Chisquare test of goodness of fit
Chisquare test of goodness of fit is a nonparametric test.
One of the principle use of $\chi^2$distribution is to test how well an observed distribution fits to a theoretical one. That is, the chisquare test of goodness of fit enables us to compare the distribution of classes of observations with an expected distribution.
In the test of hypothesis it is usually assumed that the random variable follows a particular distribution like Binomial, Poisson, Normal etc.
To test whether the assumption about the distribution is true, Chisquare test of goodness of fit is performed and the decision about whether the data follows a particular distribution will be taken.
Assumptions
 Observations are independent
 Expected frequencies must be at least 5.
The chisquare goodness of fit test is not applicable if the expected frequencies are too small ( < 5).
Step by Step procedure for Chisquare test of goodness of fit
The step by step procedure for chisquare goodness of fit test is as follows:
Step 1 : Setup the null and alternative hypothesis
The null hypothesis for test of goodness of fit is
$H_0:$ There is no significant difference between the observed and expected values. (That is, data fits well to the assumed distribution (like Binomial, Poisson, Normal, etc.))
The alternative hypothesis is
$H_1:$ There is a significant difference between the observed and expected values. (That is, data does not fit well to the assumed distribution)
Step 2 : Define the test statistic
The test statistic for testing the above hypothesis testing problem is
$$ \begin{equation*} \chi^2= \sum \frac{(O_{i} E_{i})^2}{E_{i}} \sim \chi^2_{(nk1)}\\ \end{equation*} $$
where
 $O_i=f_i$ observed frequency for $i^{th}$ category or $i^{th}$ value of $x$,
 $E_i$ expected frequencies according to the assumed distribution,
 $n$ is the number of categories after pooling,
 $k$ is the number of parameters estimated.
Step 3 : Sepcify the level of significance.
Specify the level of significance $\alpha$ ($0 < \alpha < 1$)
Step 4 : Calculate the test statistic
Estimate the parameter(s) (if any).

Assuming that the null hypothesis is true, estimate the parameter(s) of the distribution.

Using the estimated parameter(s) and the assumed distribution determine the probabilities, say, $P(X=x)$ for each value of $x$.

Compute the expected frequencies on multiplying the probabilities by total number of observations ($N$). That is $E_i=N*P(X=x_i)$.

If any of the expected frequency is less than five then pooled the frequency to adjust the categories or values of $x$. After pooling the frequency we will get the number of pooled categories $n$.
Compute the test statistic
$$ \begin{equation*} \chi^2_{obs}= \sum \frac{(O_{i} E_{i})^2}{E_{i}} \end{equation*} $$
Step 5 Critical value of Chisquare
The degrees of freedom for the chisquare test of goodness of fit is $nk1$.
The table value of $\chi^2$ for $nk1$ degrees of freedom and at $\alpha$ level of significance is $\chi^2_t=\chi^2_{nk1,\alpha}$
.
Step 6 (Traditional approach)
If $\chi^2_{obs}\leq \chi^2_t$, then fail to reject the null hypothesis $H_0$ at $\alpha$ level of significance, i.e., the data fits well to the assumed distribution, other wise reject $H_0$ at $\alpha$ level of significance.
OR
Step 6 Decision (pvalue approach)
The pvalue of the test is
$$ p = P(\chi^2_{nk1}\geq\chi^2_{obs}) $$
If $p$value of the test is less than $\alpha$, then reject the null hypothesis $H_0$ at $\alpha$ level of significance, otherwise fail to reject $H_0$ at $\alpha$ level of significance.
Chisquare test of goodness of fit Example 1
To test whether a die is fair, 60 rolls were made, and the corresponding outcomes were as follows:
Face Value  1  2  3  4  5  6 

Observed freq.  5  7  17  16  8  7 
Solution
The observed data is
Face Value  Obs. Freq.$(O_i)$  $p_i$  Expe.Freq.$(E_i)$ 

