Question

Out of 300 people sampled, 111 preferred Candidate A. Round to three decimals.Based on this estimate, what proportion (as a decimal) of the voting population (p) prefers Candidate A?Compute a 90% confidence interval, and give your answers to 3 decimal places.

Answer 1

We are given that out of 300 people, 111 preferred candidate A. The proportion that prefer candidate A is the number of people that prefer candidate A divided by the total population, like this:

$P=\frac{111}{300}$

Solving the operation:

$P=0.37$

Therefore, the proportion that prefers candidate A is 0.37.

To determine the confidence interval we use the following formula:

$CI=(P-z\sqrt{\frac{P(1-P)}{n}},P+z\sqrt{\frac{P(1-P)}{n}})$

Where:

$\begin{gathered} P=\text{ proportion} \\ n=\text{ size of the population} \\ z=\text{ critical value} \end{gathered}$

The critical value for a 90% confidence interval is:

$z=1.64$

Now, we substitute the values in the formula:

$CI=(0.37-1.64\sqrt{\frac{(0.37)(1-0.37)}{300}},0.37+1.64\sqrt{\frac{(0.37)(1-0.37)}{300}})$

Solving the operations:

$CI=(0.324,0.416)$

And thus we get a confidence interval of 90% confidence.