In a 1962 social experiment, 123 3- and 4-year-old children from poverty-level families in Ypsilanti, Michigan, were randomly assigned either to a treatment group receiving 2 years of preschool instruction or to a control group receiving no preschool. The participants were followed into their adult years. The following table shows how many in each group were arrested for some crime by the time they were 19 years old. (Time, July 29, 1991).
| Arrested | Not Arrested | Total | |
|---|---|---|---|
| Preschool | 19 | 42 | 61 |
| Control | 32 | 30 | 62 |
| Total | 51 | 72 | 123 |
Is a statistically significant difference between the rate of arrest (or no arrest) in the two treatment groups.
There are two categorical variables, each with two levels. We could either use a two sample test to compare proportions (groups: treatment, response: arrest outcome) OR we could use a chi-square test of independence. These tests will give identical results. For this example, we will use the chi-square test. State your hypotheses needed to test the question above.
This example differs from example 2 because we have data in a summarized two-way table. (Example 2 had the raw categorical variables available.) To run the chi-square test, we first must create a matrix of counts using the cbind command that binds together columns of counts:
counts <- cbind(c(19,32), c(42,30))
colnames(counts) <- c("arrested", "not arrested") # adds column names
rownames(counts) <- c("preschool", "control") # adds row names
counts arrested not arrested
preschool 19 42
control 32 30We then use this in the chisq.test command:
preschool.test <- chisq.test(counts)
preschool.test
Pearson's Chi-squared test with Yates' continuity correction
data: counts
X-squared = 4.4963, df = 1, p-value = 0.03397Answer: Yes, they are all above 5.
51/123 # overall arrest rate[1] 0.414634172/123 # overall non arrest rate[1] 0.5853659preschool.test$expected arrested not arrested
preschool 25.29268 35.70732
control 25.70732 36.29268How do the arrest rates differ for each treatment group? Compute a 95% confidence interval for the difference in arrest rates between those who had the preschool and control treatments.
prop.table(counts,1) arrested not arrested
preschool 0.3114754 0.6885246
control 0.5161290 0.4838710Answer: About 52% of the control group were arrested while only about 31% of the preschool group were arrested.
\[ \hat{p}_{control} = \dfrac{32}{62} = 0.516129, \ \ \ \hat{p}_{preschool} = \dfrac{19}{61} = 0.3114754 \]
The 95% for the difference in true arrest rates \(p_{control} - p_{preschool}\) is
\[\begin{align*} 0.516129 - 0.3114754 \pm & 1.96 \sqrt{\dfrac{0.516129(1-0.516129)}{62} + \dfrac{0.3114754(1-0.3114754)}{61}} \\ 0.2046536 \pm & 1.96(0.0868549) \\ (0.034418, 0.3748892) \\ \end{align*}\]
We are 95% confident that the true rate of arrest for the preschool treatment is 3.4 to 37.5 percentage points lower than the arrest rate for the control group. This is evidence that the preschool treatment lowered the risk of arrest.
The chisq.test command uses a test stat “correction” when both of your categorical variables have only 2 levels. With this correction, your chi-square test results won’t exactly match a two-sample test for the difference of two proportions. If you turn off the correct with correct=FALSE you will obtain identical results.
chisq.test(counts, correct=FALSE) # exact same as two-sample proportion test
Pearson's Chi-squared test
data: counts
X-squared = 5.3059, df = 1, p-value = 0.02125A survey of college graduates was done to study how frequently they exercised. The survey was completed by 470 graduates. They were asked where they lived their senior year. Use the following data to determine whether there is an association between exercise on campus and students’ living arrangements.
counts3 <- cbind(c(32, 74, 110, 39), c(30,64,25,6), c(28,42,15,5))
colnames(counts3) <- c("No regular exercise", "Sporadic exercise", "Regular exercise")
rownames(counts3) <- c("Dormitory", "On-Campus Apartment", "Off-campus Apartment", "At Home") knitr::kable(counts3)| No regular exercise | Sporadic exercise | Regular exercise | |
|---|---|---|---|
| Dormitory | 32 | 30 | 28 |
| On-Campus Apartment | 74 | 64 | 42 |
| Off-campus Apartment | 110 | 25 | 15 |
| At Home | 39 | 6 | 5 |
test3 <- chisq.test(counts3)Answer:
\[\begin{align*} H_0 &: \text{exercise and living arrangements independent of each other}\\ H_A &: \text{exercise and living arrangements dependent of each other} \end{align*}\]
The observed and expected values from the chi square test are:
test3$observed No regular exercise Sporadic exercise Regular exercise
Dormitory 32 30 28
On-Campus Apartment 74 64 42
Off-campus Apartment 110 25 15
At Home 39 6 5round(test3$expected,2) No regular exercise Sporadic exercise Regular exercise
Dormitory 48.83 23.94 17.23
On-Campus Apartment 97.66 47.87 34.47
Off-campus Apartment 81.38 39.89 28.72
At Home 27.13 13.30 9.57All of the expected counts are greater than 5.