1  5  0.1666667  10 
2  7  0.1666667  10 
3  17  0.1666667  10 
4  16  0.1666667  10 
5  8  0.1666667  10 
6  7  0.1666667  10 
Total  60  1.0000000  60 
Step 1 Setup the Null and alternative hypothesis
The null and alternative hypothesis are as follows:
$H_0:p_{1}=p_{2} =\cdots = p_{6} =\frac{1}{6}$
$H_1:$ At least one of the proportion is different from $\frac{1}{6}$.
Step 2 Test statistic
The test statistic for testing above hypothesis is
$$ \begin{equation*} \chi^2= \sum \frac{(O_{i} E_{i})^2}{E_{i}} \sim \chi^2_{nk1}\\ \end{equation*} $$
Step 3 Specify the level of significance
The level of significance be $\alpha = 0.05$.
Step 4 Calculate the Test Statistic
As there is no parameter to be estimated, $k=0$.
Under the null hypothesis $p_i=\frac{1}{6}$ for all $i$.
The expected frequencies can be calculated as
$$ \begin{aligned} E_{i} &=N*P(X=x_i)\\ &=N*p_i \end{aligned} $$
For example, $E_{1}$ is given by
$$ \begin{eqnarray*} E_{1} & = &N*p_1\\ &=& 60*0.1667\\ &=&10. \end{eqnarray*} $$
Face Value  Obs. Freq.$(O_i)$  $p_i$  Expe.Freq.$(E_i)$  $(O_iE_i)^2/E_i$ 

1  5  0.1666667  10  2.5 
2  7  0.1666667  10  0.9 
3  17  0.1666667  10  4.9 
4  16  0.1666667  10  3.6 
5  8  0.1666667  10  0.4 
6  7  0.1666667  10  0.9 
Total  60  1.0000000  60  13.2 
The test statistic is
$$ \begin{eqnarray*} \chi^2_{obs}&=& \sum \frac{(O_{i} E_{i})^2}{E_{i}} \sim \chi^2_{(k1)}\\ &=&\frac{(510)^2}{10}+\frac{(710)^2}{10}+ \frac{(1710)^2}{10}\\ & & +\frac{(1610)^2}{10}+\frac{(810)^2}{10}+ \frac{(710)^2}{10}\\ &=& 2.5 + 0.9+ 4.9+ 3.6+ 0.4 + 0.9\\ &=& 13.2. \end{eqnarray*} $$
Step 5 Critical value of Chisquare
The degrees of freedom for the chisquare test of goodness of fit is $df=nk1=6  0 1 = 5$.
The table value of $\chi^2$ for $n1$ degrees of freedom and at $\alpha$ level of significance is $\chi^2_t=\chi^2_{nk1,\alpha}=\chi^2_{5,0.05}=11.0705$
.
Step 6 (Traditional approach)
The test statistic is $\chi^2_{obs} =13.2$ which falls $inside$ the critical region bounded by the critical value $11.0705$, we $\textit{reject}$ the null hypothesis.
OR
Step 6 Decision ($p$value approach)
The estimate of $p$value is $P(\chi^2_{5}>13.2) =0.02157$.
Chisquare test of goodness of fit Example 2
A dropin auto repair shop staffs the same number of mechanics on every weekday (weekends are not counted here). One of the mechanics thinks this is a bad idea because he suspects the number of customers is not evenly distributed across these days. For a sample of 289 customers, the counts by weekday are given in the table.
Number of Customers by Day (n = 289)
Day  Monday  Tuesday  Wednesday  Thursday  Friday 

Count  51  68  57  67  46 
Test the claim that the number of customers is not evenly distributed across the five weekdays. Test this claim at the 0.05 significance level.
Solution
The observed data is
Day  Obs. Freq.$(O_i)$  Prop. $p_i$ 

