Transcript Document
CHAPTER 14
•
Chi- square goodness of fit Test “GOF”
•
Chi-square Test for Homogeneity
•
Chi-square Test for Independence DONE SAME WAY
•
Symbol for Chi-Square is
χ
2
•
What’s the difference?
Goodness of Fit…..how well does the observed data match the expected….2 rows or 2 columns • • Homogeneity…..More than one sample is taken with one categorical variable in mind
(2+ Samples, 1 category)
• • Independence/Association…..Only one sample is taken and there are two or more categories.
(1 sample, 2+ categories)
Chi-Square Curve
It is not a NORMAL CURVE!!!
It is always skewed to the right some
Is your die fair—one more time.
Roll your die 60 times.
Write down the number for every roll.
Chi-Square GOF Test
• If your die is fair you would expect to get 10 of each number in 60 rolls.
• In this test we compare the EXPECTED results vs the OBSERVED results.
Hypotheses
• Ho: The proportion of each number that occurs on my die is 1/6 • Ha: The proportion of each number that occurs on my die is different than 1/6 • There are no symbols for the chi-square. However, it is always one-sided, even though the word “ different ” is used.
Normal condition for
χ
2
• 80% of the expected cells are greater than or equal to 5.
(not observed cells —expected cells!)
Formula
Degrees of Freedom (df)
• For all chi-square tests use the following: • df = (r – 1)(c – 1) • r is the row and c is the column
Calculator steps
TI-83+ Calculator Put your observed counts in L1 and Expected in L2 Then hit 2 nd and Stat Tyoe the observed in L1 and expected in L2. Then click on the L3 heading and type the formula(then click enter), then quit out to the main screen Find the sum of L3, your answer is the chi-square statistic
Calculator steps
After your get the sum you need to obtain the p-value X 2 , UB , df This will give you the p-value
Calculator steps
TI-84 calculator does most of the work for you Make sure you have typed your observed counts in L1 and expected counts in L2 5
Demographics.
• Rancho is approximately 53.6% Hispanic, 29.2% Asian, 12.7% white and 4.5% other.
(data as of 2012-2013 school year)
• Does Mr. Pines ’ AP stats classes reflect this diversity? Run the appropriate test, verify your requirements, and write a conclusion.
2015
Demographics.
H o : The diversity in Mr. Pines classes is the same as Rancho ’s diversity H a : The diversity in Mr. Pines classes is different than Rancho ’s diversity
Assumptions
: We have an independent random sample of 159 students ethnicity. We can assume that there have been at least 1590 students in Mr. Pines classes. 100% of our expected cells are 5 or more.
Chi-square GOF test
X
2 = 12.38
P-value =.0062
df = 3
This p-value is low enough to reject Ho at the 1% level This is strong evidence to suggest that Mr. Pines class diversity may be different than Rancho ’s diversity.
Is there a difference……
• Do boys and girls prefer types of social media?
• Please choose your favorite of the three below.
One Hundred Fifty Seven students were surveyed………….
We took a sample of 83 girls and a sample of 74 boys.
Girls Boys Instagram
36 19
Snapchat
25 36
22 19
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What’s the difference?
Goodness of Fit…..how well does the observed data match the expected….2 rows or 2 columns • • Homogeneity…..More than one sample is taken with one categorical variable in mind
(2+ Samples, 1 category)
• • Independence/Association…..Only one sample is taken and there are two or more categories.
(1 sample, 2+ categories)
One Hundred Fifty Seven students were surveyed………….
Girls Boys Instagram
36 (29.076) 19 (25.924)
Snapchat
25 (32.248) 36 (28.752)
22 (21.675) 19 (19.325)
Hypotheses for Chi-Square Test for Homogeneity
H o : There is no difference between gender and social media preference H a : There is a difference between gender and social media preference OR H o : The proportions of boys and girls who prefer each type of social media are the same H a : The proportions of boys and girls who prefer each type of social media are different
H o : There is no difference between gender and social media preference H a : There is a difference between gender and social media preference Assumptions
: We have two independent random samples of students social media preferences(83 girls and 74 boys). There are obviously more than 830 girls and 740 boys in the population sampled from who use social media. 100% of our expected cells are 5 or more.
