STAT 104 - Chapter 6: Two-way tables, Chapter 25: Chi-Squared Tests

1. A table where individuals have been classified according to two variables.

The variables may be categorical or quantitative.

• 2. Row: Horizontal
•     Column: Vertical

3. Make an extra row and an extra column for the totals. Then total each column and row up. Also, get a final total of all individuals in the table to put on the bottom right corner.

1. The row totals and column totals in a two-way table give the marginal distribution of the two individual variables. 

It is clearer to present these distributions as percents of the table total. 

Marginal distribution tells us nothing about the relationship between the variables. 

2. There are two sets of conditional distributions in a two-way table. The distribution of the row variable for each fixed value of the column variable, and vice versa. 

To find the conditional distribution of the row variable for one specific value of the column variable, look only at that one column in the table. Find each entry in the column as a percent of the column total.

There is a separate conditional distribution for each value of the other variable

• 3a. Add up all the people in the table.
• 736 + 450 + 193 + 205 + 144 + 80 = 1808
• This table describes 1808 people.

• 736 + 450 + 193 = 1379
• 1379 people played video games

3b. A's and B's: 736 + 205 / 1808 = 0.5205 x 100 = 52.05%

C's: 450 + 144 / 1808 = 0.3285 x 100 = 32.85%

D's and F's: 193 + 80 / 1808 = 0.151 x 100 = 15.1%

• The complete marginal distribution for grades is:
• 

• 4. There are 1379 players (736 + 450 + 193). 
• Of these, 53.37% earned A's and B's. (736/1379) x 100

C's: 450 / 1379 x 100 = 32.63%

D's and F's: 193 / 1379 x 100 = 14.00

Now do the same process for non-games

There are 429 non-gamers (205 + 144 + 80)

• Of these, 47.79% earned A's and B's
• (205/429) x 100
• 

• Make a Label column for your response variable and put the categories in each row. Then make separate columns for the different categories of your explanatory variable, and put the data into the rows like this.
• 
• Choose 'Values from a table', then choose 'two-way table cluster'
• 
• Put your explanatory variables (all categories) into the 'graph variables' box.
• The put your response variable into the 'row labels' box.

• **Change 'table arrangement' to 'rows are outermost categories and columns are innermost'
• 
• Click ok and you'll get your graph
• 

Chi-squared test of independence (also known as the chi-squared test of association) is used to find evidence for a relationship (dependence, connection, link etc.) between two variables.

*The other test is a chi-squared test of homogeneity, but we don't touch on it in this course.

1a. The sample is representative of the population

1b. All expected counts are at least 5. A few isolated expected counts below 5 do not matter, provided that they are not all in one row or all in one column. 

*Need to check on Minitab to find out*

2. Observed counts: 736, 450, 193, 205, 144 and 80

Expected counts: 717.7, 453.1, 208.2, 223.3, 140.9 and 64.8

• 1. Observed counts are simply the numbers from a sample that are put into a table.
• 

2. The sample of teenaged boys represents the population of all teenaged boys at all similar schools.

• STATE: 
• How strong is the evidence that grades depend on gaming status

PLAN: *NO PARAMETER TO WORRY ABOUT

• Ho: Grades do not depend on gaming status
• Ha: Grades depend on gaming status

We'll use a χ² test of association because we're looking for a relationship between two variables. 

• SOLVE: 
• State the Conditions
• 1. Representative sample
• 2. All expected counts are at least 5. A few isolated expected counts below 5 do not matter, provided that they are not all in one row or column. 

• Checking the Conditions
• 1. No information is given about the selection process concerning how they chose the boys. In practice we would contact the person who collected the data to find out.

• 2. All expected counts exceed 5. For example, 717.7
• 

• Test stat: 6.739
• p-value: 0.034

*Always use pearson p-value*

• CONCLUDE: 
• Since the p-value is between 0.01 and 0.05, we have strong evidence that grades depend on gaming status in the population of all teenage boys at similar schools

• 1. Type in the observed counts into Minitab
• 
• 
• Go to 'Stats' -> 'Tables' -> 'Chi-square Test for Association'
• 
• Change drop down menu to 'summarized data in a two-way table'

• Now put in each column from the two-way table into the box called 'Columns containing the table'
• 
• Press 'ok'
• 

1. We can use quantitative or categorical variables. 

2. No, only correlation. 

3. Higher household income for the gamers could be causing them to get better grades, so we can't claim causation. Higher income households have more money for games and also for tutors to help with school

1. The expected counts are the numbers of boys we would expect in each cell if grades were the same for gamers and non-gamers. 

• If grades were the same for gamers and non-gamers, then the grade distribution for gamers would be the same as the grade distribution for non-gamers and this would be estimated by the marginal distribution of grades. 
• 

For example, 52.046% of 1379 (total number of gamers) would be expected to get A's and B's. 

The same would go for the expected counts for non-gamers. If grades were the same for non-gamers as they were for gamers, then 52.045% of 429 (non-gamers) would mean that 223.3 students would get A's and B's. 



2. The chi-square statistic compares each observed count with the corresponding expected count. 

• ex. 
• 
• Don't memorize specific formula, just have an idea how it works.

3. If the observed counts are very different from the expected counts, then the chi-squared statistic will be large and we will conclude that grades depend on gaming status. 

If the observed counts are close to the expected counts, then the chi-squared statistic will be small, and we will conclude that we can't claim that grades depend on gaming status.

1. p-value = between 0.25 and 0.05

• 2. p-value = 0.15
• 

3. In a chi-squred test involving a two-way table that has r rows and c columns, the degrees of freedom are given by:

(r-1)(c-1)

4. 2 rows, 3 columns

(2-1)(3-1) = 

1 x 2 = 

2 df

1. The numbers are counts of individuals (not percentages/summarized data)

2. Each individual is counted exactly once

3. The conditions are satisfied (representative sample and expected counts all at least 5...)