Monday  51  0.2 
Tuesday  68  0.2 
Wednesday  57  0.2 
Thursday  67  0.2 
Friday  46  0.2 
Step 1 Setup the Null and alternative hypothesis
The null and alternative hypothesis are as follows:
$H_0:p_{Mon}=p_{Tue} =p_{Wed} =p_{Thu} = p_{Fri}=1/5$
$H_1:$ The number of customers is not evenly distributed across the five weekdays.
$H_1:$ At least one of the proportion is different from $\frac{1}{5}$.
Step 2 Test statistic
The test statistic for testing above hypothesis is
$$ \begin{equation*} \chi^2= \sum \frac{(O_{i} E_{i})^2}{E_{i}} \sim \chi^2_{nk1}\\ \end{equation*} $$
Step 3 Specify the level of significance
The level of significance be $\alpha = 0.05$.
Step 4 Calculate the Test Statistic
As there is no parameter to be estimated, $k=0$.
The test statistic for testing above hypothesis is
$$ \begin{aligned} \chi^2& = \sum \frac{(O_{i} E_{i})^2}{E_{i}} \sim \chi^2_{(k1)}\\ \end{aligned} $$
The expected frequencies can be calculated as
$$ \begin{aligned} E_{i} =N*p_i \end{aligned} $$
For example, $E_{1}$ is given by
$$ \begin{aligned} E_{1} & = N*p_1\\ &= 289*0.2\\ &=57.8. \end{aligned} $$
$E_{2}$ is given by
$$ \begin{aligned} E_{2} & = N*p_2\\ &= 289*0.2\\ &=57.8. \end{aligned} $$
$E_{3}$ is given by
$$ \begin{aligned} E_{3} & = N*p_3\\ &= 289*0.2\\ &=57.8. \end{aligned} $$
$E_{4}$ is given by
$$ \begin{aligned} E_{4} & = N*p_4\\ &= 289*0.2\\ &=57.8. \end{aligned} $$
$E_{5}$ is given by
$$ \begin{aligned} E_{5} & = N*p_5\\ &= 289*0.2\\ &=57.8. \end{aligned} $$
Day  Obs. Freq.$(O_i)$  Prop. $p_i$  Expe.Freq.$(E_i)$  $(O_iE_i)^2/E_i$ 

Monday  51  0.2  57.8  0.8 
Tuesday  68  0.2  57.8  1.8 
Wednesday  57  0.2  57.8  0.011 
Thursday  67  0.2  57.8  1.464 
Friday  46  0.2  57.8  2.409 
The test statistic is
$$ \begin{aligned} \chi^2_{obs}&= \sum \frac{(O_{i} E_{i})^2}{E_{i}} \sim \chi^2_{(k1)}\\ &=\frac{(5157.8)^2}{57.8}+\frac{(6857.8)^2}{57.8}+\frac{(5757.8)^2}{57.8}\\ &\quad +\frac{(6757.8)^2}{57.8}+ \frac{(4657.8)^2}{57.8}\\ &= 0.8 +1.8 +0.011+ 1.464+2.409\\ &= 6.484. \end{aligned} $$
Step 5 Critical value of Chisquare
The degrees of freedom for the chisquare test of goodness of fit is $df=nk1=5  0 1 = 4$.
The table value of $\chi^2$ for $n1$ degrees of freedom and at $\alpha$ level of significance is $\chi^2_t=\chi^2_{nk1,\alpha}=\chi^2_{4,0.05}=9.4877$
.
Step 6 (Traditional approach)
The test statistic is $\chi^2_{obs} =6.484$ which falls $outside$ the critical region bounded by the critical value $9.4877$, we $\textit{fail to reject}$ the null hypothesis.
OR
Step 6 Decision ($p$value approach)
The estimate of $p$value is $P(\chi^2_{4}>6.484) =0.1658$.
We conclude that there is not enough data to support the claim that the number of customers is not evenly distributed across the five weekdays.
Chisquare test of goodness of fit Example 3
A doctor believes the number of births by season is uniformly distributed. To test this claim you randomly select 2246 births and record the season in which each takes palce. The results are shown below. At $\alpha = 0.01$, can you reject the claim that the distribution is uniform?
Season  Spring  Summer  Winter  Fall 

Births  564  602  555  525 
Solution
Assuming the uniform distribution the proportions are $p_{Spring}=p_{Summer}=p_{Winter}=p_{Fall}=1/4 = 0.25$
.
The observed data is
Season  Obs. Freq.$(O_i)$  Prop.$p_i$ 

