Question

Small Sample:

Answer 1

Answer:

The 80% confidence interval for the proportion of companies that are planning to increase their workforce is (0.033, 0.331).

A sample size of 8852 is needed.

Step-by-step explanation:

First question:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

Only 2 of the 11 companies were planning to increase their workforce.

This means that $n = 11, \pi = \frac{2}{11} = 0.1818$

80% confidence level

So $\alpha = 0.2$, z is the value of Z that has a pvalue of $1 - \frac{0.2}{2} = 0.9$, so $Z = 1.28$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1818 - 1.28\sqrt{\frac{0.1818*0.8182}{11}} = 0.033$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1818 + 1.28\sqrt{\frac{0.1818*0.8182}{11}} = 0.331$

The 80% confidence interval for the proportion of companies that are planning to increase their workforce is (0.033, 0.331).

Second question:

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

A poll taken in July 2010 estimates this proportion to be 0.36.

This means that $\pi = 0.36$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.01?

This is n for which M = 0.01. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.01 = 1.96\sqrt{\frac{0.36*0.64}{n}}$

$0.01\sqrt{n} = 1.96\sqrt{0.36*0.64}$

$\sqrt{n} = \frac{1.96\sqrt{0.36*0.64}}{0.01}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.36*0.64}}{0.01})^2$

$n = 8851.04$

Rounding up(as a sample of 8851 will have a margin of error slightly over 0.01):

A sample size of 8852 is needed.