Question

In a random sample of 250 people from a city, 148 of them favor apples over other fruits.

a. Use a 90% confidence interval to estimate the true proportion p of people in the population who favor apples over other fruits.

b. How large a sample would be needed to estimate p to be within .15 with 90% confidence?

Short answer

a.The true proportion of measurements in the population with characteristics A has the confidence interval (0.6429, 0.5411).

b. The sample size needed to estimate p is 30.

Step-by-step solution

Given information

The sample size is $n=250$

Let x denotes the measurements possess the characteristics of A, i.e.$x=148$

Calculating the Confidence Interval to estimate the true proportion p

a.

The 90% confidence interval to estimate the true proportion of measurements in the population with characteristics A, is as follows

Since, the proportion of the characteristics of A is$\hat{p}=\frac{x}{n}$

Therefore,

$\begin{array}{rcl}\backslash \hat{p}& =& \frac{x}{n}\\ & =& \frac{148}{250}\\ & =& 0.592\\ & & \end{array}$

Now, the 90% confidence interval for p is given as follows:

$\hat{p}\pm z_{\alpha /2}\sqrt{\frac{\hat{p}\hat{q}}{n}}$

The critical value of z at 90% confidence level, $z_{\alpha /2}=1.645$(from z-table)

Hence, the 90% confidence interval for p is calculated as follows:

$\hat{p}\pm z_{\alpha /2}\sqrt{\frac{\hat{p}\hat{q}}{n}}$

$$\begin{gathered} =0.592\pm 1.645\left(\sqrt{\frac{0.592\left(1-0.592\right)}{250}}\right) \\ =0.592\pm 1.645\times \sqrt{\frac{0.592\left(0.408\right)}{250}} \\ =0.592\pm 1.645\times \sqrt{0.000966144} \\ =0.592\pm 1.645\times 0.0310 \\ =0.592\pm 0.0509 \\ =\left(0.6429,0.5411\right) \end{gathered}$$

Hence, the true proportion of measurements in the population with characteristics A has the confidence interval (0.6429, 0.5411).

Calculating the Sample size

b.

The sampling error is $SE=0.15$

The proportion of the characteristics of A is $\hat{p}=0.592$

The sample size n is obtained by using the following formula:

$n=\frac{{\left(z_{\alpha /2}\right)}^2\left(\hat{p}\hat{q}\right)}{{\left(SE\right)}^2}$

$\begin{array}{rcl}n& =& \frac{{\left(1.645\right)}^2\left(0.592\left(1-0.592\right)\right)}{{\left(0.15\right)}^2}\\ & =& \frac{2.706\times 0.592\times 0.408}{0.0225}\\ & =& 29.04\\ & & \end{array}$

Hence, the sample size needed to estimate p is 30.