Question

In a random sample of 400 registrered voters, 120 indicated they plan to vote for Candidate A. Determine a 95% confidence interval for the proportion of all the registered voters who will vote for Candidate A.

Answer 1

Concepts and reason

Sampling Distribution of Proportion: The random sample of size n is taken from the population with sample proportion $\hat p$ .

The sampling distribution of proportion has mean $p$ and the standard deviation $\sqrt {\frac{{p\left( {1 - p} \right)}}{n}}$ . Moreover, the sample distribution of proportion follows normal distribution for large sample size n.

Confidence interval: A range of values such that the population parameter can expected to contain for the given confidence level is termed as the confidence interval. In other words, it can be defined as an interval estimate of the population parameter which is calculated for the given data based on a point estimate and for the given confidence level.

Moreover, the confidence level indicates the possibility that the confidence interval can contain the population parameter. Usually, the confidence level is denoted by . The value is chosen by the researcher. Some of the most common confidence levels are 90%, 95%, and 99%.

Fundamentals

Formula for sample proportion is $\hat p = \frac{x}{n}$ , x represents the successes and n is the sample size.

The formula for the confidence interval for the proportion is,

${\rm{Confidence interval}} = \hat p \pm {Z_{\frac{\alpha }{2}}}\sqrt {\frac{{\hat p\left( {1 - \hat p} \right)}}{n}}$

Where,

$\begin{array}{c}\\\hat p:\,\,{\rm{Sample proportion}}\\\\n:\,\,{\rm{Sample size}}\\\\{Z_{\frac{\alpha }{2}}}{\rm{:}}\,{\rm{Critical value}}\\\end{array}$

The proportion of all the registered voters who will vote for Candidate A is obtained as below:

From the given information, a random sample of 400 registered voters, 120 indicated they plan to vote for Candidate. That is, $n = 400{\rm{ and }}x = 120$ .

The required proportion is,

$\begin{array}{c}\\\hat p = \frac{{120}}{{400}}\\\\ = 0.3\\\end{array}$

The value of ${Z_{\frac{\alpha }{2}}}$ is obtained as shown below:

Consider, the confidence level is 0.95.

$\begin{array}{c}\\{\rm{For}}\left( {1 - \alpha } \right) = 0.95\\\\\alpha = 0.05\\\\\frac{\alpha }{2} = 0.025\\\\1 - \frac{\alpha }{2} = 0.975\\\end{array}$

That is, $P\left( {Z \le z} \right) = 0.975$

Procedure for finding the z-value is listed below:

1.From the table of standard normal distribution, locate the probability value of 0.975.

2.Move left until the first column is reached. Note the value as 1.9.

3.Move upward until the top row is reached. Note the value as 0.06.

4.The intersection of the row and column values gives the area to the one tail of z.

This implies that, $P\left( {Z \le 1.96} \right) = 0.975$

From standard normal table, the required ${Z_{0.05}}$ value for 95% confidence level is 1.96.

The 95% confidence interval for the proportion of all the registered voters who will vote for Candidate A is obtained below:

It is clear that, $\hat p = 0.30$ , for ${Z_{\frac{\alpha }{2}}} = 1.96$ , $n = 400$ .

The 95% confidence interval is,

$\begin{array}{c}\\{\rm{CI}} = 0.3 \pm \left( {1.96} \right)\sqrt {\frac{{0.3\left( {1 - 0.3} \right)}}{{400}}} \\\\ = 0.3 \pm \left( {1.96 \times 0.0229} \right)\\\\ = 0.3 \pm 0.0449\\\\ = \left( {0.3 - 0.0449,\,\,0.3 + 0.0449} \right)\\\end{array}$

$= \left( {0.2551,0.3449} \right)$

Ans:

The 95% confidence interval for the proportion of all the registered voters who will vote for Candidate A is $\left( {0.2551,0.3449} \right)$ .