I. (10 points) Nylon bars were tested for brittleness. Each of 280 bars was mold

Question

I. (10 points) Nylon bars were tested for brittleness. Each of 280 bars was molded under similar conditions and was tested in n=5 places. Assuming that each bar has uniform composition, the number of breaks on a given bar should be binomially distributed with five trials and an unknown probability p of failure. If the bars are all of the same uniform strength, p should be the same for all of them; if they are of different strengths, p should vary from bar to bar. Thus, the null hypothesis is that the p's are all equal. The following table summarizes the outcome of the experiment: Breaks/Bar Frequency 157 69 35 17 0 Pooling the last three cells, test the agreement of the observed frequency distribution with the binomial distribution using Pearson's chi-square test

Explanation / Answer

Question 1.

H0 : probability of failure are all equal. p = p0

Ha probability of failure of each bar is different p p0

Here first we will identify the estimate of proportion of failure p^ = 199/(280 * 5) = 0.1421

Here expected value = 280 * BIN(X ; 280; 0.1421)

X2 = 36.95

so for dF = 3 and p - value = CHITEST = 0.0000

so which is less than standard significance level alpha = 0.05 or 0.01 so we shall reject the null hypothesis and can say that the distribution doesn't follow binomial distribution.

QUestion 2

H0 : Number of accidents follow poisson distibution.

Ha : Number of accidents doesn't follow poisson distribution.

Now we have to find the poission parament lambda for number of accidents per month

Here on average accidents per month = 16/32 = 0.5

so Now expected number of months for X number of accidents = POISSON (X ; 0.5) * 32

Now observed expected table

X2 = 6.98

dF = 4 -1 = 3

so p - value = Pr (X2 > 6.93 ; 3) = 0.0724 > 0.05

so we shall not reject the null hypothesis and can conclude that the accident rate follow poisson distribution.

Breaks/bar Frequency (f) 0 157 0 1 69 69 2 35 70 3 17 51 4 1 4 5 1 5 sum 280 199

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