Past Paper Example

Q4, (Jan 2008, Q4a)

In Germany, towards the end of the nineteenth century, a study was undertaken into the distribution of the sexes in families of various sizes. The table shows some data about the number of girls in 500 families, each with 5 children. It is thought that the binomial distribution B(5, p) should model these data.

Number of girls	Number of families
0	32
1	110
2	154
3	125
4	63
5	16

i) Use this information to calculate an estimate for the mean number of girls per family of 5 children. Hence show that 0.45 can be taken as an estimate of p.

ii) Investigate at a 5% significance level whether the binomial model with p estimated as 0.45 fits the data. Comment on your findings and also on the extent to which the conditions for a binomial model are likely to be met. [12 marks]

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Solution: i)

$$\Sigma xf = 32 \times 0 + 110 \times 1 + 154 \times 2 + 125 \times 3 + 63 \times 4 + 16 \times 5 = :highlight[1125]\\ \Sigma f = 500 \quad \text{so} \quad \bar{x} = \frac{1125}{500} = :highlight[2.25]$$$$\mu = np \quad \Rightarrow \quad 2.25 = 5p \quad \text {(Since n=5 in this binomial model)}\\ \Rightarrow \quad p = \frac{2.25}{5} = :success[0.45] \quad \text{(as required)}$$

Since we have estimated one of the population parameters, this means we have one less degree of freedom. Remember this point when checking critical values from the table.

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Solution: ii)

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Step 1: State hypotheses:

• $H_0$: The proposed model fits the data well.
• $H_1$: The proposed model does not fit the data well.

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Step 2: Using the proposed model: Calculate the proportion of the total frequency associated with each outcome.

Using $X \sim B(5, 0.45)$:

$P(X = 0)$ $= 0.05033$

$P(X = 1)$ $= 0.2059$

$P(X = 2)$ $= 0.3369$

$P(X = 3)$ $= 0.2957$

$P(X = 4)$ $= 0.1128$

$P(X = 5)$ $= 0.01845$

Now, dividing the total frequency with the proportions, we get expectations:

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Expectations:

| Number of girls | Number of families | | |---|---|---|---| | 0 | 25.165 | $\leftarrow \quad 500 \times 0.05033$ | | 1 | 102.95 | $\leftarrow \quad 500 \times 0.2059$ | | 2 | 168.45 | $\leftarrow \quad 500 \times 0.3369, \ etc.$ | | 3 | 137.85 | | | 4 | 56.4 | | | 5 | 9.225 | |

Notice no expectations < 5 $\therefore$ no combining of items.

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Observations:

Number of girls	Number of families
0	32
1	110
2	154
3	125
4	63
5	16

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Step 3: Calculate $\chi^2_{\text{calc}}$ where contributions are calculated by:

$$\text{Contributions} = \frac{(O - E)^2}{E}$$$$\chi^2_{\text{calc}} = \frac{(32 - 25.165)^2}{25.165} + \frac{(110 - 102.95)^2}{102.95} + \frac{(154 - 168.45)^2}{168.45}+\frac{(125 - 137.85)^2}{137.85} + \frac{(63 - 56.4)^2}{56.4} + \frac{(16 - 9.225)^2}{9.225}$$

Note: If asked to analyze contributions, it is necessary to calculate the value of each individual contribution.

$$= 1.856 + 0.4828 + 1.240 + 1.198 + 0.7723 + 4.976$$$$\approx :highlight[10.525]$$

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Step 4: Check the critical value and conclude appropriately.

• Remember:

$$v = (\text{no. of items after combining}) - (\text{no. of estimated parameters}) - 1$$

$v = 6 - 1 - 1 = :highlight[4] \quad \text{(because we estimated } p)$

Critical Value (C.V.) from the table:

$$\text{C.V.} = :highlight[9.488]$$

Calculated value:

$$:highlight[10.525 > 9.488]$$

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Conclusion: Reject H₀.

The binomial model is not a good fit for the data.

In the proposed model, we seem to underestimate in the extremes and overestimate in the middle.

The biggest contribution is for $X = 5$ , indicating that this model is a poor fit, especially at the right-hand tail.

Within a family, the sex of one child may not be statistically independent of a previously born child. Also, the probability of giving birth to a girl is unlikely to be $0.45$ across all families. Therefore, the binomial model may not be appropriate.