In: Statistics and Probability
8) Accelerated lifetime tests are carried out for a batch of 200 thermocouples. The mean lifetime is found to be 936 h with a standard deviation of 28 h. The smallest and largest measurements in the sample are 860 and 1012 h. The measurements are divided into 9 data bins with boundaries at 859.5, 876.5, 893.5, 910.5, 927.5, 944.5, 961.5, 978.5, 995.5, 1012.5. The measurement count in bin 1 from 869.5 days to 886.5 days is 3 and the count in the other successive bins is 8, 27, 39, 46 40, 24, 11 and 2. Apply the chi-squared test to see whether the measurements fit a Gaussian distribution to a a) 99% confidence level, b) a 97.5% confidence level, and c) a 95% confidence level.
Let X denotes Lifetime for a batch of thermocouples
N = 200
E(X) = 936 and S.D(X) = 28
By central lilmit theorem

From the information
| Class Interval | Observed Frequency | Probability | Expected Frequecy = N * Probability | Approximate | 
| 859.5-876.5 | 3 | P( 859.5 < X < 876.5) = 0.01364 | 2.7294 | 3 | 
| 876.5-893.5 | 8 | P( 876.5 < X < 893.5) = 0.04773 | 9.5464 | 10 | 
| 893.5-910.5 | 27 | P(893.5 < X < 910.5) =0.11669 | 23.3396 | 23 | 
| 910.5-927.5 | 39 | P( 910.5 < X < 927.5) = 0.19950 | 39.9008 | 40 | 
| 927.5-944.5 | 46 | P(927.5 < X < 944.5) = 0.23854 | 47.7091 | 48 | 
| 944.5-961.5 | 40 | P( 944.5 < X < 961.5) = 0.19950 | 39.9008 | 40 | 
| 961.5-978.5 | 24 | P(961.5 < X < 978.5) = 0.11669 | 23.3396 | 23 | 
| 978.5-995.5 | 11 | P(978.5 < X < 999.5) = 0.04773 | 9.5464 | 10 | 
| 995.5-1012.5 | 2 | P( 995.5 < X < 1012.5) = 0.01364 | 2.7294 | 3 | 
We have whether the measurement fit a Gaussian distribution(Normal).
we use Chi-square goodness of fit test.
The value of test statistic is

Where Oi : Observed frequency , Ei - Expected Frequency and n = number of classes = 9
| Observed Frequency | Expected Frequency | (Oi-Ei)^2 /Ei | 
| 3 | 3 | 0 | 
| 8 | 10 | 0.4 | 
| 27 | 23 | 0.69565217 | 
| 39 | 40 | 0.025 | 
| 46 | 48 | 0.08333333 | 
| 40 | 40 | 0 | 
| 24 | 23 | 0.04347826 | 
| 11 | 10 | 0.1 | 
| 2 | 3 | 0.33333333 | 
| 200 | 200 | 1.6807971 | 
The value of Chi-square statistic

Critical Values for 8 degrees of freedom for different level of significance
| Confidence Interval | Level of significance | Critical Value | 
| 99% | 0.01 | 20.0902 | 
| 97.50% | 0.025 | 17.5345 | 
| 95% | 0.05 | 15.5073 | 
From the above table we see that all critical values is greater than value of test statistics. we accept Ho at 1%, 2.5% and 5% level of significance.
Conclusion : Measurement fit a Gaussian distribution.