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:


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*Use the hypothesized distribution to create expected values for each bin.

*Use the observed and expected values to create a chi-squared value. 

*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.

*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!
 A: 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.


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*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 $Np_i \geq 5$.

*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: $E_i=Np_i$.

*Now you just need to count the actual number of observations in each bin ($O_i$).

*For each bin, you want to form the statistic: $S_i=\frac{(O_i-E_i)^2}{E_i}$

*Calculate the Chi-squared statistic $\chi^2 = \sum S_i$

*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.

*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 $\chi^2_{k-1}$)

