Practice Is there a significant ( =.01) relationship between opinions about the death penalty and...

Practice• Is there a significant ( = .01) relationship between

opinions about the death penalty and opinions about the legalization of marijuana?

• 933 Subjects responded yes or no to:• “Do you favor the death penalty for persons convicted of

murder?”

• “Do you think the use of marijuana should be made legal?”

Results

Yes No

Yes 152 561

No 61 159

Marijuana ?

Dea

th

Pen

alty

?

Step 1: State the Hypothesis

• H1: There is a relationship between opinions about the death penalty and the legalization of marijuana

• H0:Opinions about the death penalty and the legalization of marijuana are independent of each other

Step 2: Create the Data Table

Yes No Total

Yes 152 561 713

No 61 159 220

Total 213 720 933

Marijuana ?

Dea

th

Pen

alty

?

Step 3: Find 2 critical

• df = (R - 1)(C - 1)

• df = (2 - 1)(2 - 1) = 1 = .01

2 critical = 6.64

Step 4: Calculate the Expected Frequencies

Yes No Total

Yes 152(162.77)

561(550.23)

713

No 61(50.23)

159(169.78)

220

Total 213 720 933

Marijuana ?

Dea

th

Pen

alty

?

Step 5: Calculate 2

O E O - E (O - E)2 (O - E)2

E152 162.77 -10.77 115.99 .71

61 50.23 10.77 115.99 2.31

561 550.23 10.77 115.99 .21

159 169.78 -10.78 115.99 .682 = 3.91

Step 6: Decision

• Thus, if 2 > than 2critical

– Reject H0, and accept H1

• If 2 < or = to 2critical

– Fail to reject H0

Step 6: Decision

• Thus, if 2 > than 2critical

– Reject H0, and accept H1

• If If 22 < or = to < or = to 22criticalcritical

– Fail to reject HFail to reject H00

2 = 3.91

2 crit = 6.64

Step 7: Put it answer into words

• H0:Opinions about the death penalty and the legalization of marijuana are independent of each other

• A persons opinion about the death penalty is not significantly (p > .01) related with their opinion about the legalization of marijuana

Effect Size

• Chi-Square tests are null hypothesis tests

• Tells you nothing about the “size” of the effect

• Phi (Ø)– Can be interpreted as a correlation coefficient.

Phi

• Use with 2x2 tables

N

2 N = sample size

Practice• Is there a significant ( = .01) relationship between

opinions about the death penalty and opinions about the legalization of marijuana?

• 933 Subjects responded yes or no to:• “Do you favor the death penalty for persons convicted of

murder?”

• “Do you think the use of marijuana should be made legal?”

Results

Yes No

Yes 152 561

No 61 159

Marijuana ?

Dea

th

Pen

alty

?

Step 6: Decision

• Thus, if 2 > than 2critical

– Reject H0, and accept H1

• If If 22 < or = to < or = to 22criticalcritical

– Fail to reject HFail to reject H00

2 = 3.91

2 crit = 6.64

Phi

• Use with 2x2 tables

06.933

91.3

Bullied Example

Height Yes No Total

Short 42 50 92

Not short 30 87 117

Total 72 137 209

Ever Bullied

2

O E O - E (O - E)2 (O - E)2

E42 31.69 10.31 106.30 3.35

50 60.30 -10.30 106.09 1.76

30 40.30 -10.30 106.09 2.63

87 76.69 10.31 106.30 1.392 = 9.13

Phi

• Use with 2x2 tables

21.209

13.9

Practice

• Practice– Page 170 #6.10– How strong is the relationship?

Results

Remed Reg

No 22 (28.37)

187 (180.62)

209

Yes 61 (12.63)

159 (80.37)

93

41 261 302

English

AD

D

X2 = 5.38

X2 crit = 3.83

Phi

• Use with 2x2 tables

13.302

38.5

Practice

• In the 1930’s 650 boys participated in the Cambridge-Somerville Youth Study. Half of the participants were randomly assigned to a delinquency-prevention pogrom and the other half to a control group. At the end of the study, police records were examined for evidence of delinquency. In the prevention program 114 boys had a police record and in the control group 101 boys had a police record. Analyze the data and write a conclusion.

