1 Test Statistic for Chi-squared Goodness of Fit Test in R

The chi-squared goodness of fit test has test statistics that takes the form:

\[\chi^2=\sum_{i}\frac{(O_{i}-E_{i})^2}{E_{i}}.\]

With \(k\) categories, when the null hypothesis is true, \(\chi^2\) follows a chi-squared distribution (\(\chi^2_{k-1}\)) with degrees of freedom, \(k-1\),

\(O_i\) is the observed frequency in category (or cell) \(i\),

\(E_i\) is the expected frequency in category (or cell) \(i\), or \(E_i = np_i\),

where \(p_i\) is the distribution proportion in category (or cell) \(i\) and,

\(n\) is the total number of observations in all categories (or cells).

See also chi-squared contigency table tests.

2 Simple Chi-squared Goodness of Fit Test in R

Using an observed distribution for 187 randomly sampled sales, test the claim that there are the same amounts of sales in each weekday.

For the following null hypothesis \(H_0\), and alternative hypothesis \(H_1\), with the level of significance \(\alpha=0.05\).

\(H_0:\) the counts in each cell are equal.

\(H_1:\) the counts in at least one cell is different from the others.

For goodness of test, the chisq.test() function has the default proportions as equal, hence, you do not need to specify the "p" argument in this case.

chisq.test(c(34, 42, 33, 37, 41),
           p = c(1/5, 1/5, 1/5, 1/5, 1/5))

Or:

chisq.test(c(34, 42, 33, 37, 41))

    Chi-squared test for given probabilities

data:  c(34, 42, 33, 37, 41)
X-squared = 1.7433, df = 4, p-value = 0.7828

The test statistic, \(\chi^2_4\), is 1.7433,

the degree of freedom is \(k-1=4\),

the p-value, \(p\), is 0.7828.

Interpretation:

• P-value: With the p-value (\(p = 0.7828\)) being greater than the level of significance 0.05, we fail to reject the null hypothesis that the counts in each cell are equal.

• \(\chi^2_4\) T-statistic: With test statistics value (\(\chi^2_4 = 1.7433\)) being less than the critical value, \(\chi^2_{4,\alpha}=\text{qchisq(0.95, 4)}=9.487729\) (or not in the shaded region), we fail to reject the null hypothesis that the counts in each cell are equal.

x = seq(0.01, 18, 1/1000); y = dchisq(x, df=4)
plot(x, y, type = "l",
     xlim = c(0, 18), ylim = c(-0.02, min(max(y), 1)),
     main = "Chi-squared Test Goodnes of Fit Test
Shaded Region for Simple Test",
     xlab = "x", ylab = "Density",
     lwd = 2, col = "blue")
abline(h=0)
# Add shaded region and legend
point = qchisq(0.95, 4)
polygon(x = c(x[x >= point], 18, point),
        y = c(y[x >= point], 0, 0),
        col = "blue")
legend("topright", c("Area = 0.05"),
       fill = c("blue"), inset = 0.01)
# Add critical value and chi-value
arrows(10, 0.1, 1.7433, 0)
text(10, 0.11, "chi-squared = 1.7433")
text(9.487729, -0.01, expression(chi[4][','][alpha]^2==9.487729))

Chi-squared Test Goodness of Fit Test Shaded Region for Simple Test in R

See line charts, shading areas under a curve, lines & arrows on plots, mathematical expressions on plots, and legends on plots for more details on making the plot above.

3 Chi-squared Goodness of Fit Test Critical Value in R

To get the critical value for a chi-squared goodness of fit test in R, you can use the qchisq() function for chi-squared distribution to derive the quantile associated with the given level of significance value \(\alpha\).

The critical value is qchisq(\(1-\alpha\), df).

Example:

For \(\alpha = 0.05\), and \(\text{df} = 5\).

qchisq(0.95, 5)

[1] 11.0705

4 Chi-squared Goodness of Fit Test for Weights in R

Using an observed distribution for 534 randomly sampled students, test whether the proportion of the total students in the senior classes (Sen) doubles that of the total students in the junior classes (Jun), while the proportions are equal among the senior classes and equal among the junior classes.

For the following null hypothesis \(H_0\), and alternative hypothesis \(H_1\), with the level of significance \(\alpha=0.1\).

\(H_0:\) the proportion in each category fits that in the proposed distribution.

\(H_1:\) the proportion in each category does not fit that in the proposed distribution.

chisq.test(c(61, 75, 52, 102, 109, 135),
           p = c(1, 1, 1, 2, 2, 2),
           rescale.p = TRUE)

    Chi-squared test for given probabilities

data:  c(61, 75, 52, 102, 109, 135)
X-squared = 10.466, df = 5, p-value = 0.06305

Interpretation:

• P-value: With the p-value (\(p = 0.06305\)) being less than the level of significance 0.1, we reject the null hypothesis that the proportion in each category fits that in the proposed distribution.