Chi-square test of homogeneity
X
2 = 6.96
P-value =.0307
df = 2
This p-value is low enough to reject Ho at the 5% level This is evidence to suggest that there may be a difference between gender and social media preference.
Period 1
Period 3
Referrals vs Days of week
Monday
12
Tuesday
5
Wednesday
9
Thursday
4
Friday
15 The table shows the number of students referred for disciplinary reasons to the principals office, broken down by day of the week.
Are referrals related to the day of the week?
BIRTHDAYS.
• Are Mr. Pines students birth months distributed in proportion to the number of days in each month?
• We can run a chi-square GOF based on the # of days in each month n = 153
C/O 2015 per 1 only H 0: Mr. Pines students birthday months are in proportion to the number of days in each month.
H a : Mr. Pines students birthday months are different than the proportion of the number of days in each month.
We have an independent sample of 53 students birth months. All births is obviously more than 10x our sample. 100% of our expected counts are 5 or more.
X
2 = 9.81
Chi-square GOF Test This p-value is too high to reject H o P-value = .5475
df = 11 Based on this sample, there is NOT enough evidence to suggest that Mr. Pines students birthday months are different than the proportion of the number of days in each month.
11 12 12 17
O
9 7 12 11 14 16 20 12
E
n = 153
BIRTHDAYS
To figure out the expected we need to think about the number of days in each month.
Jan 31 July 31 Feb 28 Aug 31 Mar 31 Sep 30 Apr 30 Oct 31 May 31 Nov 30 June 30 Dec 31 Total = 365 days
C/O 2014 H 0: Births are uniformly distributed by the # of days in each month H a : Births are not uniformly distributed by the # of days in each month CONDITIONS df = 11 We have an independent sample of 129 students birth months. All births is obviously more than 10x our sample. 100% of our expected counts are 5 or more.
X
2 = 7.75
P = .7355
This p-value is too high to reject H o .
There is not enough evidence to suggest that births are not uniformly distributed by the # of days in each month.
O E
12 10.956
5 9.896
12 10.956
12 10.603
13 10.956
7 10.603
9 10.956
14 10.956
11 10.603
11 10.956
8 10.603
15 10.956
Chi-Square Test for Homogeneity
• Data is given in a 2-way table • Expected counts are found by using a matrix on your calculator or by multiplying the (ROW TOTAL)(COLUMN TOTAL)/GRAND TOTAL • Conditions and df are the same as GOF test
Is there a difference……
• Do boys and girls prefer different video game consoles?
• Please choose your favorite console out of the 3?
Hypotheses for Chi-Square Test for Homogeneity
Remember Ha is always means different!
H o : There is no difference between gender and video game console preference H a : There is a difference between gender and video game preference OR H o : The proportions of boys and girls who prefer each type of console are the same H a : The proportions of boys and girls who prefer each type of console are different
Why is this a Homogeneity Test?
• Two samples were taken separately • Boys console preference • Girls console preference • There is ONE category of interest.
Two-Way Table
2013 G E N D E R
Girls Boys Total VIDEO GAME CONSOLE PREFERENCE Wii Xbox 360 PS3 27 3 30 16 33 49 20 32 52 63 68 131
Two-Way Table
2014 G E N D E R
Girls Boys Total VIDEO GAME CONSOLE PREFERENCE Wii Xbox Playstation 19 6 25 10 21 31 32 34 66 61 61 122
Hit 2 nd Matrix, go to EDIT Quit to home screen, go to test menu Set the appropriate matrix size Should be already setup if you used A and B Enter observed counts in matrix
You need the expected counts….so go back to 2 nd matrix. Use NAMES and go down to Matrix B, calculator generates them after you run the test You will most likely have to scroll to the right to see all of the expected counts
REMEMBER!....EXPECTED COUNTS MUST BE ON YOUR PAPER!
College Students’ Drinking
In 1987, a random sample of undergraduate students at Rutgers University was sent a questionnaire that asked about their alcohol drinking habits. Here are the results displayed in a two-way table.
Live on Campus Total
Abstain Light Drinker Light Moderate Moderate Heavy Heavy Total 46 71 55 63 67 302
Live Off Campus(Not with Parents)
17 38 34 24 28 141
Live Off Campus with Parents
43 42 26 15 17 143 106 151 115 102 112 586
Chi-Square Test for Independence
There was one sample taken and then data was broken down into different categories.
