Loading in 5 sec....

For testing significance of patterns in qualitative dataPowerPoint Presentation

For testing significance of patterns in qualitative data

- 112 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' For testing significance of patterns in qualitative data' - herrod-acosta

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

- 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

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

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.

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 side while in fact the null hypothesis is true and sample frequencies are not significantly apart from the ideal frequencies?of the table is used
- nondirectional test
- the statistic has no sign

Observed Expected while in fact the null hypothesis is true and sample frequencies are not significantly apart from the ideal frequencies?

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 = while in fact the null hypothesis is true and sample frequencies are not significantly apart from the ideal frequencies?

number of terms -1

2 x 2 contingency tables while in fact the null hypothesis is true and sample frequencies are not significantly apart from the ideal frequencies?

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 while in fact the null hypothesis is true and sample frequencies are not significantly apart from the ideal frequencies?

notdef.

total

def

Operator

A

100

900

1000

B

60

440

500

total

160

1340

1500

Result while in fact the null hypothesis is true and sample frequencies are not significantly apart from the ideal frequencies?

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 while in fact the null hypothesis is true and sample frequencies are not significantly apart from the ideal frequencies?

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)

not while in fact the null hypothesis is true and sample frequencies are not significantly apart from the ideal frequencies?def.

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 while in fact the null hypothesis is true and sample frequencies are not significantly apart from the ideal frequencies?

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 while in fact the null hypothesis is true and sample frequencies are not significantly apart from the ideal frequencies?
- 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)

Download Presentation

Connecting to Server..