• \(\chi^2_5\) T-statistic: With test statistics value (\(\chi^2_5 = 10.466\)) being in the critical region (shaded area), that is, \(\chi^2_5 = 10.466\) greater than \(\chi^2_{5, \alpha}=\text{qchisq(0.9, 5)}=9.2363569\), we reject the null hypothesis that the proportion in each category fits that in the proposed distribution.

x = seq(0.01, 25, 1/1000); y = dchisq(x, df=6)
plot(x, y, type = "l",
     xlim = c(0, 25), ylim = c(-0.01, min(max(y), 1)),
     main = "Chi-squared Goodness of Fit Test for Weights
Shaded Region",
     xlab = "x", ylab = "Density",
     lwd = 2, col = "blue")
abline(h=0)
# Add shaded region and legend
point = qchisq(0.9, 5)
polygon(x = c(x[x >= point], 25, point),
        y = c(y[x >= point], 0, 0),
        col = "blue")
legend("topright", c("Area = 0.1"),
       fill = c("blue"), inset = 0.01)
# Add critical value and chi-value
arrows(15, 0.05, 10.466, 0)
text(15, 0.055, "chi-squared = 10.466")
text(9.236357, -0.006, expression(chi[5][','][alpha]^2==9.236357))

Chi-squared Goodness of Fit Test for Weights Shaded Region for in R

5 Chi-squared Goodness of Fit Test for Proportions in R

Using an observed distribution for 226 randomly sampled students in a club, test whether the proportions of participating students by year equal the proposed or expected proportions.

For the following null hypothesis \(H_0\), and alternative hypothesis \(H_1\), with the level of significance \(\alpha=0.1\).

\(H_0:\) the proportion in each category fits that in the proposed distribution.

\(H_1:\) the proportion in each category does not fit that in the proposed distribution.

chisq.test(c(88, 65, 55, 18),
           p = c(4/10, 3/10, 2/10, 1/10))

    Chi-squared test for given probabilities

data:  c(88, 65, 55, 18)
X-squared = 3.2404, df = 3, p-value = 0.356

Interpretation:

• P-value: With the p-value (\(p = 0.356\)) being greater than the level of significance 0.1, we fail to reject the null hypothesis that the proportion in each category fits that in the proposed distribution.

• \(\chi^2_3\) T-statistic: With test statistics value (\(\chi^2_3 = 3.2404\)) being less than the critical value, \(\chi^2_{3,\alpha}=\text{qchisq(0.9, 3)}=6.2513886\) (or not in the shaded region), we fail to reject the null hypothesis that the proportion in each category fits that in the proposed distribution.

x = seq(0, 15, 1/1000); y = dchisq(x, df=3)
plot(x, y, type = "l",
     xlim = c(0, 15), ylim = c(-0.015, min(max(y), 1)),
     main = "Chi-squared Goodness of Fit Test for Proportions
Shaded Region",
     xlab = "x", ylab = "Density",
     lwd = 2, col = "blue")
abline(h=0)
# Add shaded region and legend
point = qchisq(0.9, 3)
polygon(x = c(x[x >= point], 15, point),
        y = c(y[x >= point], 0, 0),
        col = "blue")
legend("topright", c("Area = 0.1"),
       fill = c("blue"), inset = 0.01)
# Add critical value and chi-value
arrows(7.5, 0.15, 3.2404, 0)
text(7.5, 0.16, "chi-squared = 3.2404")
text(6.251389, -0.01, expression(chi[3][','][alpha]^2==6.251389))

Chi-squared Goodness of Fit Test for Proportions Shaded Region for in R

6 Chi-squared Goodness of Fit Test: Test Statistics, P-value & Degree of Freedom in R

Here for a chi-squared goodness of fit test, we show how to get the test statistics (or chi-squared value), p-values, expected values, and degrees of freedom from the chisq.test() function in R, or by written code.

chsq_object = chisq.test(c(32, 35, 28, 31),
                         p = c(1/6, 2/6, 2/6, 1/6))
chsq_object

    Chi-squared test for given probabilities

data:  c(32, 35, 28, 31)
X-squared = 16.357, df = 3, p-value = 0.000958

To get the test statistic or chi-squared value; observed and expected values:

\[\chi^2=\sum_{i}\frac{(O_i-E_i)^2}{E_i},\]

chsq_object$statistic

X-squared 
 16.35714 

# to remove name X-squared
unname(chsq_object$statistic)

[1] 16.35714

chsq_object$observed

[1] 32 35 28 31

chsq_object$expected

[1] 21 42 42 21

Same as:

obs = c(32, 35, 28, 31)
p = c(1/6, 2/6, 2/6, 1/6)
n = sum(obs)
exp = n*p
chi = sum(((obs-exp)^2)/exp)
chi

[1] 16.35714

obs

[1] 32 35 28 31

exp

[1] 21 42 42 21

To get the p-value:

The p-value is, \(P \left(\chi^2_{df}> \text{observed} \right)\)

chsq_object$p.value

[1] 0.0009579516

Same as:

Note that the p-value depends on the \(\text{test statistics}\) (\(\chi^2_3 = 16.35714\)), \(\text{degrees of freedom}\) (3). We also use the distribution function pchisq() for the chi-squared distribution in R.

1-pchisq(16.35714, 3)

[1] 0.0009579529

To get the degrees of freedom:

The degree of freedom is \(k-1\).

chsq_object$parameter

df 
 3 

# to remove name df
unname(chsq_object$parameter)

[1] 3

Same as:

k = length(obs)
k-1

[1] 3