The test statistics is calculated as:
\[\begin{align*} \chi^2 & = \sum \frac{(O- E)^2}{E}\\ &= \frac{(32 - 48.83)^2}{48.83} + \frac{(30 - 23.94)^2}{23.94} + \frac{(28 - 17.23)^2}{17.23} +\\ & \qquad \frac{(74 - 97.66)^2}{97.66} + \frac{(64 - 47.87)^2}{47.87} + \frac{(42 - 34.47)^2}{34.47} + \\ & \qquad \frac{(110 - 81.38)^2}{81.38} + \frac{(25 - 39.89)^2}{39.89} + \frac{(15 - 28.72)^2}{28.72} + \\ & \qquad \frac{(39 - 27.13)^2}{27.13} + \frac{(6 - 13.30)^2}{13.30} + \frac{(5 - 9.57)^2}{9.57} \\ &= 5.80 + 1.53 + 6.73 + 5.73 + 5.44 + 1.64 + 10.06 + 5.56 + 6.55 + 5.19 + 4.01 + 2.18\\ &= 60.42 \end{align*}\]
(32 - 48.83)^2/48.83 + (30 - 23.94)^2/23.94 + (28 - 17.23)^2/17.23 + (74 - 97.66)^2/97.66 + (64 - 47.87)^2/47.87 + (42 - 34.47)^2/34.47 + (110 - 81.38)^2/81.38 + (25 - 39.89)^2/39.89 + (15 - 28.72)^2/28.72 + (39 - 27.13)^2/27.13+(6 - 13.30)^2/13.30+(5 - 9.57)^2/9.57[1] 60.438855.80 + 1.53 + 6.73 + 5.73 + 5.44 + 1.64 + 10.06 + 5.56 + 6.55 + 5.19 + 4.01 + 2.18[1] 60.42The degree of freedom of \(\chi^2\) is \(df = (4-1)*(3-1) = 6\).
test3
Pearson's Chi-squared test
data: counts3
X-squared = 60.439, df = 6, p-value = 3.664e-11The p-value can also be calculated as
1 - pchisq(60.43, df = 6)[1] 3.680733e-11There is significant evidence of an association between the exercise and living arrangements (\(\chi^2 = 60.43\), df=6, p-value \(\approx\) 0).
Consider survey questions about political comfort level and religion. We want know if the response to the comfort level question is associated with their religious practice. To test this question about two categorical variables with one variable containing at least 3 levels, we must conduct a chi-square test for association.
State the hypotheses for this test.
Answer: The null can be stated a couple of equivalent ways: There is no association between religion and comfort level; the variables comfort level and religion are independent of one another; the distribution of comfort level is the same for all three religion types.
The alternatives are just “not the null” statements: There is an association between religion and comfort level; the variables comfort level and religion are dependent; the distribution of comfort level is the different for at least one religion type.
Does the data suggest that there is an association between comfort level and religion?
library(dplyr)
library(ggplot2)
# read the data
survey <- read.csv("https://raw.githubusercontent.com/deepbas/statdatasets/main/Survey.csv")
# and drop the rows containing missing values using the tidyr package
survey <- survey %>% tidyr::drop_na()
# rename comfort level using fct_recode() from the forcats package
survey <- survey %>% mutate(comfortness = forcats::fct_recode(Question.9,
`rarely` = "rarely, if ever, comfortable",
`sometimes` = "sometimes comfortable",
`almost always` = "almost always comfortable"),
comfortness = forcats::fct_relevel(comfortness,
"almost always",
"sometimes",
"rarely"))
# rename comfort level using fct_recode() from the forcats package
survey <- survey %>%mutate(religiousness = forcats::fct_recode(Question.8,
`not religious` = "not religious",
`religious not active` = "religious but not actively practicing",
`religious active` = "religious and actively practicing my religion"),
religiousness = forcats::fct_relevel(religiousness,
"not religious",
"religious not active",
"religious active"))# Make a two way table
library(kableExtra)
counts <- table(survey$religiousness, survey$comfortness)
prop.table(counts,1) # dist of comfort level given religious level
almost always sometimes rarely
not religious 0.53092784 0.39175258 0.07731959
religious not active 0.39393939 0.41414141 0.19191919
religious active 0.31578947 0.42105263 0.26315789kableExtra::kable(table(survey$religiousness, survey$comfortness),
caption = "A two way table of religious preference and political comfortness") %>%
kable_styling(position = "center")| almost always | sometimes | rarely | |
|---|---|---|---|
| not religious | 103 | 76 | 15 |
| religious not active | 39 | 41 | 19 |
| religious active | 18 | 24 | 15 |
ggplot(survey, aes(x=religiousness, fill=comfortness)) +
geom_bar(position="fill") +
labs(fill = "Comfort level", x = "Religious level", y = "proportion",
title="Comfort level by religious level") +
scale_x_discrete(labels = function(x) stringr::str_wrap(x, width = 16))
Compute the expected number of “not religious” people who are “almost always comfortable”.
