Question

A random sample of thirty firms dealing in wireless products were selected to determine the proportion of such firms that have implemented new software to improve productivity. It turned out that eight of the thirty have implemented such software. Compute a 95% confidence interval on p, the true proportion of such firms that have implemented new software. In other situation, suppose there is concern about whether or not the point estimate of p = 8/30 is accurate enough because the confidence interval around p is not sufficiently narrow. By using p as your estimate of p, calculate how many companies would need to be sampled to have a 95% confidence interval with a width of only 0.05.
(10 marks)

Answer 1

The point estimate of $p$ is $\hat{p}={8 \over30}$.

If $\hat{p}$ is the proportion of successes in a random sample of size $n$ and $\hat{q}=1-\hat{p},$ then an approximate $100(1-\alpha)\%$ confidence interval ( CI ), for the binomial parameter $\hat{h}$ is given by

$$\hat{p}\pm z_{\alpha/2}\cdot \sqrt{{\hat{p}(1-\hat{p}) \over n}}$$

where $\hat{p}$ is the sample proportion, $n$ is the sample size, and $z_{\alpha/2}$is the appropriate value from the standard normal distribution for our desired confidence interval.

Given that $\hat{p}=8/30, n=30, \alpha=0.05.$

Check

$$n\hat{p}=30\cdot{8 \over30}=8>5$$ $$n(1-\hat{p})=30\cdot(1-{8 \over30})=22>5$$

Determine $z_{\alpha/2}$

$$z_{\alpha/2}=z_{0.05/2}=1.96$$

Compute a 95% confidence interval on p

$$CI=(\hat{p}- z_{\alpha/2}\cdot \sqrt{{\hat{p}(1-\hat{p}) \over n}}, \hat{p}+ z_{\alpha/2}\cdot \sqrt{{\hat{p}(1-\hat{p}) \over n}})$$

$$CI=\bigg({8 \over30}- 1.96\sqrt{{{8 \over30}(1-{8 \over30}) \over 30}}, {8 \over30}+ 1.96 \sqrt{{{8 \over30}(1-{8 \over30}) \over 30}}\bigg)$$

$$CI=(0.108, 0.425)$$

We shall calculate how many companies would need to be sampled to have a 95% confidence interval with a width $w$ of only 0.05. 

We must choose $n$ such that

$$z_{\alpha/2}\cdot \sqrt{{\hat{p}(1-\hat{p}) \over n}}\leq w$$

Solve for $n$

$${\hat{p}(1-\hat{p}) \over n}\leq{(z_{\alpha/2})^2 \over w^2}$$

$$n\geq{(z_{\alpha/2})^2 \hat{p}(1-\hat{p})\over w^2}$$

Given that

$$\hat{p}={8 \over 30}, \alpha=0.05, z_{\alpha/2}=z_{0.05/2}=1.96, w=-.05$$

Substitue

$$n\geq{(1.96)^2 ({8 \over 30})(1-{8 \over 30})\over (0.05)^2}$$

$$n\geq301$$

We need to sample at least 301 companies to have a 95% confidence interval with a width of only 0.05.