Lecture 6 - Analysis of Qualitative Data

Question 1

In this lecture, we are moving from continuous data to ____ data where we CANNOT use t-tests to evaluate the data.

Answer 1

count

Question 2

What is Analysis of Qualitative Data ?

Answer 2

For evaluating qualitative frequency data (nominal or ordinal data) for “goodness of fit” (with an hypothesized underlying distribution), or for “association” between multiple measures.

Question 3

Describe the Mendelian theory

Answer 3

25% = Homozygous Dominant

50% = Heterozygous Dominant

25% = Homozygous Recessive

*So 75% of people will express the dominant gene (unaffected) and 25% of people will be express the recessive gene (affected)

Question 4

So the observed values would be ?

Answer 4

0. 75 x n = Unaffected

0. 25 x n = Affected

Question 5

How do you determine the goodness of fit?

Answer 5

You use Chi Square

Question 6

Chi-square statistic has ____ values only

Answer 6

positive

Question 7

Therefore only one parameter is needed to specify any chi-squared distribution. What parameter ?

Answer 7

degrees of freedom

Question 8

For goodness of fit/chi square, what is the formula for degrees of freedom ?

Answer 8

df = number of categories - 1

Question 9

What is required if df = 1 ?

Answer 9

Yate’s correction (this is the -0.5) thing

Question 10

What is not required if df > 1 ?

Answer 10

Yate’s correction is not required! (So you DO NOT - 0,5)

Question 11

For a chi square test, we should do an ____ tail probability test.

Answer 11

upper

Question 12

If the X^2 value < Chi square critical value, ?

Answer 12

Accept Ho - no significant differences

Question 13

If the X^2 value > Chi square critical value, ?

Answer 13

Reject Ho - there is a significant difference

Question 14

Assumptions of the Goodness of Fit test ?

Answer 14

1) Random sampling: Not essential for calculation of chi-square, but may obviously impact the interpretation (bias)
2) Observations are independent: (A paired variation is available)
3) Categories must be mutually exclusive
4) Expected frequencies: In all the cells, there must be at least 1, with no more than 20% of cells with an expected frequency <5 for the resulting theoretical distribution to be reasonably accurate
5) The distribution of chi-square: Depends on the number of treatments being compared and the number of outcomes. This dependency is quantified in a degrees of freedom parameter v equal to the number of rows in the table minus 1, times the number of columns in the table minus 1

v = (r-1)(c-1)

6) The Yates Correction: For v=1 contingency tables, computing chi square using the standard formula leads to p-values smaller than they ought to be; results are biased towards Ha. The Yates correction is used when v = 1.

Question 15

How do you calculate expected frequency of a 2x2 contingency table ?

Answer 15

EF = Row total x Column total / Grand total

EF = Expected Frequency

Question 16

For a 2x2 contingency table, what is the Ho and Ha ?

Answer 16

Ho = Treatment and outcome are independent

Ha = Treatment and outcome are associated

*Association does not mean causation or correlation

Question 17

Chi-square with multiple outcomes:

Prob won’t have to do ____ Correction

Answer 17

Yates

Question 18

Describe what a post hoc evaluation is

Answer 18

So you can decompose the 32 table into 22 tables for postdoc evaluation to explain significant findings.

*Basically you have to choose which 2 groups make sense to lump together as 1 group.

You use a Bonferroni correction (alpha/2) so 0.05/2 = 0.025 –> this gives you a different value from the Chi Square test.