Template:Chi-squared test for grouped data

Chi-Squared Test for Grouped Data

A Chi-Squared goodness-of-fit test is used to test the null hypothesis that the Crow-AMSAA reliability model adequately represents a set of grouped data. The expected number of failures in the interval from [math]\displaystyle{ {{T}_{i-1}} }[/math] to [math]\displaystyle{ {{T}_{i}} }[/math] is approximated by:

[math]\displaystyle{ {{\widehat{\theta }}_{i}}=\hat{\lambda }\left( T_{i}^{{\hat{\beta }}}-T_{i-1}^{{\hat{\beta }}} \right) }[/math]

For each interval, [math]\displaystyle{ {{\widehat{\theta }}_{i}} }[/math] shall not be less than 5 and, if necessary, adjacent intervals may have to be combined so that the expected number of failures in any combined interval is at least 5. Let the number of intervals after this recombination be [math]\displaystyle{ d }[/math] and let the observed number of failures in the [math]\displaystyle{ {{i}^{th}} }[/math] new interval be [math]\displaystyle{ {{N}_{i}} }[/math] and let the expected number of failures in the [math]\displaystyle{ {{i}^{th}} }[/math] new interval be [math]\displaystyle{ {{\widehat{\theta }}_{i}} }[/math] . Then the following statistic is approximately distributed as a Chi-Squared random variable with degrees of freedom [math]\displaystyle{ d-2 }[/math] .

[math]\displaystyle{ {{\chi }^{2}}=\underset{i=1}{\overset{d}{\mathop \sum }}\,\frac{{{({{N}_{i}}-{{\widehat{\theta }}_{i}})}^{2}}}{{{\widehat{\theta }}_{i}}} }[/math]

The null hypothesis is rejected if the [math]\displaystyle{ {{\chi }^{2}} }[/math] statistic exceeds the critical value for a chosen significance level. This means that the hypothesis that the Crow-AMSAA model adequately fits the grouped data shall be rejected. Critical values for this statistic can be found in tables of the Chi-Squared distribution.

Example 6
An aircraft has scheduled inspections at intervals of 20 flight hours. Table 5.4 gives the data set from the first 100 hours of flight testing.

Table 5.4 - Grouped test data

Start Time	End Time	Number of Failures
0	20	13
20	40	16
40	60	5
60	80	8
80	100	7

1) Estimate the parameters of the Crow-AMSAA model using maximum likelihood estimation.

2) Evaluate the goodness-of-fit.

Solution

1) Obtain the estimator of [math]\displaystyle{ \beta }[/math] using Eqn. (vv). Using RGA the value of [math]\displaystyle{ \widehat{\beta } }[/math] is 0.75285. Now plug this value into Eqn. (vv1) and [math]\displaystyle{ \widehat{\lambda } }[/math] is:

[math]\displaystyle{ \widehat{\lambda }=1.52931 }[/math]

2) There are a total of observed failures from [math]\displaystyle{ d=5 }[/math] intervals. Table 5.5 shows that those adjacent intervals do not have to be combined after applying Eqn. (seta) to the original intervals.

Table 5.5 - Observed vs. Expected Number of Failures for Grouped data

Start Time	End Time	Observed Number of Failures	Expected Number of Failures
0	20	13	14.59
20	40	16	9.99
40	60	5	8.77
60	80	8	8.07
80	100	7	7.58

To test the model's goodness-of-fit, a Chi-Squared statistic of 5.45 is compared to the critical value of 7.8 corresponding to 3 degrees of freedom and a 0.05 significance level. Since the statistic is less than the critical value, the applicability of the Crow-AMSAA model is accepted.