test of goodness of fit - lecture 42 section...

Test ofGoodness of

Fit

Robb T.Koether

HomeworkReview

TheGoodness-of-FitTest

TheChi-SquareStatistic

Goodness-of-Fit Test on theTI-83

Male vs.Female BirthsAgain

Assignment

Test of Goodness of FitLecture 42

Section 14.3

Robb T. Koether

Hampden-Sydney College

Fri, Nov 14, 2008

Test ofGoodness of

Fit

Robb T.Koether

HomeworkReview





Assignment

Outline

1 Homework Review

2 The Goodness-of-Fit Test

3 The Chi-Square Statistic

4 Goodness-of-Fit Test on the TI-83

5 Male vs. Female Births Again

6 Assignment

Test ofGoodness of

Fit

Robb T.Koether

HomeworkReview





Assignment

Homework Review

Homework ReviewSix basic questions.

Are we testing hypotheses or finding a confidenceinterval?Does the problem concern means or proportions?Is there one sample or are there two samples?

If the problem concerns means, then the next twoquestions are:

Is σ known?Is the population normal?

Finally,Is the sample size large?

Test ofGoodness of

Fit

Robb T.Koether

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Assignment

Homework Review

Exercise 11.28, page 712.

A toothpaste manufacturer claims that childrenbrushing their teeth daily with his company’s newtoothpaste product will have fewer cavities than childrenusing a competitor’s brand.In a carefully supervised study in which children wererandomly assigned to one of the two brands oftoothpaste for a 2-year period, the number of cavitiesfor children using the new brand was compared withthe number of cavities for children using thecompetitor’s brand.

Test ofGoodness of

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Robb T.Koether

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Assignment

Homework Review

Exercise 11.28, page 712.

The results are as follows:New Brand: 2 1 1 2 3 1 2 3 4 1 2

Competitor Brand: 3 1 1 4 0 7 1 1 4 6 1

Test the manufacturer’s claim using a significance levelof 0.01.

Test ofGoodness of

Fit

Robb T.Koether

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Assignment

Homework Review

SolutionAre we testing hypotheses or finding a confidenceinterval?

Z-Test ZIntervalT-Test TInterval2-SampZTest 2-SampZInt2-SampTTest 2-SampTInt1-PropZTest 1-PropZInt2-PropZTest 2-PropZInt

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Homework Review

SolutionDoes the problem concern means or does it concernproportions?

Z-Test 1-PropZTestT-Test 2-PropZTest2-SampZTest2-SampTTest

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Homework Review

SolutionDoes the problem involve one sample or two samples?

Z-Test 2-SampZTestT-Test 2-SampTTest

Test ofGoodness of

Fit

Robb T.Koether

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Homework Review

SolutionTo choose between 2-SampZTest and 2-SampTTest,we need to know

Are σ1 and σ2 not known?Are the populations normal?Are the sample sizes small?

2-SampZTest 2-SampTTest

Test ofGoodness of

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Robb T.Koether

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Assignment

Homework Review

SolutionTo choose between 2-SampZTest and 2-SampTTest,we need to know

Are σ1 and σ2 not known? YESAre the populations normal? YES (QQ Plot)Are the sample sizes small? YES

2-SampZTest 2-SampTTest

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Fit

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Assignment

The Hypotheses

Example (Goodness-of-Fit Test)We are testing the hypothesis that the distributions ofcolors in plain M&Ms is

HypotheticalColor ProportionsBlue 24%Orange 20%Green 16%Yellow 14%Brown 13%Red 13%

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The Hypotheses

Example (Goodness-of-Fit Test)From two packages of plain M&Ms, we obtained thefollowing data.

ObservedColor CountsBlue 20Orange 24Green 25Yellow 23Brown 12Red 10

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The Hypotheses

The null hypothesis specifies the probability (orproportion) for each color.The null hypothesis is

H0 : p1 = 0.24, p2 = 0.20, p3 = 0.16,p4 = 0.14, p5 = 0.13, p6 = 0.13.

The alternative hypothesis will always be a simplenegation of H0:

H1 : H0 is false.Let α = 0.05.

