For testing significance of patterns in qualitative data

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# For testing significance of patterns in qualitative data - PowerPoint PPT Presentation

Chi-squared Tests. For testing significance of patterns in qualitative data Test statistic is based on counts that represent the number of items that fall in each category Test statistics measures the agreement between actual counts and expected counts assuming the null hypothesis.

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Presentation Transcript
Chi-squared Tests
• For testing significance of patterns in qualitative data
• Test statistic is based on counts that represent the number of items that fall in each category
• Test statistics measures the agreement between actual counts and expected counts assuming the null hypothesis
Chi-squared Distribution

The chi-square distribution can be used to see whether or not an observed counts agree with an expected counts.Let

O = observed count and

E = Expected count

Testing if Observed Counts

are in Agreement with Known Percentages

Consider items of a population distributed over k categories in in proportions

If H0 is true then we expect

Ei = n , expected frequency

for the ith category as opposed to Oi, observed frequency.

An Example

Biased Coin?

Observed Expected

Frequency Frequency

H 40 50

T 60 50

sum 100 100

degrees of freedom = (R –1)(C – 1)

R = number of rows

C = number of columns

Is our chi square value an extreme outcome just by chance while in fact the null hypothesis is true and sample frequencies are not significantly apart from the ideal frequencies?

Note that chi-squared statistic is a positive number

only the right-hand sideof the table is used
• nondirectional test
• the statistic has no sign
Observed Expected

Die Frequency Frequency

1 4 10

2 6 10

3 17 10

4 16 10

5 8 10

6 9 10

sum 60 60

degrees of freedom =

number of terms -1

2 x 2 contingency tables

Chi-squared test for independence

Var B

total

b1

b2

Var A

a1

a2

total

Ho : The two variable are independent

Ha : The two variables are associated

Result

notdef.

total

def

Operator

A

100

900

1000

B

60

440

500

total

160

1340

1500

Result

notdef.

total

def

Operator

A

100

900

1000

B

60

440

500

total

160

1340

1500

Total number of items=1500

Total number of defective items=160

Overall defective rate =160/1500=0.1067

Now, apply this rate to the number of items produced by each operator.

Result

notdef.

total

def

Operator

A

100

900

1000

B

60

440

500

total

160

1340

1500

Expected defective from Operator A

= 1000 * 0.1067 = 106.7

(expected not defective=1000-106.7=893.3)

Expected defective from Operator B

= 500 * 0.1067 = 53.3

(expected not defective=500-53.3=446.7)

notdef.

total

def

Operator

1000

A

100

900

B

60

440

500

total

160

1340

1500

Expected

notdef.

total

def

Operator

A

106.7

893.3

B

53.3

446.7

total

Result

r x c contingency tables

SA A NO D SD

Gr 1 12 18 4 8 12

Gr2 48 22 10 8 10

Gr3 10 4 12 10 12

use when you have categorical data
• measure the difference between actual counts and expected counts
• test the independence of two variables
• Assumptions:data set is a random sampleyou have at least 5 counts in each category
• degrees of freedom =(categories var1 -1)(categories var2 -1)