Spring  564  0.25 
Summer  602  0.25 
Winter  555  0.25 
Fall  525  0.25 
Step 1 Setup the Null and alternative hypothesis
The null and alternative hypothesis are as follows:
$H_0:p_{Spring}=p_{Summer}=p_{Winter}=p_{Fall}=1/4 = 0.25$
(i.e., The number of births by season is uniformly distributed.)
$H_1:$ The number of births by season is not uniformly distributed.
Step 2 Test statistic
The test statistic for testing above hypothesis is
$$ \begin{equation*} \chi^2= \sum \frac{(O_{i} E_{i})^2}{E_{i}} \sim \chi^2_{nk1}\\ \end{equation*} $$
Step 3 Specify the level of significance
The level of significance be $\alpha = 0.01$.
Step 4 Calculate the Test Statistic
As there is no parameter to be estimated, $k=0$.
The test statistic for testing above hypothesis is
$$ \begin{aligned} \chi^2& = \sum \frac{(O_{i} E_{i})^2}{E_{i}} \sim \chi^2_{(k1)}\\ \end{aligned} $$
The expected frequencies can be calculated as
$$ \begin{aligned} E_{i} =N*p_i \end{aligned} $$
For example, $E_{1}$ is given by
$$ \begin{aligned} E_{1} & = N*p_1\\ &= 2246*0.25\\ &=561.5. \end{aligned} $$
$E_{2}$ is given by
$$ \begin{aligned} E_{2} & = N*p_2\\ &= 2246*0.25\\ &=561.5. \end{aligned} $$
$E_{3}$ is given by
$$ \begin{aligned} E_{3} & = N*p_3\\ &= 2246*0.25\\ &=561.5. \end{aligned} $$
$E_{4}$ is given by
$$ \begin{aligned} E_{4} & = N*p_4\\ &= 2246*0.25\\ &=561.5. \end{aligned} $$
Season  Obs. Freq.$(O_i)$  Prop. $p_i$  Expe.Freq.$(E_i)$  $(O_iE_i)^2/E_i$ 

Spring  564  0.25  561.5  0.011 
Summer  602  0.25  561.5  2.921 
Winter  555  0.25  561.5  0.075 
Fall  525  0.25  561.5  2.373 
The test statistic is
$$ \begin{aligned} \chi^2_{obs}&= \sum \frac{(O_{i} E_{i})^2}{E_{i}} \sim \chi^2_{(k1)}\\ &=\frac{(564561.5)^2}{561.5}+\frac{(602561.5)^2}{561.5}\\ &\quad +\frac{(555561.5)^2}{561.5}+ \frac{(525561.5)^2}{561.5}\\ &= 0.011 +2.921 +0.075+ 2.373\\ &= 5.38. \end{aligned} $$
Step 5 Critical value of Chisquare
The degrees of freedom for the chisquare test of goodness of fit is $df=nk1=4  0 1 = 3$.
The table value of $\chi^2$ for $nk1$ degrees of freedom and at $\alpha$ level of significance is $\chi^2_t=\chi^2_{nk1,\alpha}=\chi^2_{3,0.01}=11.3449$
.
Step 6 (Traditional approach)
The test statistic is $\chi^2_{obs} =5.38$ which falls $outside$ the critical region bounded by the critical value $11.3449$, we $\textit{fail to reject}$ the null hypothesis.
OR
Step 6 Decision ($p$value approach)
The estimate of $p$value is $P(\chi^2_{3}>5.38) =0.14599$.
We conclude that the number of births by season is uniformly distributed at $\alpha = 0.01$ level of significance.
Chisquare test of goodness of fit Example 4
Three similar coins are tossed 100 times and the number of heads are recorded as follows:
Number of heads  Frequency 

0  32 
1  42 
2  20 
3  6 
Fit a binomial distribution to the above data and test its goodness of fit.
Solution
Step 1 : Setup the null and alternative hypothesis
The null hypothesis for test of goodness of fit is
$H_0:$ The number of heads follows a Binomial distribution.
The alternative hypothesis is
$H_1:$ The number of heads do not follows a Binomial distribution.
Step 2 : Define the test statistic
We use the chisquare test of goodness of fit for testing hypothesis testing problem. The chisquare test statistic for testing above hypothesis testing problem is
$$ \begin{equation*} \chi^2= \sum \frac{(O_{i} E_{i})^2}{E_{i}} \sim \chi^2_{(nk1)}\\ \end{equation*} $$
where
 $O_i=f_i$ observed frequency for $i^{th}$ category or $i^{th}$ value of $x$,
 $E_i$ expected frequencies according to the assumed distribution,
 $n$ is the number of categories after pooling,
 $k$ is the number of parameters estimated.
Step 3 : Sepcify the level of significance.
The level of significance $\alpha=0.05$.
Step 4 : Calculate the test statistic
x  Freq.$(f)$  f*x 