Answer: There are 194 “not religious” respondents and the overall rate (ignoring religion) of “almost always” comfortable is about 45.7%. If the null is true (and religion doesn’t relate to comfort level), the expected number is about
\[194 \times \dfrac{160}{350} = 88.686\]
table(survey$religiousness)
not religious religious not active religious active
194 99 57 table(survey$comfortness)
almost always sometimes rarely
160 141 49 What is the contribution to the chi-square test statistic from the “not religious”/“almost always comfortable” cell?
Answer: The contribution to the chi-square test stat from this category is 2.31.
\[ \dfrac{(103 - 88.6857143)^2}{88.6857143} = 2.31 \]The chisq.test(x,y) can be used to give chi-square test results. For this version, x and y are categorical variables from a data set.
ComfortReligion <- chisq.test(survey$religiousness, survey$comfortness)
ComfortReligion
Pearson's Chi-squared test
data: survey$religiousness and survey$comfortness
X-squared = 19.33, df = 4, p-value = 0.0006768What is your conclusion for this test?
Are the expected counts large enough to use the chi-square distribution to compute the p-value?
ComfortReligion$expected survey$comfortness
survey$religiousness almost always sometimes rarely
not religious 88.68571 78.15429 27.16
religious not active 45.25714 39.88286 13.86
religious active 26.05714 22.96286 7.98If you were concerned that the expected counts weren’t large enough to trust using a chi-square distribution to compute a p-value, you can add a simulate.p.value = TRUE argument to use a randomization distribution to compute the p-value:
chisq.test(survey$religiousness, survey$comfortness, simulate.p.value = TRUE)
Pearson's Chi-squared test with simulated p-value (based on 2000
replicates)
data: survey$religiousness and survey$comfortness
X-squared = 19.33, df = NA, p-value = 0.0009995The p-value is slightly different, but your conclusion should be the same.
Use the grouped bar graph and conditional percents from part (b) to describe the association you (should have) found in part (f). To help quantify differences, compute a 95% confidence interval for the difference in the true proportions of “rarely comfortable” people in the not religious and actively religious groups.
Answer: The largest test stat contributions comes from the not religious/rarely comfortable group and the active religious/rarely comfortable group. We can see that the not religious respondents have a low “rarely” comfortable level compared to religious groups (7.7% vs. 26.3% for active and 19.2% for not active) and they have a very high almost always comfortable level compared to religious groups (53.1% vs. 31.6% for active and 39.4% for not active).
If \(p_{not.rel}\) and \(p_{active.rel}\) denote the true proportions of “rarely comfortable” for the not religious and active religious groups. We want a 95% CI for \(p_{not.rel} - p_{active}\). The sample proportions are computed from the counts table (or the prop.table output). Of the 194 “not religious” respondents, 15 are rarely comfortable so \[\hat{p}_{not.rel} = \frac{15}{194} = 0.077\] \[\hat{p}_{active.rel} = \frac{15}{57} = 0.263\]
So a 95% CI for the difference in the true rates of rarely comfortable is \[\begin{align*} CI &= (0.077 - 0.263) \pm 1.96\cdot\sqrt{\frac{0.077(1-0.077)}{194} + \frac{0.263(1-0.263)}{57}}\\ &= (-0.306, -0.066) \end{align*}\]
round((0.077 - 0.263) + c(-1,1)* 1.96* sqrt(0.077*(1-0.077)/194 + 0.263*(1-0.263)/57),3)[1] -0.306 -0.066