Exam-Style Problem

9231 P23 - Jun 2018 - Q11O - 14 marks

6199

Question 11 OR alternative.

A scientist carries out an experiment to investigate the quantity \(X\), which takes the values \(0,1,2,3,4,5\) or \(6\). He believes that the values taken by \(X\) follow a binomial distribution. He conducts \(250\) trials. His results are summarised in the following table.

\(x\)	0	1	2	3	4	5	6
Observed frequency	22	83	72	53	17	3	0

(i) Show that unbiased estimates of the mean and variance for these results are \(1.876\) and \(1.266\) respectively, correct to 3 decimal places. By evaluating the mean and variance of the distribution \(B(6,0.313)\), explain why \(X\) could have this distribution.

The expected frequencies corresponding to the distribution \(B(6,0.313)\) are shown in the following table.

\(x\)	0	1	2	3	4	5	6
Observed frequency	22	83	72	53	17	3	0
Expected frequency	26.3	71.9	81.8	49.7	17.0	3.1	0.2

(ii) Show how the expected frequency for \(x=4\) is calculated.

(iii) Test at the \(5\%\) significance level whether the scientist's belief is correct.

Solution Checked by expert

Answer: The sample estimates are \(1.876\) and \(1.266\), the expected frequency for \(x=4\) is about \(17.0\), and the test does not reject the binomial model.

(i) First calculate the sample mean. The total of \(x\times\text{frequency}\) is

\[0(22)+1(83)+2(72)+3(53)+4(17)+5(3)+6(0)=469.\]

There are \(250\) trials, so

\[\bar x=\frac{469}{250}=1.876,\]

as required.

Next calculate

\[\sum x^2f=0^2(22)+1^2(83)+2^2(72)+3^2(53)+4^2(17)+5^2(3)+6^2(0)=1195.\]

The unbiased variance estimate is

\[s^2=\frac{1195-469^2/250}{249}=1.266\ldots.\]

So the unbiased estimate of the variance is \(1.266\), correct to 3 decimal places.

For \(X\sim B(6,0.313)\),

\[E(X)=np=6(0.313)=1.878,\]

and

\[\operatorname{Var}(X)=np(1-p)=6(0.313)(0.687)=1.290\ldots.\]

These are close to the sample mean \(1.876\) and sample variance \(1.266\). Therefore \(B(6,0.313)\) is a reasonable possible model for \(X\).

(ii) For a binomial distribution with \(n=6\) and \(p=0.313\),

\[P(X=4)=\binom64(0.313)^4(0.687)^2.\]

There are \(250\) observations, so the expected frequency for \(x=4\) is

\[250\binom64(0.313)^4(0.687)^2.\]

This is

\[250(0.06795\ldots)=16.99\ldots,\]

so the expected frequency is approximately \(17.0\).

(iii) Test

\[H_0:\text{the binomial distribution }B(6,0.313)\text{ fits the data}.\]

The expected frequencies for \(x=5\) and \(x=6\) are less than \(5\), so combine the last three cells into \(x\geq4\). This gives

\[\begin{array}{c|ccccc} \text{class} & 0 & 1 & 2 & 3 & \geq4\\ \hline O & 22 & 83 & 72 & 53 & 20\\ E & 26.3 & 71.9 & 81.8 & 49.7 & 20.3 \end{array}\]

Now calculate

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

\[\chi^2=\frac{(22-26.3)^2}{26.3}+\frac{(83-71.9)^2}{71.9}+\frac{(72-81.8)^2}{81.8}+\frac{(53-49.7)^2}{49.7}+\frac{(20-20.3)^2}{20.3}.\]

Therefore

\[\chi^2=3.81\quad\text{approximately}.\]

There are \(5\) combined classes, and one parameter has been estimated, so the degrees of freedom are

\[5-1-1=3.\]

At the \(5\%\) significance level,

\[\chi^2_{3,0.95}=7.815.\]

Since

\[3.81<7.815,\]

we do not reject \(H_0\). There is no evidence against the binomial model, so the scientist's belief is consistent with the data at the \(5\%\) significance level.