# 8.3 Confidence interval for a population proportion (modified r.  (Page 2/7)

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However, in the error bound formula, we use $\sqrt{\frac{p\text{'}\cdot q\text{'}}{n}}$ as the standard deviation, instead of $\sqrt{\frac{p\cdot q}{n}}$

In the error bound formula, the sample proportions $p\text{'}$ and $q\text{'}$ are estimates of the unknown population proportions $p$ and $q$ . The estimated proportions $p\text{'}$ and $q\text{'}$ are used because $p$ and $q$ are not known. $p\text{'}$ and $q\text{'}$ are calculated from the data. $p\text{'}$ is the estimated proportion of successes. $q\text{'}$ is the estimated proportion of failures.

For the normal distribution of proportions, the z-score formula is as follows.

If $P\text{'}$ ~ $N\left(p,\sqrt{\frac{p\cdot q}{n}}\right)$ then the z-score formula is $z=\frac{p\text{'}-p}{\sqrt{\frac{p\cdot q}{n}}}$

Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. 500 randomly selected adult residents this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes - they own cell phones. Using a 95% confidence level, compute a confidence interval estimate for the trueproportion of adults residents of this city who have cell phones.

Let $X$ = the number of people in the sample who have cell phones. $X$ is binomial. $X$ ~ $B\left(500,\frac{421}{500}\right)$ .

To calculate the confidence interval, you must find $p\text{'}$ , $q\text{'}$ , and $\text{EBP}$ .

$n=500\phantom{\rule{20pt}{0ex}}x$ = the number of successes $=421$

$p\text{'}=\frac{x}{n}=\frac{421}{500}=0.842$

$p\text{'}=0.842$ is the sample proportion; this is the point estimate of the population proportion.

$q\text{'}=1-p\text{'}=1-0.842=0.158$

Since $\text{CL}=0.95$ , then $\alpha =1-\text{CL}=1-0.95=0.05\phantom{\rule{20pt}{0ex}}\frac{\alpha }{2}=0.025$ .

${z}_{\frac{\alpha }{2}}={z}_{.025}=1.96\phantom{\rule{20pt}{0ex}}$

Use the TI-83, 83+ or 84+ calculator command invnorm(.975,0,1) to find ${z}_{.025}$ . Remember that the area to the right of ${z}_{.025}$ is 0.025 and the area to the left of ${z}_{.025}$ is 0.975. This can also be found using appropriate commands on other calculators, using a computer, or using a Standard Normal probability table.

$\text{EBP}={z}_{\frac{\alpha }{2}}\cdot \sqrt{\frac{p\text{'}\cdot q\text{'}}{n}}=1.96\cdot \sqrt{\left[\frac{\left(.842\right)\cdot \left(.158\right)}{500}\right]}=0.032$

$p\text{'}-\text{EBP}=0.842-0.032=0.81$

$p\text{'}+\text{EBP}=0.842+0.032=0.874$

The confidence interval for the true binomial population proportion is $\left(p\text{'}-\text{EBP},p\text{'}+\text{EBP}\right)=$ $\left(0.810,0.874\right)$ .

## Interpretation

We estimate with 95% confidence that between 81% and 87.4% of all adult residents of this city have cell phones.

## Explanation of 95% confidence level

95% of the confidence intervals constructed in this way would contain the true value for the population proportion of all adult residents of this city who have cell phones.

For a class project, a political science student at a large university wants to determine the percent of students that are registered voters. He surveys 500students and finds that 300 are registered voters. Compute a 90% confidence interval for the true percent of students that are registered voters and interpret the confidenceinterval.

$x=300$ and $n=500$ . Using a TI-83+ or 84 calculator, the 90% confidence interval for the true percent of students that are registered voters is (0.564, 0.636).

$p\text{'}=\frac{x}{n}=\frac{300}{500}=0.600$

$q\text{'}=1-p\text{'}=1-0.600=0.400$

Since $\text{CL}=0.90$ , then $\alpha =1-\text{CL}=1-0.90=0.10\phantom{\rule{20pt}{0ex}}\frac{\alpha }{2}=0.05$ .

${z}_{\frac{\alpha }{2}}={z}_{.05}=1.645\phantom{\rule{20pt}{0ex}}$

Use the TI-83, 83+ or 84+ calculator command invnorm(.95,0,1) to find ${z}_{.05}$ . Remember that the area to the right of ${z}_{.05}$ is 0.05 and the area to the left of ${z}_{.05}$ is 0.95. This can also be found using appropriate commands on other calculators, using a computer, or using a Standard Normal probability table.

$\text{EBP}={z}_{\frac{\alpha }{2}}\cdot \sqrt{\frac{p\text{'}\cdot q\text{'}}{n}}=1.645\cdot \sqrt{\left[\frac{\left(.60\right)\cdot \left(.40\right)}{500}\right]}=0.036$

$p\text{'}-\text{EBP}=0.60-0.036=0.564$

$p\text{'}+\text{EBP}=0.60+0.036=0.636$

## Interpretation:

• We estimate with 90% confidence that the true percent of all students that are registered voters is between 56.4% and 63.6%.
• Alternate Wording: We estimate with 90% confidence that between 56.4% and 63.6% of ALL students are registered voters.

## Explanation of 90% confidence level

90% of all confidence intervals constructed in this way contain the true value for the population percent of students that are registered voters.

## Calculating the sample size

If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size.

The error bound formula for a proportion is $\text{EBP}={z}_{\frac{\alpha }{2}}\cdot \sqrt{\frac{p\text{'}\cdot q\text{'}}{n}}$ . Solving for $n$ gives you an equation for the sample size:

$n=\frac{z^{2}\mathrm{p\text{'}}\mathrm{q\text{'}}}{\mathrm{EBP}^{2}}$ , where $z={z}_{\frac{\alpha }{2}}$

Suppose a mobile phone company wants to determine the current percentage of customers aged 50+ that use text messaging on their cell phone. How many customers aged 50+ should the company survey in order to be 90% confident that the estimated (sample) proportion is within 3 percentage points of the true population proportion of customers aged 50+ that use text messaging on their cell phone.

From the problem, we know that EBP=0.03 (3%=0.03) and ${z}_{\frac{\alpha }{2}}={z}_{.05}=1.645$ because the confidence level is 90%

However, in order to find n , we need to know the estimated (sample) proportion p'. Remember that q'=1-p'. But, we do not know p' yet. Since we multiply p' and q' together, we make them both equal to 0.5 because p'q'= (.5)(.5)=.25 results in the largest possible product. (Try other products: (.6)(.4)=.24; (.3)(.7)=.21; (.2)(.8)=.16 and so on). The largest possible product gives us the largest n. This gives us a large enough sample so that we can be 90% confident that we are within 3 percentage points of the true population proportion. To calculate the sample size n, use the formula and make the substitutions.

$n=\frac{z^{2}\mathrm{p\text{'}}\mathrm{q\text{'}}}{\mathrm{EBP}^{2}}$ gives $n=\frac{\mathrm{1.645}^{2}\mathrm{\left(.5\right)}\mathrm{\left(.5\right)}}{\mathrm{.03}^{2}}$ =751.7

Round the answer to the next higher value. The sample size should be 758 cell phone customers aged 50+ in order to be 90% confident that the estimated (sample) proportion is within 3 percentage points of the true population proportion of all customers aged 50+ that use text messaging on their cell phone.

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