• Chi Square = 1.17

• Chi Square observed = 3.84

• Phi = .04– Note the results go in the opposite direction

that was expected!

2 as a test for goodness of fit

• But what if:

• You have a theory or hypothesis that the frequencies should occur in a particular manner?

Example

• M&Ms claim that of their candies:

• 30% are brown

• 20% are red

• 20% are yellow

• 10% are blue

• 10% are orange

• 10% are green

Example

• Based on genetic theory you hypothesize that in the population:

• 45% have brown eyes

• 35% have blue eyes

• 20% have another eye color

To solve you use the same basic steps as before (slightly

different order)• 1) State the hypothesis

• 2) Find 2 critical

• 3) Create data table

• 4) Calculate the expected frequencies

• 5) Calculate 2

• 6) Decision

• 7) Put answer into words

Example

• M&Ms claim that of their candies:

• 30% are brown• 20% are red• 20% are yellow• 10% are blue• 10% are orange• 10% are green

Example

• Four 1-pound bags of plain M&Ms are purchased

• Each M&Ms is counted and categorized according to its color

• Question: Is M&Ms “theory” about the colors of M&Ms correct?

Observed

Brown 602

Red 396

Yellow 379

Blue 227

Orange 242

Green 235

Total 2081

Step 1: State the Hypothesis

• H0: The data do fit the model

– i.e., the observed data does agree with M&M’s theory

• H1: The data do not fit the model

– i.e., the observed data does not agree with M&M’s theory

– NOTE: These are backwards from what you have done before

Step 2: Find 2 critical

• df = number of categories - 1

Step 2: Find 2 critical

• df = number of categories - 1

• df = 6 - 1 = 5 = .05

2 critical = 11.07

Observed

Brown 602

Red 396

Yellow 379

Blue 227

Orange 242

Green 235

Total 2081

Step 3: Create the data table

Observed ExpectedProp.

Brown 602 .30

Red 396 .20

Yellow 379 .20

Blue 227 .10

Orange 242 .10

Green 235 .10

Total 2081

Step 3: Create the data tableAdd the expected proportion of each category

Observed ExpectedProp.

Brown 602 .30

Red 396 .20

Yellow 379 .20

Blue 227 .10

Orange 242 .10

Green 235 .10

Total 2081

Step 4: Calculate the Expected Frequencies

Observed ExpectedProp.

ExpectedFreq

Brown 602 .30

Red 396 .20

Yellow 379 .20

Blue 227 .10

Orange 242 .10

Green 235 .10

Total 2081

Step 4: Calculate the Expected Frequencies

Expected Frequency = (proportion)(N)

Observed ExpectedProp.

ExpectedFreq

Brown 602 .30 624.30

Red 396 .20

Yellow 379 .20

Blue 227 .10

Orange 242 .10

Green 235 .10

Total 2081

Step 4: Calculate the Expected Frequencies

Expected Frequency = (.30)(2081) = 624.30

Observed ExpectedProp.

ExpectedFreq

Brown 602 .30 624.30

Red 396 .20 416.20

Yellow 379 .20

Blue 227 .10

Orange 242 .10

Green 235 .10

Total 2081

Step 4: Calculate the Expected Frequencies

Expected Frequency = (.20)(2081) = 416.20

Observed ExpectedProp.

ExpectedFreq

Brown 602 .30 624.30

Red 396 .20 416.20

Yellow 379 .20 416.20

Blue 227 .10

Orange 242 .10

Green 235 .10

Total 2081

Step 4: Calculate the Expected Frequencies

Expected Frequency = (.20)(2081) = 416.20

Observed ExpectedProp.