Test ofGoodness of

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Expected Counts

The test statistic will involve the observed and theexpected counts.To find the expected counts, we apply the hypotheticalproportions to the sample size.For example, the hypothetical proportion for red is 24%,so we compute 24% of 114:

0.24× 114 = 27.36.

Do not round the values off to whole numbers.

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Assignment

The Test Statistic

Make a chart showing both the observed counts andthe expected counts (in parentheses).

Color Blue Orange Green Yellow Brown RedObserved 20 24 25 23 12 10(Expected) (27.36) (22.80) (18.24) (15.96) (14.82) (14.82)

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The Test Statistic

Denote the observed counts by O and the expectedcounts by E.Define the chi-square (χ2) statistic to be

χ2 =∑

all cells

(O− E)2

E.

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The Value of the Test Statistic

Clearly, if all of the deviations O− E are small, then χ2

will be small.But if even a few the deviations O− E are large, then χ2

will be large.

Test ofGoodness of

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The Value of the Test Statistic

Now calculate χ2.

χ2 =(20− 27.36)2

27.36+

(24− 22.80)2

22.80+

(25− 18.24)2

18.24

+(23− 15.96)2

15.96+

(12− 14.82)2

14.82+

(10− 14.82)2

14.82= 1.9799 + 0.0632 + 2.5054

+3.1054 + 0.5366 + 1.5676

= 9.7581.

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Robb T.Koether

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Assignment

Compute the p-Value

The p-value is the likelihood of observing a χ2 value aslarge at 9.7581.To find that value, we need to know something aboutthe distribution of χ2.

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Fit

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Assignment

Chi-Square Degrees of Freedom

The χ2 distribution has an associated degrees offreedom, just like the t distribution.Each χ2 distribution has a slightly different shape,depending on the number of degrees of freedom.For example, we let χ2

5 denote the chi-square statisticwith 5 degrees of freedom.

Definition (χ2 degrees of freedom)

In a goodness-of-fit test, the number of degrees of freedomis one less than the number of cells.

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The Graph of χ21

2.5 5 7.5 10 12.5 15

0.05

0.1

0.15

0.2

0.25

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The Graph of χ22

2.5 5 7.5 10 12.5 15

0.05

0.1

0.15

0.2

0.25

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Assignment


The Graph of χ23

2.5 5 7.5 10 12.5 15

0.05

0.1

0.15

0.2

0.25

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Fit

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Assignment


The Graph of χ24

2.5 5 7.5 10 12.5 15

0.05

0.1

0.15

0.2

0.25

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Assignment


The Graph of χ25

2.5 5 7.5 10 12.5 15

0.05

0.1

0.15

0.2

0.25

Test ofGoodness of

Fit

Robb T.Koether

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Assignment


The Graph of χ26

2.5 5 7.5 10 12.5 15

0.05

0.1

0.15

0.2

0.25

Test ofGoodness of

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Assignment


The Graph of χ27

2.5 5 7.5 10 12.5 15

0.05

0.1

0.15

0.2

0.25

Test ofGoodness of

Fit

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Assignment


The Graph of χ28

2.5 5 7.5 10 12.5 15

0.05

0.1

0.15

0.2

0.25

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Assignment

Properties of χ2

The chi-square distribution with df degrees of freedomhas the following properties.

χ2 ≥ 0.It is unimodal.It is skewed right (not symmetric!)µχ2 = df .σχ2 =

√2df .

If df is large, then χ2df is approximately normal with

mean df and standard deviation√

2df .

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Assignment

Chi-Square vs. Normal

The graph of χ28 vs. N(8, 4)

-5 5 10 15 20

0.02

0.04

0.06

0.08

0.1

0.12

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10 20 30 40 50 60

0.01

0.02

0.03

0.04

0.05

0.06

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80 100 120 140 160 180

0.005

0.01

0.015

0.02

0.025

0.03

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Assignment



450 500 550 600

0.002

0.004

0.006

0.008

0.01

0.012

0.014

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Assignment

TI-83 - Chi-Square Probabilities

TI-83 Chi-square ProbabilitiesPress 2nd DISTR.Select χ2cdf.Enter the lower endpoint, the upper endpoint, and thedegrees of freedom.Press ENTER. The probability appears in the display.

