Stats chapter 14

Chapter 14

Inference for Distributions of Categorical Variables: Chi-Square Procedures

14.1 TEST FOR GOODNESS OF FIT

The problem

• Suppose we open a bag of M&M’s and count the number of M&M’s of each color.

• How would we know if our color counts are at normal levels?

• How would we know if our color counts were abnormal?

Chi-Square Distribution

• When we want to test the proportion of many counts (i.e. a two-way table or an array), we need to use a new distribution-

• The Chi-Square Distribution (Chi = = “KAI”)• As you might suspect, this is another (the last

of the year) PHANTOMS procedure.• The 2 distribution is found at table D and the

[2nd] -> [Vars] (DIsT) menu on your calculator

Chi-Square Distribution

• When we want to test the proportion of many counts (i.e. a two-way table or an array), we need to use a new distribution-

• The Chi-Square Distribution (Chi = = “KAI”)• As you might suspect, this is another (the last

of the year) PHANTOMS procedure.• The 2 distribution is found at table D and the

[2nd] -> [Vars] (DIsT) menu on your calculator

The 2 distribution

• Like the t-distribution, the 2 distribution is variable. i.e. the distribution also has degrees of freedom.

• It is single peaked, right skewed.• As the df increases, the peak decreases in

height, moves to the right and becomes more symmetric/Normal.

• As df increases, the 2 statistic needed for statistically significant results also increases

The 2 distribution

Chi-Square Goodness of Fit

• When we want to check whether a distribution fits a hypothesized distribution, we use the “2 goodness of fit test”

• This is procedure is frequently used to see if a distribution is not in equal proportions

• No, this will not be much different than what we have already been doing for the last 3 chapters.

2 GOF Test

ParameterUnlike previous tests, you will not need to state a or a p.You need to state where the distribution come from.EXWe are investigating the proportions of all 15 oz. bags ofchocolate M&M’s of M&M’s

2 GOF Test

HypothesesThere are two styles for stating hypothesis

Style 1In this style, you will refer to a written table-or- state that all proportions are “equal”H0: the proportions of M&M’s are the same as the table providedHa: at least one color count is different than the table

H0: the proportions of accidents for each day is equalHa: at least one day has a count that is not equal

2 GOF Test

Hypotheses (cont.)Style 2In this style, you will write out the expected proportionsH0: pred = pblue = pyel = pbrn = pgrn = porg = 1/6Ha: at least one probability is different that stated above.

2 GOF Test

Hypotheses (cont)Notice that the alternative hypothesis in each case is that at least one proportion is different than hypothesized

2 GOF Test

Assumptions1. All expected cell counts are greater than 12. No more than 20% of the cell counts is less than 5(that’s a whole lot easier, yeah?)

Name of the Test“2 Goodness Of Fit Test”

2 GOF Test

Test StatisticObserved Count (O) is the count for each cell that we observed. The sum of each observed count is ‘n’

Expected Count (E) is the expected frequency of each cell times the sample size ‘n’

2 GOF Test

Test Statistic (cont)If we opened up a bag of M&M’s and found the following count:

Red Blue Brwn Yel Grn Orng

O : 5 3 10 6 4 3 n = 31E: 5.17 5.17 5.17 5.17 5.17 5.17

Note: expected counts are all equal to 31/6We are testing to see if M&M’s come in equal proportions

2 GOF Test

Test Statistic (cont)The test statistic is 2 (“kai squared”):

Degrees of freedom (df) = # of classes – 1

2 2 2 2

1 1 2 2 3 32

1 2 3

2

... n n

n

O E O E O E O E

E E E E

O E

E

2 GOF Test

Test Statistic (cont.)

2 2 2 2

2

2 2

5 5.17 3 5.17 10 5.17 6 5.17

5.17 5.17 5.17 5.17

4 5.17 3 5.17

5.17 5.17

2 6.739

6 1 5df

2 GOF Test

P Valuep val = P(2(df) > test statistic )on the calculator, [2nd] -> [VARS] (DIST) -> 2-cdfUsage: “2-cdf( lower, upper, df )

pval = P(2(5) > 6.739)

2 GOF Test

P Valuep val = P(2(df) > test statistic )on the calculator, [2nd] -> [VARS] (DIST) -> 2-cdfUsage: “2-cdf( lower, upper, df )

pval = P(2(5) > 6.739)

2 GOF Test

P Valuep val = P(2(df) > test statistic )on the calculator, [2nd] -> [VARS] (DIST) -> 2-cdfUsage: “2-cdf( lower, upper, df )

pval = P(2(5) > 6.739)

2 GOF Test

P Valuep val = P(2(df) > test statistic )on the calculator, [2nd] -> [VARS] (DIST) -> 2-cdfUsage: “2-cdf( lower, upper, df )

pval = P(2(5) > 6.739)pval = 0.2409

2 GOF Test

DecisionSimilarly to the other tests, reject the null hypothesis when the p-value is below the accepted level