ExpectedFreq

Brown 602 .30 624.30

Red 396 .20 416.20

Yellow 379 .20 416.20

Blue 227 .10 208.10

Orange 242 .10 208.10

Green 235 .10 208.10

Total 2081

Step 4: Calculate the Expected Frequencies

Expected Frequency = (.10)(2081) = 208.10

Step 5: Calculate 2

O = observed frequency

E = expected frequency

2

O E O - E (O - E)2 (O - E)2

E

2

O E O - E (O - E)2 (O - E)2

E602 624.30

396 416.20

379 416.20

227 208.10

242 208.10

235 208.10

2

O E O - E (O - E)2 (O - E)2

E602 624.30 -22.3

396 416.20 -20.2

379 416.20 -37.2

227 208.10 18.9

242 208.10 33.9

235 208.10 26.9

2

O E O - E (O - E)2 (O - E)2

E602 624.30 -22.3 497.29

396 416.20 -20.2 408.04

379 416.20 -37.2 1383.84

227 208.10 18.9 357.21

242 208.10 33.9 1149.21

235 208.10 26.9 723.61

2

O E O - E (O - E)2 (O - E)2

E602 624.30 -22.3 497.29 .80

396 416.20 -20.2 408.04 .98

379 416.20 -37.2 1383.84 3.32

227 208.10 18.9 357.21 1.72

242 208.10 33.9 1149.21 5.22

235 208.10 26.9 723.61 3.48

2

O E O - E (O - E)2 (O - E)2

E602 624.30 -22.3 497.29 .80

396 416.20 -20.2 408.04 .98

379 416.20 -37.2 1383.84 3.32

227 208.10 18.9 357.21 1.72

242 208.10 33.9 1149.21 5.22

235 208.10 26.9 723.61 3.48

15.52

Step 6: Decision

• Thus, if 2 > than 2critical

– Reject H0, and accept H1

• If 2 < or = to 2critical

– Fail to reject H0

Step 6: Decision

• Thus, if Thus, if 22 > than > than 22criticalcritical

– Reject HReject H00, and accept H, and accept H11

• If 2 < or = to 2critical

– Fail to reject H0

2 = 15.52

2 crit = 11.07

Step 7: Put it answer into words

• H1: The data do not fit the model

• M&M’s color “theory” did not significantly (.05) fit the data

Practice

• Among women in the general population under the age of 40:

• 60% are married• 23% are single• 4% are separated• 12% are divorced• 1% are widowed

Practice

• You sample 200 female executives under the age of 40

• Question: Is marital status distributed the same way in the population of female executives as in the general population ( = .05)?

Observed

Married 100

Single 44

Separated 16

Divorced 36

Widowed 4

Total 200

Step 1: State the Hypothesis

• H0: The data do fit the model

– i.e., marital status is distributed the same way in the population of female executives as in the general population

• H1: The data do not fit the model

– i.e., marital status is not distributed the same way in the population of female executives as in the general population

Step 2: Find 2 critical

• df = number of categories - 1

Step 2: Find 2 critical

• df = number of categories - 1

• df = 5 - 1 = 4 = .05

2 critical = 9.49

Step 3: Create the data table

Observed ExpectedProp.

Married 100 .60

Single 44 .23

Separated 16 .04

Divorced 36 .12

Widowed 4 .01

Total 200

Step 4: Calculate the Expected Frequencies

Observed ExpectedProp.

ExpectedFreq.

Married 100 .60 120

Single 44 .23 46

Separated 16 .04 8

Divorced 36 .12 24

Widowed 4 .01 2

Total 200

Step 5: Calculate 2

O = observed frequency

E = expected frequency

2

O E O - E (O - E)2 (O - E)2

E100 120 -20 400 3.33

44 46 -2 4 .09

16 8 8 64 8

36 24 12 144 6

4 2 2 4 2

19.42

Step 6: Decision

• Thus, if 2 > than 2critical

– Reject H0, and accept H1

• If 2 < or = to 2critical

– Fail to reject H0

Step 6: Decision

• Thus, if Thus, if 22 > than > than 22criticalcritical

– Reject HReject H00, and accept H, and accept H11

• If 2 < or = to 2critical

– Fail to reject H0

2 = 19.42

2 crit = 9.49

Step 7: Put it answer into words

• H1: The data do not fit the model

• Marital status is not distributed the same way in the population of female executives as in the general population ( = .05)

Practice

• In the past you have had a 20% success rate at getting someone to accept a date from you.