Test ofGoodness of

Fit

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Assignment

TI-83 - Chi-Square Probabilities

Practice

Find P(χ2 > 6) with df = 3.Find P(20 < χ2 < 30) with df = 25.Find P(χ2 < 10) with df = 6.

Test ofGoodness of

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The Goodness-of-Fit Test

In our example, we found χ2 = 9.7581.There are 6 categories (colors), so there are 5 degreesof freedom.The p-value is

χ2cdf(9.7581,E99,5) = 0.0823.

That is greater than α, so we accept H0.We conclude that the colors fit the distribution given bythe Mars Candy Company.

Test ofGoodness of

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Robb T.Koether

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The Goodness-of-Fit Test

Example (The Goodness-of-Fit Test)(1) H0 : p1 = 0.24, p2 = 0.20, p3 = 0.16, p4 = 0.14,

p5 = 0.13, p6 = 0.13H1 : H0 is not true.

(2) α = 0.05.

(3) χ2 =∑

all cells(O−E)2

E .

(4)Color Blue Orange Green Yellow Brown RedObserved 20 24 25 23 12 10(Expected) (27.36) (22.80) (18.24) (15.96) (14.82) (14.82)

χ2 = 9.7581.(5) p-value = χ2cdf(9.7581,E99,5) = 0.0823.(6) Accept H0.(7) The color distribution in plain M&Ms is what the Mars

Candy Company advertises it is.

Test ofGoodness of

Fit

Robb T.Koether

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Assignment

Goodness-of-Fit Test on the TI-83

Be careful when using the TI-83!There is a function called χ2-Test, but it does notperform this test.Be careful! On the TI-83 there is a function calledχ2-Test, but it does not perform this test.Some TI-84s have a GOF-Test function.The GOF-Test function does perform this test.

Test ofGoodness of

Fit

Robb T.Koether

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Assignment

Goodness-of-Fit Test on the TI-83

TI-83 Goodness-of-fit testPut the observed counts in list L1.Put the hypothetical proportions in list L2.Multiply L2 by the sample size and store as L2. Theseare the expected counts.Calculate (L1-L2)2/L2.Go to LIST > MATH and select sum (item #5).Enter Ans and press ENTER. The value of χ2 appears.Then use χ2cdf to find the p-value.

Test ofGoodness of

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Example - Male vs. Female Births

Example (TI-83 Goodness-of-fit test)Suppose we observe 1000 births and find that 520 aremale and 480 are female.Does this indicate that male births and female birthsare not equally likely?

Test ofGoodness of

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Assignment


Example (TI-83 Goodness-of-fit test)(1) Let p1 = proportion of male births.

Let p2 = proportion of female births.H0 : p1 = 0.50, p2 = 0.50H1 : H0 is not true.

(2) α = 0.05.(3) The test statistic is

χ2 =∑

all cells

(O− E)2

E.

Test ofGoodness of

Fit

Robb T.Koether

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Assignment


Example (TI-83 Goodness-of-fit test)(4) We have the table

Male FemaleObserved 520 480(Expected) (500) (500)

Calculate

χ2 =(520− 500)2

500+

(480− 500)2

500= 0.8 + 0.8

= 1.6

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Assignment


Example (TI-83 Goodness-of-fit test)(5) The p-value is

p-value = χ2cdf(1.6,E99,1) = 0.2059.

(6) Accept H0.(7) The proportion of male births is 50%.

Test ofGoodness of

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Robb T.Koether

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Assignment


Perform the above test as a two-tailed one-proportion ztest.That is, let the alternative hypothesis be

H1 : p1 6= p2.

What is the p-value?What is the value of the test statistic z?Square that number. What do you get?

Test ofGoodness of

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Robb T.Koether

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Assignment

Assignment

HomeworkRead Sections 14.1 - 14.3, pages 921 - 935.Let’s Do It! 14.2, 14.3.Exercises 6 - 11, 14, 15, page 935.

test of goodness of fit - lecture 42 section...

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