Question

Please only do questions e and f

2. Many electrical components were packed up for shipment overseas, with 5 items per box. Unfortunately the shipment was exposed to a fairly high dose of ionising radiation. The owners were concerned that some of the components could have been damaged. On arrival at their destination a random sample of 60 boxes was taken. Each box in the sample was opened and the number of defective components was tested. The results are given in the table below. Number of defectives 0 1 2 3 4 5 Total Number of boxes (fi) Pi 0.0102 0.0768 0.2304 0.3456 0.2592 0.0778 1 please do not write here Answer the following questions to test whether the number of defective com- ponents per box, X, could have a binomial distribution with n = 5 and some (unknown) parameter p, which is the probability that a randomly selected com- ponent was damaged during shipment. (a) List the conditions required for a random variable to have a binomial dis- tribution. (b) Calculate the sample mean and sample standard deviation for the number of defective components per box. (c) Estimate the parameter(s) of the binomial distribution that best fits these data. (Hint: remember that the 60 boxes orginally contained 300 undam- aged electrical components in total.) (d) Write down the formula to calculate the binomial probability P(X = 3) = 0.3456, as given in the table. Show clearly how to use the answer to (c) to calculate this probability. (e) The Pi-values given in the table were obtained as in part (d). Use those Pi-values to: i. calculate the odds that (X = 3); iii. calculate the corresponding expected values (ei). Write the expected (f) Now do a goodness-of-fit test on these data, using a 5% level of signifi- cance, to test whether it is plausible that the number of defective compo- nents per box could have a binomial distribution. (25 marks)

Answer 1

2.

Number of defectives (x)	Number of boxes ($\small f_i$)	Expected probability ($\small p_i$)	Expected frequency ($\small e_i$)
0	1	0.0102	1
1	6	0.0768	4
2	14	0.2304	14
3	16	0.3456	21
4	17	0.2592	15
5	6	0.0778	5
Total	60	1.0000	60

(e)

(i)

Odds that (X=3) is given by

P(X 3) 0,3456 - 0.5281 P(X 3) 1 0.3456

(ii)

P(X 3) 0.3456 P(X 3X> 2)= 0.506299 P(X 2) 0.6826

$\small \therefore P(X\neq3|X>2)=1-P(X=3|X>2)=0.493701$

Odds that (X=3) given that (X>2)

$\small =\frac{P(X=3|X>2)}{P(X\neq3|X>2)}=1.0255$

Or, it could be calculated directly as

$\small \frac{P(X=3)}{P(X=4)+P(X=5)}=1.0255$

(iii)

Expected frequencies are written in the last column of the table by multiplying 60 with corresponding expected probabilities and then rounding to integral values (keeping in mind that the sum of expected frequencies must be 60).

(f)

We perform chi-squared test for goodness of fit.

We have to test for null hypothesis $\small H_0:$ The data follows binomial distribution

against the alternative hypothesis $\small H_1:$ The data does not follow binomial distribution

Degrees of freedom, v = Number of values of x - 1 = 6-1 = 5    [Here 6 as X takes 6 different values 0, 1, 2, 3, 4 & 5]

Test statistic is given by

   $\small \chi^2=\sum_{i=0}^{5}\frac{\left ( f_i-e_i \right )^2}{e_i}$

$\small \therefore \chi^2_{calculated}=2.657$

Level of significance $\small \alpha=0.05$

Critical value $\small \chi^2_{critical}=\chi^2_{\alpha,v}=\chi^2_{0.05,5}=11.070$

We reject null hypothesis if $\small \chi^2_{calculated}>\chi^2_{critical}$ .

But here we see, $\small \chi^2_{calculated}=2.657\ngtr 11.070\chi^2_{critical}$ and so we do not reject the null hypothesis.

Based on given data, we conclude that the given set of data follows binomial distribution.