Management collected the number of carryout orders at Ashoka

Management collected the number of carryout orders at Ashoka Curry Express on Friday nights. They are interested in the null hypothesis that the number of carryout orders follows a normal distribution, and decide to use a Chi-Square goodness of fit test to check this hypothesis. The data was placed into one of eight different bins (each having the same probability), the sample mean and standard deviation were computed, the observed and expected frequencies were computed, and preliminary computations for the Chi-Square test statistic were performed..
Bins   Max   Observed Frequency   Expected Frequency   (f-e)^2/e
1 19.8 6 5.25 0.11
2   24.5   6   5.25   0.11
3   28.9   6   5.25   0.11
4   31.2   3   5.25   0.96
5   34.3   5   5.25   0.01
6   37.8   6   5.25   0.11
7   42.5   3   5.25   0.96
8   99999999   7   5.25   0.58
Calculate the Chi-Square critical value. Assume that alpha is equal to 0.05 and round your answer to the nearest whole number.

Solution

^2 Test Statistic = 2.95
^2 degrees of freedom = k-1 = 8 - 1 = 7, ^2 tab at 0.05 LOS is = 14.07
^2 p_value =0.8893      
| ^2 tab | > | ^2 cal | i.e. Do n\'t Reject Ho  

[ANSWER] Chi-Square critical value = 14.07