When only one sample is taken we are doing a Chi-Square Test for Independence/Association
Hypotheses
This is a chi-square test for Independence/Association, you have a few options for writing the hypotheses
H o : There is no association between students ’ residence type and drinking habits.
H a There is an association between students ’ residence type and drinking habits.
OR
H o : Student drinking habits and residence type are independent H a Student drinking habits and residence type are not independent
Full Moon
Some people believe that a full moon elicits unusual behavior in people. The table shows the number of arrests made in a small town during the weeks of six full moons and six other randomly selected weeks in the same year. Is there evidence of a difference in the types of illegal activity that takes place.
This is a chi-square test for Homogeneity
Violent (murder,assault,rape, etc.) Property(burglary, vandalism, etc.) Drugs/Alcohol Domestic Abuse Other offenses
Full Moon
2 17 27 11 9
Not Full
3 21 19 14 6
Thanks To: Grace Montgomery
THANKS TO: Amy Nguyen
Testing M&M
’
s
• The Mars company has always claimed that the color distribution of their M&M ’ s follow a certain proportion as follows:
Brown
13%
Red
13%
Yellow
14%
Green
16%
Orange
20%
Blue
24% Check the M&M ’ s that were given to you. How many of each color do you have? We will run a Chi-Square GOF test to see if their claim is accurate.
Do not eat your M&M ’ s until we have all observed and expected counts completed!
Hypotheses
• Ho: My bag of M&M ’ s follow the same color distribution as the Mars company claim.
• Ha: My bag of M&M ’ s follows a different color distribution as the Mars company claim.
Assumptions/Conditi ons
• ___% of expected counts >5 • My bag of M&M ’ s can be considered an independent random sample of M&M ’ s
Colors
Claim % Expected Observed
M&M Combined Results
Brown
13% 728.78
606
Red
13% 728.78
586
Yellow
14% 784.84
803
Green
16% 896.96
1101
Orange
20% 1121.2
1335
Blue
24% 1345.44
1175
There were a total of 5606 M&M
’
s sampled.
We have a chi-square statistic of 157.85 which gives a P-value of 0.
Mr. Pines Poker Chips
• • • 44 white chips = 20pts 5 blue chips = 30pts 1 red chip = 50pts
Mr. Pines Poker Chips
• • • 44 white chips = 20pts 5 blue chips = 30pts 1 red chip = 50pts There have been 135 attempts at randomly choosing poker chips out of the bag. A White chip has been pulled 301 times, a Blue chip 47 times, and the Red chip 16 times. Has this followed the expected probabilities? Run a chi-square GOF Test.
Baseball Bats
• There have been some major bat changes for the 2011 season. Aluminum baseball bats have been regulated so that they meet certain safety standards. After 5 games this season, coach Pines has noticed significant reductions in power numbers such as 2B ’ s, 3B ’ s, and HR ’ s…..Of course he would like to test his hypothesis.
Baseball Bats
• Run a Chi-Squared two-way table test to see if there is an association between the power numbers and types of bats.
• Also run a 2-Prop. Z Test between the types of bats used.
• If these are done correctly, Z 2 = X 2
Hypotheses
• Ho: There is no association between type of bats and extra base hits • Ha: There is an association between type of bats and extra base hits
Assumptions/Conditi ons
• E---All expected counts > 5 • S----We have a random sample of 23 schools hitting stats for the first 5 games of the 2010 and 2011 baseball seasons • I----We can assume that all stats are independent of other teams stats
Observed and Expected Counts
Singles Extra Base Hits
2010(BESR Bats)
672 (695.26) 313 (289.74)
2011(BBCOR Bats)
703 (679.74) 260 (283.26)
•
X
2 = 5.35
•
P-value = .0207
• This p-value is low enough to reject at the 5% level.
• There is evidence to suggest that there may be an association between the types of bats and extra base hits
Tootsie Pop Wrappers
We are interested in whether or not the designs on the wrappings on Tootsie Roll Pops are independent of the flavor of the pop.
Hypotheses
This is a chi-square test for Independence/Association, you have a few options for writing the hypotheses
H o : There is no association between pop flavor and designs on the wrapper.
H a There is an association between pop flavor and designs on the wrapper.
OR
H o : Pop flavor and wrapper designs are independent.
H a Pop flavor and wrapper designs are not independent.