13.5.103 chi squared test with data

Perform a​ chi-square homogeneity test. An independent simple random sample of residents in three regions gave the data on race shown in the table. At the 1 % significance​ level, do the data provide sufficient evidence to conclude that a difference exists in race distributions among the three​ regions?


Name of variables

  • row total: rTotal

  • column total: cTotal

  • total numbers of data: total

  • original data: dframe1

  • expected frequencies data: dframe2

What are the null and alternative​ hypotheses?

Since the question asks “that a difference exists in race distributions among the three​ regions,” the correct hypothesis is

\(H_0:\) The racial distribution is the same in each of the three regions.

\(H_a:\) The racial distribution is not the same in each of the three regions.

Find the test statistic.,\(\chi^2\)

First we need to get the data from the question. (We can import it from Excel)

data <- read.csv("https://raw.githubusercontent.com/sileaderwt/MTH1320-UMSL/main/Image%2BData/13.5.103/13.5.103.csv")
data
##        X. White Black Other
## 1    East    99    10     7
## 2 Midwest   124    18    10
## 3    West   120    11    20

We store data into 2 different frame. The first one shows original data. The second shows expected frequency.

dframe1 = data.frame(data)
dframe2 = data.frame(data)

Conditions to run a chi-square test

  • All expected frequencies are 1 or greater

  • At most 20% of the expected frequencies are less than 5

  • The sample is a simple random sample

  • The sample is an independent sample

    First approach: using chisq() in R which professor Covert introduces in the video lectures.(Recommended)

This approach is cleaner and faster

First we need to create a new data frame which does not contain a column of names so our data frame only contains data.

We can use subset to drop column of name.

dframe3 = subset(dframe1, select = -c(X.))
dframe3
##   White Black Other
## 1    99    10     7
## 2   124    18    10
## 3   120    11    20

To see the difference between dframe1 and dframe3, we can run

dframe1
##        X. White Black Other
## 1    East    99    10     7
## 2 Midwest   124    18    10
## 3    West   120    11    20

We run chisq.test()

chisq.test(dframe3, correct=FALSE)
## 
##  Pearson's Chi-squared test
## 
## data:  dframe3
## X-squared = 7.2825, df = 4, p-value = 0.1217

Since df = 4, to find critical value for \(\alpha = .01\)

alpha = .01
round(qchisq(1- alpha,4),3)
## [1] 13.277
Since \(\chi^2\) is right-tailed test by nature, our test statistic does not lie in rejected region 7.282 < 13.277 , we do not have enough evidence to reject the hypothesis

If the question ask about expected frequencies, we can find it by running

chisq.test(dframe3, correct=FALSE)$expected
##       White    Black    Other
## 1  94.95943 10.79714 10.24344
## 2 124.42959 14.14797 13.42243
## 3 123.61098 14.05489 13.33413

Second approach using formular

First we need to find row total and column total.

cTotal = rep(0, nrow(dframe2))
rTotal = c()
for (i in 2:ncol(dframe2)){
  cTotal = cTotal + dframe2[,i]
  rTotal = c(rTotal, sum(dframe2[,i]))
}
total = sum(cTotal)
cTotal
## [1] 116 152 151
rTotal
## [1] 343  39  37
total
## [1] 419

We can find the test statistic \(\chi^2\) by using the formula \(\chi^2=\sum{\frac{(O-E)^2}{E}}\)

for (i in 2:ncol(dframe2)){
  dframe2[,i] <- sum(dframe2[,i])*cTotal/total
  print(dframe2[,i])
}
## [1]  94.95943 124.42959 123.61098
## [1] 10.79714 14.14797 14.05489
## [1] 10.24344 13.42243 13.33413
dframe2
##        X.     White    Black    Other
## 1    East  94.95943 10.79714 10.24344
## 2 Midwest 124.42959 14.14797 13.42243
## 3    West 123.61098 14.05489 13.33413

Find the test statistic.,\(\chi^2\)

chi = 0
for (i in 2:ncol(dframe2)){
  chi = chi + sum((dframe1[,i]-dframe2[,i])^2/dframe2[,i])
}
chi
## [1] 7.282498

Round to three decimal places

round(chi,3)
## [1] 7.282

Find the critical value.\(\chi_{\alpha}^2\)

We find degree of freedom by using the formula \(df=(r-1)(c-1)\)

Since dframe2 has 4 variables and the first one is x, so ncol(dframe2) returns 1 more than actual value. We find degree of freedom by running

df = (ncol(dframe2)-2) * (nrow(dframe2)-1)
df
## [1] 4

Our significant level is 1%, so \(\alpha = .01\)

alpha = .01
round(qchisq(1- alpha,df),3)
## [1] 13.277
Since \(\chi^2\) is right-tailed test by nature, our test statistic does not lie in rejected region 7.282 < 13.277 , we do not have enough evidence to reject the hypothesis

Hope that helps!