Riemann surface
matlab

Exercise 2.1 Cultivated surface area
We want to estimate the surface area cultivated on the farms of a rural township. Of the $N=2010$ farms that comprise the township, we select 100 using simple random sampling. We measure $y_k$, the surface area cultivated on the farm $k$ in hectares, and we find
$$\sum_{k \in S} y_k=2907 \text { ha and } \sum_{k \in S} y_k^2=154593 \mathrm{ha}^2 .$$

Give the value of the standard unbiased estimator of the mean
$$\bar{Y}=\frac{1}{N} \sum_{k \in U} y_k$$

Give a $95 \%$ confidence interval for $\bar{Y}$.

Solution
In a simple design, the unbiased estimator of $\bar{Y}$ is
$$\widehat{\bar{Y}}=\frac{1}{n} \sum_{k \in S} y_k=\frac{2907}{100}=29.07 \text { ha. }$$
The estimator of the dispersion $S_y^2$ is
$$s_y^2=\frac{n}{n-1}\left(\frac{1}{n} \sum_{k \in S} y_k^2-\widehat{\bar{Y}}^2\right)=\frac{100}{99}\left(\frac{154593}{100}-29.07^2\right)=707.945 .$$
82 Simple Random Sampling
The sample size $n$ being ‘sufficiently large’, the $95 \%$ confidence interval is estimated in hectares as follows:
\begin{aligned} {\left[\hat{\bar{Y}} \pm 1.96 \sqrt{\frac{N-n}{N} \frac{s_y^2}{n}}\right] } & =\left[29.07 \pm 1.96 \sqrt{\frac{2010-100}{2010} \times \frac{707.45}{100}}\right] \ & =[23.99 ; 34.15] \end{aligned}

Exercise 2.2 Occupational sickness
We are interested in estimating the proportion of men $P$ affected by an occupational sickness in a business of 1500 workers. In addition, we know that three out of 10 workers are usually affected by this sickness in businesses of the same type. We propose to select a sample by means of a simple random sample.

What sample size must be selected so that the total length of a confidence interval with a 0.95 confidence level is less than 0.02 for simple designs with replacement and without replacement ?

What should we do if we do not know the proportion of men usually affected by the sickness (for the case of a design without replacement) ?
To avoid confusions in notation, we will use the subscript $W R$ for estimators with replacement, and the subscript $W O R$ for estimators without replacement.

Solution

a) Design with replacement.
If the design is of size $m$, the length of the (estimated) confidence interval at a level $(1-\alpha)$ for a mean is given by
$$\mathrm{CI}(1-\alpha)=\left[\hat{\bar{Y}}-z_{1-\alpha / 2} \sqrt{\frac{\tilde{s}y^2}{m}}, \hat{\bar{Y}}+z{1-\alpha / 2} \sqrt{\frac{\tilde{s}y^2}{m}}\right],$$ where $z{1-\alpha / 2}$ is the quantile of order $1-\alpha / 2$ of a random normal standardised variate. If we denote $\widehat{P}{W R}$ as the estimator of the proportion for the design with replacement, we can write \begin{aligned} & C I(1-\alpha)=\left[\widehat{P}{W R}-z_{1-\alpha / 2} \sqrt{\frac{\widehat{P}{W R}\left(1-\hat{P}{W R}\right)}{m-1}} .\right. \ & \left.\widehat{P}{W R}+z{1-\alpha / 2} \sqrt{\frac{\widehat{P}{W R}\left(1-\widehat{P}{W R}\right)}{m-1}}\right] \ & \end{aligned}

Exercise 2.2
9
Indeed, in this case,
$$\widehat{\operatorname{var}}\left(\widehat{P}{W R}\right)=\frac{\widehat{P}{W R}\left(1-\widehat{P}{W R}\right)}{(m-1)} .$$ So that the total length of the confidence interval does not exceed 0.02 , it is necessary and sufficient that $$2 z{1-\alpha / 2} \sqrt{\frac{\hat{P}{W R}\left(1-\widehat{P}{W R}\right)}{m-1}} \leq 0.02 .$$
By dividing by two and squaring, we get
$$z_{1-\alpha / 2}^2 \frac{\widehat{P}{W R}\left(1-\widehat{P}{W R}\right)}{m-1} \leq 0.0001,$$
which gives
$$m-1 \geq z_{1-\alpha / 2}^2 \frac{\widehat{P}{W R}\left(1-\widehat{P}{W R}\right)}{0.0001}$$
For a $95 \%$ confidence interval, and with an estimator of $P$ of 0.3 coming from a source external to the survey, we have $z_{1-\alpha / 2}=1.96$, and
$$m=1+1.96^2 \times \frac{0.3 \times 0.7}{0.0001}=8068.36 .$$
The sample size $(m=8069)$ is therefore larger than the population size, which is possible (but not prudent) since the sampling is with replacement.

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