Exam-Style Problem

9231 P41 - Nov 2022 - Q2 - 8 marks

6780

An organisation runs courses to train students to become engineers. These students are taught in groups of 8. The director of the organisation claims that on average 60% of the students in a group achieve a pass. A random sample of 150 groups of 8 students is chosen. The following table shows the observed frequencies together with some of the expected frequencies using the appropriate binomial distribution.

Number of passes per group	0	1	2	3	4	5	6	7	8
Observed frequency	0	0	8	24	45	36	26	10	1
Expected frequency	\(p\)	1.180	6.193	18.579	34.836	\(q\)	\(r\)	13.437	2.519

(a) Find the values of \(p\), \(q\) and \(r\), giving your answers correct to 3 decimal places.

(b) Carry out a goodness of fit test, at the 10% significance level, to test whether there is evidence to reject the director's claim.

Solution

Answer: (a) \(p=0.098\), \(q=41.804\), \(r=31.353\).

(b) Using a \(\chi^2\) goodness-of-fit test with suitable grouping, the test statistic is \(\chi^2=7.84\) with \(5\) degrees of freedom. Since \(7.84 \lt 9.236\) at the 10% significance level, we do not reject \(H_0\).

There is insufficient evidence to say the data does not fit \(\mathrm{B}(8,0.6)\), so the director's claim is consistent with the observations.

(a) The number of passes in a group of 8 is modelled by \(X\sim \mathrm{B}(8,0.6)\).

For 150 groups, the expected frequency for \(X=r\) is

\(150\times \Pr(X=r)=150\times \binom{8}{r}(0.6)^r(0.4)^{8-r}\).

\(p=150\times \Pr(X=0)=150\times (0.4)^8=150\times 0.00065536=0.098304\).
Hence \(p=0.098\) to 3 d.p.
\(q=150\times \Pr(X=5)=150\times \binom{8}{5}(0.6)^5(0.4)^3\)
\(=150\times 56\times 0.07776\times 0.064=41.803776\).
Hence \(q=41.804\).
\(r=150\times \Pr(X=6)=150\times \binom{8}{6}(0.6)^6(0.4)^2\)
\(=150\times 28\times 0.046656\times 0.16=31.352832\).
Hence \(r=31.353\).

(b) We test

\(H_0\): the data follow a binomial distribution \(\mathrm{B}(8,0.6)\).

\(H_1\): the data do not follow a binomial distribution \(\mathrm{B}(8,0.6)\).

The observed and expected frequencies are:

Number of passes	0	1	2	3	4	5	6	7	8
Observed	0	0	8	24	45	36	26	10	1
Expected	0.098	1.180	6.193	18.579	34.836	41.804	31.353	13.437	2.519

For a valid \(\chi^2\) test, expected frequencies should not be too small, so we combine the first three categories and the last two categories:

Grouped category	0,1,2	3	4	5	6	7,8
Observed	8	24	45	36	26	11
Expected	\(0.098+1.180+6.193=7.471\)	18.579	34.836	41.804	31.353	\(13.437+2.519=15.956\)

Now calculate

\(\chi^2=\sum \frac{(O-E)^2}{E}\).

This gives

\(\chi^2=\frac{(8-7.471)^2}{7.471}+\frac{(24-18.579)^2}{18.579}+\frac{(45-34.836)^2}{34.836}+\frac{(36-41.804)^2}{41.804}+\frac{(26-31.353)^2}{31.353}+\frac{(11-15.956)^2}{15.956}\)

\(=0.0374+1.5817+2.9655+0.8058+0.9139+1.5394\)

\(=7.8437\approx 7.84\).

There are \(6\) grouped categories, and no parameters have been estimated from the data, so

degrees of freedom \(=6-1=5\).

At the 10% significance level, the critical value for \(\chi^2\) with \(5\) degrees of freedom is \(9.236\).

Since \(7.84 \lt 9.236\), we do not reject \(H_0\).

So there is insufficient evidence to conclude that the distribution of passes differs from \(\mathrm{B}(8,0.6)\). The director's claim is consistent with the data.