Chapter 10 - The CHI-Squared Distribution Notes

CH 10: The CHI-Squared Distribution We consider only three applicationg of the chi-squared distribution "goodness of fit tests tests of association in contingency trables", and tests and confidence intervals for sample variance Although the rationale behind the first two opplications is completely different, the calculation of the test statistic is almost identical. Goodness of fit tests To do the two-sample t-test, we found we had to make the assumption that the variances were equal Another underlying assumption for the t-test was that the samples were drawn from populations which had normal distributions. There is therefore a need to be able to test this assumption Not only can we test if the process that generated the data has a normal distribution, we con test whether data fits any specified distribution Example random In printing industry, misprints are thought to occur at (independently of one another ) Thus the number of misprints can be expected to abey a poisson distribution with some parameter A To test this, 200 pages are examined and number of misprints on each page are noted A 5% significance level is used. Number of misprints Observed number of pages. 1 43 2 69 3 93 4 21 5 In order to test whether the poisson distribution fits this data, the first problem 16 to decide what value to use for the parameter 2 of the Poisson Distribution. This can either be specified by the null hypothesis, or the data can be used to estimate 2. We treat these two situations seperately. Let us first consider the test of the null hypothesis that the data can be thought of as a sample from a Poisson Distribution with parameter A=1-2 misprints per page against the alternative that the data comes from some other distribution. Ha Data comes from a poisson distribution with 2=12 ② H1 The distribution is not poisson with 12 3 significance level: 5% + First we need to compute the expected (theoretical) frequencies, assuming that the null hypothesis is true. If misprints are occuring in accordance with a Poisson distribution with rate 12 then the probability that a page contains x misprints is p(x)= Thus the probability of no misprints is p(o) e 0 3012 A sample of 200 pages can therefore be expected to have 60.24 pages with no misprints (30.12% of 200). This theoretical frequency is to be compared with an observed frequency of 43. Similarly p(1) e 1.2 0.3619 Thus the expected frequency of one error 15 200 x 0.361472, 28. Compare this to the observed 69 pages with one error no of misprints (x) observed (O) theoretical probability p(x) expected (Ei) 1 43 12° xe/0! = 0-3012 12 xe/11 03614 60-29 2 69 53 12 xe /2! = 0.2169 12 1.2 xe/31 = 0.0867 72 28 3 21 e²/4! 1.2 xe *0.0260 43 38 17.34 8 1- Σχο ρ(x)=0.0078 5.20 5 or more 6 1-56 Even if Ho is true, ive anticipate that the observed and expected frequencies will not be exactly equal We would clearly like to reject Ho, however, if the difference between the observed frequencies and the expected frequencies is "too large We need to find a test statistic which is a function of the differences between observed and expected frequencies and which has a known sampling distribution The sum of the differences is of no use because they sum to zero We divide the squared differences by their expected frequency. (oi - Ei) 02- Σ E The Statistic D has approximately a chi-squared (x²) distribution. It has degrees of freedom attached to it. The correct degrees of freedom is given by k-1, where 'k" is the number of cells into which the data are categorized Here k 6 and therefore of for X is 5. Thus we reject Ho if D exceeds the 5% point of X's distribution From the table 11-071 (005) 0.151 f(x) 0-10 0.05 0 Σ (oi-Ei) 60-24 22 17 If D exceeds 11.071 then the observed and expected frequencies are too far opart for thear differences to be explained by chance. sampling fluctuations alone. (43-60.24) (69-72.28) 72.28 (6-1.56) 1-56 This value lies in the rejection region. We thus reject the null hypothesis that the samples, follow a distribution with = 1-2 Poisson Remember D' has approximately a x² distribution of the expected frequencies exceed about 5. In the table above we see that "5 or more' does not fit this condition and so we amalgamate the adjoining cases, it is now #4 or more We repeat steps 4 and 5 with the new figures and conclude based off of the correct darta. In this case, the conclusion remains the same Let's use the same data to estimate 2 this makes to the test. and see what difference Ho the dota fits some Poisson distribution ② H1 the data fits a distribution other than the Poisson distribution ③significance level 5% to find we need to estimate the rate at which misprints occurred in our Sample data. The total number of misprints that occurred was (0x43) + (1x69) + (2x53)+ +(sx) 300 misprints 300 in 200 pages implies misprints accur at 1.5 per page Degrees of freedom k-d-1 where k is the number of cells and d is the number of parameters estimated from data. Here k=5 and d=1, because we estimated just one parameter, 2, from the data. Thus we must use x with 6-1-13 degrees of freedom The 5% point of x 15 7.815 We reject Ho if D exceeds 7.815 Notice that goodness of fit tests are intrinsically one-sided - we reject Ho if D' is too large If D is small, it means that the distribution specified by Ho fits the data very well 0-1 (0;-E:) Σ (43-44.62) (14-13.14) 13.74 1.01 6. 1.01 lies in acceptance region, we can't reject Ho. It is reasonable to conclude Poisson distribution with 2=15