0  32  0 
1  42  42 
2  20  40 
3  6  18 
Total  100  100 
First estimate the parameter of Poisson distribution. The population mean of Binomial distribution with number of trials $n$ and probability of success (Head) $p$ is $E(X)=n\times p$. The sample mean of observed data is $\overline{X}=\frac{1}{N}\sum f_ix_i$.
The sample mean is
$$ \begin{aligned} \overline{X}&=\frac{1}{\sum f_i}\sum f_ix_i\\ &=\frac{100}{100}\\ &=1 \end{aligned} $$
By the method of moments, the parameter $p$ can be estimated as
$$ \begin{aligned} \hat{p} &= \frac{\overline{X}}{n}\\ &=\frac{1}{4}\\ &=0.25 \end{aligned} $$
The probability mass function of Binomial distribution is given by
$$ \begin{aligned} P(X=x)&= \binom{n}{x}p^xq^{1x},\\ &\quad x=0,1,2,\cdots, n;\\ &\quad 0 < p < 1, q=1p \end{aligned} $$
We have the estimated value of $p$ as $\hat{p}=0.25$.
The expected probabilities for different values of $x$ can be obtained on substituting $n=4$ and $\hat{p}=0.25$ in above probability mass function.
For example, the expected probability for $x=0$ is
$$ \begin{aligned} P(X=0)&= \binom{4}{0}0.25^{0}0.75^{30},\\ &=0.3164 \end{aligned} $$
Similarly the expected probabilities for other values of $x$ can be calculated using the probability mass function.
The expected frequency for $x=0$ can be calculated as
$$ \begin{aligned} E_{1} &=N*P(X=0)\\ &=100* 0.3164\\ &=31.64. \end{aligned} $$
Similarly the other expected frequencies can be calculated using the formula $E_i= N*P(X=x_i)$.
The expected probabilities and expected frequencies are as follows:
x  Freq.$(O_i)$  $P(X=x)$  Expe.Freq.$(E_i)$  $(O_iE_i)^2/E_i$ 

0  32  0.3164  31.64  0.004 
1  42  0.4219  42.19  0.001 
2  20  0.2109  21.09  0.056 
3  6  0.0469  4.69  0.366 
Compute the test statistic as
$$ \begin{aligned} \chi^2_{obs}&= \sum \frac{(O_{i} E_{i})^2}{E_{i}}\\ &=\frac{(3231.64)^2}{31.64}+\frac{(4242.19)^2}{42.19}+\frac{(2021.09)^2}{21.09}+ \frac{(64.69)^2}{4.69}\\ &= 0.004 +0.001 +0.056+ 0.366\\ &= 0.427. \end{aligned} $$
Step 5 Critical value of Chisquare
The degrees of freedom for the chisquare test of goodness of fit is $df=nk1=4  1 1 = 2$.
The table value of $\chi^2$ for $nk1$ degrees of freedom and at $\alpha$ level of significance is $\chi^2_t=\chi^2_{nk1,\alpha}=\chi^2_{2,0.05}=5.9915$
.
Step 6 (Traditional approach)
The test statistic is $\chi^2_{obs} =0.427$ which falls $outside$ the critical region bounded by the critical value $5.9915$, we $\textit{fail to reject}$ the null hypothesis.
OR
Step 6 Decision (pvalue approach)
The pvalue is $P(\chi^2_{2} > 0.427) =0.80775$.
As the pvalue $0.8078$ is $\textit{greater than}$ the significance level of $\alpha = 0.05$, we $\textit{fail to reject}$ the null hypothesis.
Interpretation
We conclude that given data fits well to the Binomial distribution.
Chisquare test of goodness of fit Example 5
The following table contains data on number of complaints received per day at a major retail bank's branches:
No. of Complaints  Frequency 

0  270 
1  140 
2  65 
3  14 
4 +  5 
Fit a Poisson distribution and test to see if it is consistent with the data.
Solution
Step 1 : Setup the null and alternative hypothesis
The null hypothesis for test of goodness of fit is
$H_0:$ The number of complaints follows a Poisson distribution.
The alternative hypothesis is
$H_1:$ The number of complaints do not follows a Poisson distribution.
Step 2 : Define the test statistic
We use the chisquare test of goodness of fit for testing hypothesis testing problem. The chisquare test statistic for testing above hypothesis testing problem is
$$ \begin{equation*} \chi^2= \sum \frac{(O_{i} E_{i})^2}{E_{i}} \sim \chi^2_{(nk1)}\\ \end{equation*} $$
where
 $O_i=f_i$ observed frequency for $i^{th}$ category or $i^{th}$ value of $x$,
 $E_i$ expected frequencies according to the assumed distribution,
 $n$ is the number of categories after pooling,
 $k$ is the number of parameters estimated.
Step 3 : Sepcify the level of significance.
The level of significance $\alpha=0.05$.
Step 4 : Calculate the test statistic
x  Freq.$(f)$  f*x 