SummaryUse the same 3 part summary:1) Interpret the p value w.r.t. sampling distribution2) Make decision with reference to an alpha level3) Summarize the results in context of the problem

2 GOF Test

Summary (cont.)“The given proportions in a sample of 31 would appear in approximately 24% of all random samples.”“Because this p value is greater than any acceptable alpha levels, we fail to reject the null hypothesis.”“We do not have sufficient evidence to conclude that the color distribution in M&M’s is not equally distributed”

Calculator methods

TI83/84

Calculator methods

TI83/84Begin by storing the observed counts in “L1”Store the expected counts in “L2”

Calculator methods

TI83/84Begin by storing the observed counts in “L1”Store the expected counts in “L2”

Calculator methods

TI83/84Begin by storing the observed counts in “L1”Store the expected counts in “L2”From the Home Screen evaluate:“sum((L1 – L2)2/L2)”

Calculator methods

TI83/84Begin by storing the observed counts in “L1”Store the expected counts in “L2”From the Home Screen evaluate:“sum((L1 – L2)2/L2)”

Calculator methods

TI83/84Begin by storing the observed counts in “L1”Store the expected counts in “L2”From the Home Screen evaluate:“sum((L1 – L2)2/L2)”

Calculator methods

TI83/84Begin by storing the observed counts in “L1”Store the expected counts in “L2”From the Home Screen evaluate:“sum((L1 – L2)2/L2)”This is the value of 2.

Calculator methods

TI83/84Begin by storing the observed counts in “L1”Store the expected counts in “L2”From the Home Screen evaluate:“sum((L1 – L2)2/L2)”This is the value of 2.Use the 2-cdf from the “Dist Menu” to find p-value“2-cdf (lower, upper, df)

Calculator methods

TI83/84Begin by storing the observed counts in “L1”Store the expected counts in “L2”From the Home Screen evaluate:“sum((L1 – L2)2/L2)”This is the value of 2.Use the 2-cdf from the “Dist Menu” to find p-value“2-cdf (lower, upper, df)

14.2 INFERENCE FOR TWO-WAY TABLES

Comparing two-groupsWine No Music French Music Italian Music Total

French 30 39 30 99

Italian 11 1 19 31

Other 43 35 35 113

Total 84 75 84 243

•The table above compares the background music with the # of bottles of wine purchased.•Not that information is presented in a two-way table with marginal distributions•Is there a relationship between these two categorical variables??

Comparing two-groups

• The test for relationship presented in the preceding page is a 2 test.

• In particular, this is a 2 test for homogeneity. It measures whether any one expected cell count is drastically different than the observed cell count.

Expected cell count for 2-way tables

column totalExpected Cell count row total

total

Expected cell count for 2-way tables

column totalExpected Cell count row total

total

% of population that are in the column

Expected cell count for 2-way tables

column totalExpected Cell count row total

total

Count of cell if the rows “obeyed”the column percentages

Expected cell count for 2-way tables

column totalExpected Cell count row total

total

Even for a small table, these calculations get cumbersome

Expected CountsWine No Music French Music Italian Music Total

French 39 30

Italian 11 1 19 31

Other 43 35 35 113

Total 75 84

30

84 243

99

Expected =x Column TotalRow total

Total

Expected CountsWine No Music French Music Italian Music Total

French 39 30

Italian 11 1 19 31

Other 43 35 35 113

Total 75 84

30

84 243

99

99Expected =

x 84

243

Expected CountsWine No Music French Music Italian Music Total

French 39 30

Italian 11 1 19 31

Other 43 35 35 113

Total 75 84

30

84 243

99

99Expected =

x 84

243= 34.22

Expected CountsWine No Music French Music Italian Music Total

French 39 30

Italian 11 1 19 31

Other 43 35 35 113

Total 75 8484 243

99

99Expected =

x 84

243= 34.22

34.44

Expected CountsWine No Music French Music Italian Music Total

French 39 30

Italian 11 1 19 31

Other 43 35 35 113

Total 75 8484 243

99

99Expected =

x 84

243= 34.22

34.44

Expected CountsWine No Music French Music Italian Music Total

French 39 30

Italian 11 1 19 31

Other 43 35 35 113

Total 75 8484 243

99

99Expected =

x 84

243= 34.22

34.44

Let’s start with the PHANTOMS procedure

2 Test for Homogeneity

ParameterState where each proportion comes from and what each count represents

“We are investigating the proportions of customers in the store who purchase French, Italian or other wine while listening to French, Italian or other music.”

2 Test for Homogeneity

HypothesesThe null hypothesis is always “the distributions of (group A) are the same in all population of (group B)”The alternative hypothesis is always “the distribution of (group A) are not all the same

“H0: the distributions of wine types are the same in all populations of music typesHa: the distributions of wine types are not all the same”

2 Test for Homogeneity

Assumptions(1) No more than 20% of the expected cell counts are less than 5(2) All expected cell counts are > 1(3) In a 2 x 2 table, all expected counts are greater than 5

2 Test for Homogeneity

• “All expected cell counts are greater than 5”

Wine No Music French Music Italian Music Total

French 34.22 30.56 34.22 99

Italian 10.72 9.57 10.72 31

Other 39.06 34.88 39.06 113

Total 84 75 84 243

2 Test for Homogeneity

Test Statistic

2

2 O E

E

2 2 2

2 30 34.22 39 30.56 35 39.06...

