Question: A random sample of size was drawn from a population of size . The following measurements were obtained:

a. Estimate with an approximate 95% confidence interval.

b. Estimate p, the proportion of population measurements greater than 30, with an approximate 95% confidence interval.

The given data is as follows,

Also, the sample size n=30,

The population size N=300,

Mean can be obtained using the following formula,

$\overline{x}=\frac{1}{n}\left[x_1+x_2+...+x_{10}\right]$

Therefore

$$\begin{gathered} \overline{x}=\frac{1}{30}\left[21+33+...+42\right] \\ =\frac{1081}{30} \\ =36.0333 \end{gathered}$$

Now, the standard deviation is calculated using the following formula,

$s=\sqrt{\frac{{\sum }_{i=1}^{30}{\left(x-\overline{x}\right)}^2}{n-1}}$

Therefore,

$$\begin{gathered} =\sqrt{\frac{\left[{\left(21-36.0333\right)}^2+{\left(33-36.0333\right)}^2+...+{\left(42-36.0333\right)}^2\right]}{30-1}} \\ =9.8172 \end{gathered}$$

Let’s calculate the value of$\frac{n}{N}$. If it is greater than 0.05, then the finite population correction factor should be included in the standard error calculation.

$$\begin{gathered} \frac{n}{N}=\frac{30}{300} \\ =0.1 \end{gathered}$$

Here,$\frac{n}{N}=0.1>0.05$the finite population correction factor should be included in the standard error calculation.\

The 95% confidence interval is obtained by using the following formula:

$\overline{x}\pm 2\frac{s}{\sqrt{n}}\sqrt{\frac{N-n}{N}}$

Therefore,

$$\begin{gathered} \overline{x}\pm 2\frac{s}{\sqrt{n}}\sqrt{\frac{N-n}{N}}=36.0333\pm 2\times \frac{9.8172}{\sqrt{30}}\times \sqrt{\frac{300-30}{300}} \\ =36.0333\pm 3.40 \\ =\left(32.63,39.43\right) \end{gathered}$$

Hence, the required confidence interval is (32.63, 39.43).

Let x denotes the number of measurements that are greater than 30. That is

The proportion of measurements that are greater than 30 is

$$\begin{gathered} \hat{p}=\frac{21}{30} \\ =0.7 \end{gathered}$$

The 99% confidence interval for p is obtained by using the following formula:

$$\begin{gathered} \pm 2\times \sqrt{\frac{\hat{p}\left(1-\right)}{n}}\times \sqrt{\frac{N-n}{N}}=0.7\pm 2\times \sqrt{\frac{0.7\left(1-0.7\right)}{30}}\times \sqrt{\frac{300-30}{300}} \\ =0.7\pm 0.159 \\ =\left(0.541,0.859\right) \end{gathered}$$

Hence, the required confidence interval is (0.541,0.859).