# Using Chi-Square to test normality. This is a sample question we received. I can't really figure out how to statistically show that this data is normally distributed. We are to used the chi-square method and these are the steps we are to follow:

1. Use the hypothesized distribution to create expected values for each bin.
2. Use the observed and expected values to create a chi-squared value.
3. Compare this computed chi-squared value with the chi-squared value from a table, using the appropriate number of degrees of freedom for this problem and the significance level.
4. Based on the relationship between the computed chi-squared and the table's chi-squared values, accept or reject the hypothesis.

I have never used this method and was wondering if someone could show me how to work through a problem like this. Thanks!

The chi-squared test is pretty easy to use:

You are told that the engineer has determined that the mean is 10mm and stdev is 0.1 mm. This is all the information you need to specify a normal distribution.

1. The chi-squared test is based on turning your null hyopthesis into a multinomial distribution, which is a generalization of the binomial distribution (where there are only 2 outcomes). Here, the possibilities are your "bins", where the probability assigned to each "bin" $p_i$ is equal to the probability that a normally distirbuted random variable's value would fall in that bin. As a rule of thumb, if you have N data points, you want to size your bins such tat $Np_i \geq 5$.
2. Once you have your bins, you calculate the expected number of observations you would have in each bin $E_i$, assuming that your data actually come from a $Normal(10mm,(0.1mm)^2)$ distribution: $E_i=Np_i$.
3. Now you just need to count the actual number of observations in each bin ($O_i$).
4. For each bin, you want to form the statistic: $S_i=\frac{(O_i-E_i)^2}{E_i}$
5. Calculate the Chi-squared statistic $\chi^2 = \sum S_i$
6. Now here comes the theoretical part: It turns out that when the data actually do come from the multinoimial distribution you constructed (using the underlying hypothesized normal distribution) then the distribution of $\chi^2$ is asymptotically $\chi^2_{k-1}$ where k is the number of bins.
7. Now, just compare the value of the $\chi^2$ to the $1-\alpha$ percentile of the $\chi^2_{k-1}$ distirution to see if you will reject (i.e., $\chi^2 > 1-\alpha$ percentile of the $\chi^2_{k-1}$)
• Thank you for the help. I have done everything up to point 7. I am not sure where to find the 95th percentile of the X^2(k-1) distribution. I have my X^2 but do not know what to compare it to. Thanks again! – user126582 Feb 10 '14 at 17:09
• @user126582 Tables or computer options are readily available: keisan.casio.com/exec/system/1180573197 – user76844 Feb 10 '14 at 17:32