Enter data into columns and provide headersscore epitools app

BIOSTATS 640 - Spring 2023 Unit 4. Categorical Data Analysis (2 of 2) R and Online Apps

Unit 4 – Categorical Data Analysis
Practice Problems (2 of 2)

This exercise gives you practice performing a2 x K test of trend for contingency table data.

The Western Collaborative Group Study (WCGS) was a prospective study of 3,154 men, all initially disease-free, who were followed for events of coronary heart disease (CHD). At the eight years follow-up mark, there were 257 events of CHD. Several potential predictors of CHD were of interest, including: age (at enrollment), cholesterol, systolic blood pressure, hematocrit, ECG status, smoking, and relative weight.

	35-40	41-45	Age at Enrollment (years)		55-60	Total =
	35-40	41-45	46-50	51-55	55-60	Total =
no	512	1036	680	463	206	2987
yes	31	55	70	65	36	257
Total =	543	1091	750	528	242	3154

sol_categorical 2 of 2.docx Page 1 of 8

BIOSTATS 640 - Spring 2023 Unit 4. Categorical Data Analysis (2 of 2) R and Online Apps

	chd	not chd
1	31	512
2	55	1036
3	70	680
4	65	463
5	36	206

In EpiTools App, paste data into data box. Click SUBMIT

BIOSTATS 640 - Spring 2023 Unit 4. Categorical Data Analysis (2 of 2) R and Online Apps

Question #1B
HO: No association of interval of age with event of CHD
HA: Monotone increasing trend in event of CHD with increasing interval of age, one-sided

cat("\nQuestion 1\nWCGS Data\n\n")

##

## CHD 35-40 41-45 46-50 51-55 55-60
## No 512 1036 680 463 206
## Yes 31 55 70 65 36

Inspect Data
library(gmodels) # Attach package {gmodels}

BIOSTATS 640 - Spring 2023 Unit 4. Categorical Data Analysis (2 of 2) R and Online Apps

Q1a) Test of General Association

## Null: No Association CHD w Interval Age

chisq.test(tableq1)

## X-squared = 46.653, df = 4, p-value = 0.000000001801

HO: No association of interval of age with event of CHD

unlikely result. Conclude these data provide statistically significant evidence of an association.

Q1b) Test of Trend

## Chi Square Test of Trend

## Null: No Monotone Association CHD w Interval Age

## data: tableq1

## Z = -6.3852, dim = 5, p-value = 0.00000000008561 Z-score=-6.3852 -> Chi Square = [-6.3852]2

p-value <<< .0001

Reject the null hypothesis. Assumption of the null hypothesis and application to the data has led to an extremely

#2. Source: Triola MM and Triola MF. Biostatistics for the Biological and Health Sciences Boston: Pearson Addison Wesley, John Wiley, 2006. Chapter 10, Section 10-2. page 491.

This exercise gives you practice performing a chi square goodness of fit (GOF) test.

	0	1	2					7	8	9
	0	1	2	3	4	5	6	7	8	9
Frequency	35	0	2	1	4	24	1	4	7	2

2b. How many degrees of freedom does your chi square statistic have? Degrees of freedom = (# intervals) - 1
= 9

2c. What is the value of your chi square statistic?
156.5

BIOSTATS 640 - Spring 2023 Unit 4. Categorical Data Analysis (2 of 2) R and Online Apps

Online Apps Solution

sol_categorical 2 of 2.docx Page 6 of 8

q2_df
## digit observed_n expected_percent
## 1 0 35 0.1
## 2 1 0 0.1
## 3 2 2 0.1
## 4 3 1 0.1
## 5 4 4 0.1
## 6 5 24 0.1
## 7 6 1 0.1
## 8 7 4 0.1
## 9 8 7 0.1
## 10 9 2 0.1

Q2 – Chi Square GOF Test, minimal

## Chi Square GOF Test

## Null: Equal Frequencies (10%) of Each Last Digit

digit <- c(0,1,2,3,4,5,6,7,8,9) # Last digit
obs <- c(35,0,2,1,4,24,1,4,7,2) # observed counts
q2_full <- data.frame(digit,obs) # combine to make data frame n_size <- sum(obs) # total sample size

q2_full$exp <-rep(n_size/10,times=10) # null hypothesis expected counts

BIOSTATS 640 - Spring 2023 Unit 4. Categorical Data Analysis (2 of 2) R and Online Apps

# Output Stuff

cat("\np-value =", pvalue)
## p-value = 0

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