Categorial Data Analysis

statistics

multinomial : qualitative data with more than two outcomes

classes, categories, cells : names for the outcome of a multinomial experiment

one-way table : a table which summarizes the results of a multinomial experiment for a single qualitative variable

chi-square test : a test to compares the frequency distribution of categorical data against an expectation (see: chi-square procedure)

Properties of a multinomial experiment:

1. There are $n$ identical trials.
2. There are $k$ possible outcomes of each trial.
3. The probabilities of the $k$ outcomes sum to 1 and remain the same from trial to trial.
4. The trials are independant
5. The random variables of interest are the “cell counts” of the number of observations that fall into each of the $k$ categories.

Test of hypothesis about multinomial probabilities: one-way table

$H_0=p_1=p_{1,0},p_2=p_{2,0},\dots ,p_k=p_{k,0}$ where the $p_{1,0}$ etc are hypothesized values for the probabilities. $H_a$: At least one of the probabilities doesn’t equal the hypothesis

Test stat: ${\chi }_c^2=\Sigma \frac{[n_i-E_i{]}^2}{E_i}$ where $E_i=n{p}_{i,0}$ is the expected cell count. Rejection region: ${\chi }_c^2>{\chi }_{\alpha }^2$ where ${\chi }_{\alpha }^2$ has $k-1$ degrees of freedom. p-value: $P({\chi }_c^2>{\chi }_{\alpha }^2)$

This requires the following conditions:

1. multinomial experiment has been conducted
2. the sample size $n$ will be large enough so that for every cell, $E_i$ will be equal to 5 or more.

Chi-Square test

$$\begin{array}{r}{\chi }^2=\frac{[n_1-E_1{]}^2}{E_1}+\frac{[n_2-E_2{]}^2}{E_2}+\cdots +\frac{[n_n-E_n{]}^2}{E_n}\end{array}$$

To do this on the calculator, input the observed values vs theoretical (e.g. $np$) in two diffrent lists and reference them in the Chi-Square GOF (Goodness of Fit) function.

Test of independence

Tests whether one variable is independent from another (e.g. does your hogwarts house influence if you like pizza on pineapple).

	gryffindor	hufflepuff	ravenclaw	slytherin
no	79	122	204	74	479
yes	82	130	240	69	521
	161	252	444	143	1000

df = (2-1) * (4-1) = 3

from here, we calculate the probabilities of each of these, then run the results through the GoF function.

Example 11.6 from Introductory Statistics book

In a volunteer group, adults 21 and older volunteer from one to nine hours each week to spend time with a disabled senior citizen. The program recruits among community college students, four-year college students, and nonstudents. In Table 11.15 is a sample of the adult volunteers and the number of hours they volunteer per week.

Is the number of hours volunteered independent of the type of volunteer?

H_0: hours worked are independent of type H_a: hours works are dependent on type

Actually observed:

Type	1-3hr	4-6	7-9	total
community college	111	96	48	255
four-year college students	96	133	61	290
non-students	91	150	53	294
total	298	379	162	839

Expected results, based on taking row_total * column_total / overall_total:

Type	1-3hr	4-6	7-9
community college	255*298/839	255*379/839	255*162/839
four-year college students	290*298/839	290*379/839	290*162/839
non-students	294*298/839	294*379/839	294*162/839

which is:

Type	1-3hr	4-6	7-9
community college	90.572110	115.19070	49.237187
four-year college students	103.00358	131.00119	55.995232
non-students	104.42431	132.80810	56.767580

Then we check expected vs observed via Pearson Residuals:

Type	1-3hr	4-6	7-9
community college	(90.572110-111)^2/111	(115.19070-96)^2/96	(49.237187-48)^2/48
four-year college students	(103.00358-96)^2/96	(131.00119-133)^2/133	(55.995232-61)^2/61
non-students	(104.42431-91)^2/91	(132.80810-150)^2/150	(56.767580-53)^2/53

which is:

Type	1-3hr	4-6	7-9
community college	3.7594477	3.8362809	0.031888160
four-year college students	0.51093888	0.030039409	0.41061808
non-students	1.9803527	1.9704095	0.26782376
	6.2507393	5.8367298	0.71033	12.797799

${\chi }^2=12.80$ Degrees of Freedom = 4 b/c (3col-1) * (3col-1) = 2*2 = 4 p-value = cdf(12.8, infinity, df=4) = .0123 $\alpha =0.05$ a > p-value, so reject the null hypothesis

Test of homogeneity

Tests if two samples came from the same population

$E_{r,c}=\frac{n_r\cdot n_c}{n}$ where $n_r$ is the number in the row, $n_c$ is the number for that column and $n$ is the number total.