• What is the probability that at least 2 of the next 10 people you ask out will accept?

Practice

• p zero will accept = .11

• p one will accept = .27

• p zero OR one will accept = .38

• p two or more will accept = 1 - .38 = .62

Practice

• IQ– Mean = 100– SD = 15

• What is the probability that the stranger you just bumped into on the street has an IQ between 95 and 110?

Step 1: Sketch out question

-3 -2 -1 1 2 3

11095

?

Step 2: Calculate Z scores for both values

• Z = (X - ) /

• Z = (95 - 100) / 15 = -.33

• Z = (110 - 100) / 15 = .67

Step 3: Look up Z scores

-3 -2 -1 1 2 3

.67-.33

.1293 .2486

Step 4: Add together the two values

-3 -2 -1 1 2 3

.67-.33

.3779

Practice

• A professor would like to determine if there has been a change in grading practices over the years. In the past, the overall grade distribution was 14% As, 26% Bs, 31% Cs, 19% Ds, and 10% Fs.

• A sample of 200 students this years had the following grades

Practice

• A = 32• B = 61• C = 64• D = 31• F = 12

• Do the data indicate a significant change in the grade distribution? Test at the .05 level.

Step 1: State the Hypothesis

• H0: The data do fit the model

– i.e., the grades are distributed the same

• H1: The data do not fit the model

– i.e., the grades are not distributed the same

Practice

• A = 32 28• B = 61 52• C = 64 62• D = 31 38• F = 12 20

• Chi square = 6.68• Critical Chi square (4) = 9.49

Step 6: Decision

• Thus, if 2 > than 2critical

– Reject H0, and accept H1

• If If 22 < or = to < or = to 22criticalcritical

– Fail to reject HFail to reject H00

2 = 6.68

2 crit = 9.49

Step 7

• H0: The data do fit the model

– i.e., the grades are distributed the same

• There is no evidence that the grades have changed

Practice

• An early hypothesis of schizophrenia was that it has a simple genetic cause. In accordance with the theory 25% of the offspring of a selected group of parents would be expected to be diagnosed as schizophrenic. Suppose that of 140 offspring, 19.3% were schizophrenic. Test this theory.

• Goodness of fit chi-square

• Make sure you compute the Chi square with the frequencies.

• Chi square = 2.439• Observed = 3.84

• These data are consistent with the theory!

Practice

• In 1693, Samuel Pepys asked Isaac Newton whether it is more likely to get at least one ace in 6 rolls of a die or at least two aces in 12 rolls of a die. This problems is known a Pepys' problem.

Binomial Distribution

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0 1 2 3 4 5 6Aces

p

p = .67

Binomial Distribution

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0 1 2 3 4 5 6 7 8 9 10 11 12Aces

p

p = .62

Practice

• In 1693, Samuel Pepys asked Isaac Newton whether it is more likely to get at least one ace in 6 rolls of a die or at least two aces in 12 rolls of a die. This problems is known a Pepys' problem.

• It is more likely to get at least one ace in 6 rolls of a die!

Practice

• Which is more likely: at least one ace with 4 throws of a fair die or at least one double ace in 24 throws of two fair dice? This is known as DeMere's problem, named after Chevalier De Mere.

• Blaise Pascal later solved this problem. 

Binomial Distribution

)04(0 8333.1667.)!04(!0

!4)(

Xp

p = .482 of zero aces

1 - .482 = .518 at least one ace will occur

Binomial Distribution

)024(0 9722.0278.)!024(!0

!24)(

Xp

p = .508 of zero double aces

1 - .508 = .492 at least one double ace will occur

Practice

• Which is more likely: at least one ace with 4 throws of a fair die or at least one double ace in 24 throws of two fair dice? This is known as DeMere's problem, named after Chevalier De Mere.

• More likely at least one ace with 4 throws will occur