0  270  0 
1  140  140 
2  65  130 
3  14  42 
4  5  20 
Total  494  332 
First estimate the parameter of Poisson distribution. The population mean of Poisson distribution with parameter $\lambda$ is $E(X)=\lambda$. The sample mean of observed data is $\overline{X}=\frac{1}{N}\sum f_ix_i$. By the method of moments, the parameter $\lambda$ can be estimated as
$$ \begin{aligned} \hat{\lambda} &= \overline{X}\\ &=\frac{1}{\sum f_i}\sum f_ix_i\\ &=\frac{332}{494}\\ &=0.6721 \end{aligned} $$
The probability mass function of Poisson distribution with parameter $\lambda$ is given by
$$ \begin{aligned} P(X=x)&= \frac{e^{\lambda}\lambda^x}{x!},\\ &\quad x=0,1,2,\cdots; \lambda > 0 \end{aligned} $$
We have the estimated value of $\lambda$ is $\hat{\lambda}=0.6721$.
The expected probabilities for different values of $x$ can be obtained on substituting $\hat{\lambda}=0.6721$ in above probability mass function.
For example, the expected probability for $x=0$ is
$$ \begin{aligned} P(X=0)&= \frac{e^{0.6721}0.6721^0}{0!},\\ &=0.5106 \end{aligned} $$
Similarly the expected probabilities for other values of $x$ can be calculated using the probability mass function.
The expected frequency for $x=0$ can be calculated as
$$ \begin{aligned} E_{1} &=N*P(X=0)\\ &=494* 0.5106\\ &=252.2364. \end{aligned} $$
Similarly the other expected frequencies can be calculated using the formula $E_i= N*P(X=x_i)$.
The expected probabilities and expected frequencies are as follows:
x  Freq.$(O_i)$  $P(X=x)$  Expe.Freq.$(E_i)$  $(O_iE_i)^2/E_i$ 

0  270  0.5106  252.2364  1.251 
1  140  0.3432  169.5408  5.147 
2  65  0.1153  56.9582  1.135 
3  14  0.0258  12.7452  0.124 
4  5  0.0043  2.1242  3.893 
Total  494  0.9992  493.6048  11.550 
Compute the test statistic as
$$ \begin{aligned} \chi^2_{obs}&= \sum \frac{(O_{i} E_{i})^2}{E_{i}}\\ &=\frac{(270252.2364)^2}{252.2364}+\frac{(140169.5408)^2}{169.5408}+\frac{(6556.9582)^2}{56.9582}\\ &\quad + \frac{(1412.75)^2}{12.75}+ \frac{(52.12)^2}{2.12}\\ &= 1.251 +5.147 +1.135+ 0.124+ 3.893\\ &= 11.55. \end{aligned} $$
Step 5 Critical value of Chisquare
The degrees of freedom for the chisquare test of goodness of fit is $df=nk1=4  1 1 = 2$.
The table value of $\chi^2$ for $nk1$ degrees of freedom and at $\alpha$ level of significance is $\chi^2_t=\chi^2_{nk1,\alpha}=\chi^2_{2,0.05}=5.9915$
.
Step 6 (Traditional approach)
The test statistic is $\chi^2_{obs} =11.55$ which falls $inside$ the critical region bounded by the critical value $5.9915$, we $\textit{reject}$ the null hypothesis.
OR
Step 6 Decision (pvalue approach)
The pvalue is $P(\chi^2_{2} > 11.55) =0.0031$.
As the pvalue $0.0031$ is $\textit{less than}$ the significance level of $\alpha = 0.05$, we $\textit{reject}$ the null hypothesis.
Conclusion
We conclude that the number of complaints do not follows a Poisson distribution.
Endnote
In this tutorial, you learned the chisquare test of goodness of fit. You also learned about the step by step procedure to apply chisquare test of goodness of fit. The step by step solved examples on chisquare test of goodness of fit helps you understand the chisquare test of goodness of fit.
To learn more about other parametric and nonparametric test please refer to the following tutorials:
Let me know in the comments if you have any questions on chisquare test for goodness of fit and your thought on this article.