The number of accidents Y per week at an intersection was...

The number of accidents Y per week at an intersection was checked for n = 50 weeks, with the results as shown in the table below. Test the hypothesis that the random variable Y has a Poisson distribution where ?= 0.48, assuming the observations to be independent. Use ? = 0.05. y | Frequency 0 | 32 1 | 12 2 or more | 6

Transcript

00:01 Hello students, in order to test the hypothesis, the random variable y has a poisson distribution with mean lambda which is 0 .48 and we use a chi -square test for goodness of fit and then our hypothesis assume that the observation frequencies follows the poisson distribution with mean lambda is equals to 0 .48 and the alternative hypothesis assumes that the observation frequencies do not follow the poisson distribution with a specified mean.

00:29 So, let us conduct the chi -square distribution and h0 as already said it states that lambda is equals to 0 .48 and follows poisson distribution and h1 states that lambda is equals to 0 .48, but does not follow poisson distribution.

00:53 So, in the next step let us find the significance level alpha is equals to 0 .05 which is 5 % los and now let us calculate the expected frequencies.

01:06 For y is a poisson random variable with mean 0 .48, so we calculate the expected frequencies for each category that is for 0, 1, 2 or more.

01:18 So, we have the formula for that as p of y is equals to k is equals to e power minus lambda into lambda to the power k by k factorial.

01:33 So, for this formula we keep expected value as y is equals to 0, y is equals to 1 and 2 or more.

01:41 So, for p of y is equals to 0 is equals to e power minus 0 .48 into 0 .48 to the power 0 by 0 factorial.

01:59 So, this gives 0 .6201 as the answer.

02:05 So, in the same way we need to calculate for y is equals to 1, 2 and 2 or more and this value gives 0 .2976 and for probability of y greater than or equals to 2 this is 1 minus p of y is equals to 0 minus p of y is equals to 1.

02:29 So, this gives 0 .0823 as the answer.

02:34 So, now let us calculate the expected frequencies by multiplying the probabilities by the total number of weeks which is n is equals to 50.

02:44 So, the expected frequencies are for y of 0 it is 31 .005 and for y of 1 it is 14 .880 and for y of 2 it is 2 or more it is 4 .115.

03:09 So, let us set up the observed and expected frequencies and find the chi -square test...

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