34.22 30.56 39.06

2 18.279

# rows - 1 # columns - 1df

3 - 1 3 - 1 4df

2 Test for Homogeneity

P Value

Decision

2 2P Value = (test statistic)P df df

2P Value = 4 18.279P

2 Test for Homogeneity

P Value

Decision

2 2P Value = (test statistic)P df df

2P Value = 4 18.279P

2 Test for Homogeneity

P Value

DecisionReject null hypothesis

2 2P Value = (test statistic)P df df

2P Value = 4 18.279P

P Value =0.00109

2 Test for Homogeneity

SummaryApproximately 0.1% of the time, a random sample of 243 will produce the distribution given.Because the p value is less than an of 0.05, we will reject the null hypothesis.We have sufficient evidence at the 5% significance level to conclude that the distribution of wine types purchased is not the same in all music types.

Calculator Methods

Methods on the TI84

Calculator Methods

Methods on the TI84Before you begin the test, you must enter the “observed counts” into MATRIX [A][2ND] -> [x-1] (MATRIX) -> “EDIT” -> [1]

Calculator Methods

Methods on the TI84Before you begin the test, you must enter the “observed counts” into MATRIX [A][2ND] -> [x-1] (MATRIX) -> “EDIT” -> [1]

Calculator Methods

Methods on the TI84Before you begin the test, you must enter the “observed counts” into MATRIX [A][2ND] -> [x-1] (MATRIX) -> “EDIT” -> [1]Input the correct matrix size and cell counts(Use [ENTER] or the Cursor Keys to switch between fields.)

Calculator Methods

Methods on the TI84Before you begin the test, you must enter the “observed counts” into MATRIX [A][2ND] -> [x-1] (MATRIX) -> “EDIT” -> [1]Input the correct matrix size and cell counts(Use [ENTER] or the Cursor Keys to switch between fields.)

Calculator Methods

Methods on the TI84 (cont.)IMPORTANT: after inputting the observed matrix, quit and go to the home screen[STAT] -> “TESTS” -> “2 Test”

Calculator Methods

Methods on the TI84 (cont.)IMPORTANT: after inputting the observed matrix, quit and go to the home screen[STAT] -> “TESTS” -> “2 Test”

Calculator Methods

Methods on the TI84 (cont.)IMPORTANT: after inputting the observed matrix, quit and go to the home screen[STAT] -> “TESTS” -> “2 Test” Ensure that “Observed” is set to [A] and“Expected” is set to [B]“Calculate”

Calculator Methods

Methods on the TI84 (cont.)IMPORTANT: after inputting the observed matrix, quit and go to the home screen[STAT] -> “TESTS” -> “2 Test” Ensure that “Observed” is set to [A] and“Expected” is set to [B]“Calculate”

Calculator Methods

Methods on the TI84 (cont.)IMPORTANT: after inputting the observed matrix, quit and go to the home screen[STAT] -> “TESTS” -> “2 Test” Ensure that “Observed” is set to [A] and“Expected” is set to [B]“Calculate”

Calculator Methods

Methods on the TI84 (cont.)IMPORTANT: after inputting the observed matrix, quit and go to the home screen[STAT] -> “TESTS” -> “2 Test” Ensure that “Observed” is set to [A] and“Expected” is set to [B]“Calculate”The expected cell counts will be calculated and stored in Matrix [B] (go back to the Matrix menu to see the expected Counts)

2 Tests

• Occasionally, you will be asked to find the cell that “contributed the most to the 2 statistic.”

• When this is asked, you must calculate the 2 statistic by hand and find the largest value of(O – E)2 / E.

• This is usually the cell that differs the most from the expected count

• Since this is a percent calculation, it is not always predictable.

2 Test for Independence

• A similar test for two way tables is the “2 Test for Independence” sometimes called“2 Test for Association”

• This test is asks the question, “do the two variables influence each other?”

• When there is no association, the observed two-way table is close to the expected table

2 Test for Independence

This test really only differs from the test for homogeneity in the hypotheses and the conclusion.

HypothesesThe null hypothesis is “there is no association between (group 1) and (group 2)”The alternative hypothesis is “there is an association between (group 1) and (group 2)”

2 Test for Independence

ConclusionPhrase your conclusions similar to the ones we have been constructing.When failing to reject H0:After interpreting the p value and comparing the p value to alpha, state that there is “no evidence to conclude that an association exists between (group 1) and (group 2)”

Likewise, when rejecting H0, state that “there is sufficient evidence to conclude that an association exists between (group 1) and (group 2)”

Assignment 14.2

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