13.2.18 chi squared​test

A distribution and the observed frequencies of the values of a variable from a simple random sample of the population are provided below. Use the​ chi-square goodness-of-fit test to​ decide, at the specified significance​ level, whether the distribution of the variable differs from the given distribution.

Distribution: 0.1875, 0.1875, 0.3125, 0.3125 Observed frequencies: 20, 22, 20, 34 Significance level = 0.05

Determine the null and alternative hypotheses. Choose the correct answer below.

Since the question asks whether the distribution of the variable differs from the given distribution, it is a two-tailed test

\(H_0:\) The distribution of the variable is the same as the given distribution

\(H_a:\) The distribution of the variable differs from given distribution

Compute the value of the test​ statistic,\(\chi^2\)

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

distribution <- c(0.1875, 0.1875, 0.3125, 0.3125)
obFrequency <- c(20, 22, 20, 34)

First approach, we can use chisq.test()

chisq.test(obFrequency,p=distribution, correct=FALSE)
## 
##  Chi-squared test for given probabilities
## 
## data:  obFrequency
## X-squared = 4.9778, df = 3, p-value = 0.1734

Round to 3 decimal places

print(chisq.test(obFrequency,p=distribution, correct=FALSE),6)
## 
##  Chi-squared test for given probabilities
## 
## data:  obFrequency
## X-squared = 4.978, df = 3, p-value = 0.173

Second approach using formular

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

Expected frequency = sample size * distribution

We can find the sample size by using sum(obFrequency) in R

expFrequency = sum(obFrequency)*distribution
sum((obFrequency-expFrequency)^2/expFrequency)
## [1] 4.977778

Round to three decimal places

round(sum((obFrequency-expFrequency)^2/expFrequency),3)
## [1] 4.978

\(\chi^2\) is right-tailed test by nature

Since we have \(\alpha=.05\) and there are 4 possible values for the variable, so the degree of freedom df = 4 -1 = 3

\(\chi_\alpha^2\) has \(\alpha\) is the area to the right under \(\chi\) curve

We can get \(\chi^2\) value by using the table or we can run qchisq()

qchisq() takes in the area to the left and degree of freedom

alpha = .05
df =  4 - 1
qchisq(1-alpha,df)
## [1] 7.814728

Round to three decimal places

round(qchisq(1- alpha,df),3)
## [1] 7.815
Since \(\chi^2\) is right-tailed test by nature, our test statistic does not lie in rejected region 4.978 < 7.815 , we do not have enough evidence to reject the hypothesis

Hope that helps!