for

independent samples. Before you do the t-test, you use F to help decide whether to assume equal variances or unequal variances in the samples.Excel: FTEST, FDIST, FINV, F-Test: Two-Sample for Variances

Do nowFarKlempt Robotics produces 10 parts with Machine 1 and finds

a sample variance of .60 cm2. They produce 15 parts with Machine 2 and

find a sample variance of .44

cm

2

. Are these variances same?

Data are in the examples2.xlsx | Machine_varSlide41
Slide42

Estimate variances from data using VAR.

FDIST value is exactly ½ of FTEST. (FDIST is one-tailed, FTEST is two-tailed)Slide43

It finds a critical value.

Probability is 0.025, because two tailed test is with

α

= 0.05.Slide44
Slide45

For future

useSlide46

http://www.intuitor.com/statistics/CurveApplet.html

α

β

powerSlide47

Theoretically, when you test a null hypothesis versus an alternative hypothesis, each hypothesis corresponds to a separate sampling distribution.

When you do a hypothesis test, you never know which distribution produces the results. You work with a sample mean - a point on the horizontal axis. It’s your job to decide which distribution the sample mean is part of. You set up a

critical value

- a decision criterion. If the sample mean is on one side of the critical value, you reject Ho. If not, you don’t.