# 7.3 Confidence interval for a population proportion  (Page 2/7)

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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.

The confidence interval can only be used if the number of successes $np\text{'}$ and the number of failures $nq\text{'}$ are both larger than 5.

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 in 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.

## Solution

• You can use technology to directly calculate the confidence interval.
• The first solution is step-by-step (Solution A).
• The second solution uses a function of the TI-83, 83+ or 84 calculators (Solution B).

## Using a function of the ti-83, 83+ or 84 calculators:

Press STAT and arrow over to TESTS .
Arrow down to A:1-PropZint . Press ENTER .
Arrow down to $x$ and enter 421.
Arrow down to $n$ and enter 500.
Arrow down to C-Level and enter .95.
Arrow down to Calculate and press ENTER .
The confidence interval is (0.81003, 0.87397).

For a class project, a political science student at a large university wants to estimate 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.

• You can use technology to directly calculate the confidence interval.
• The first solution is step-by-step (Solution A).
• The second solution uses a function of the TI-83, 83+ or 84 calculators (Solution B).

## Solution a

$x=300$ and $n=500$ .

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

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

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(0.95,0,1) 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.

$\text{EBP}={z}_{\frac{\alpha }{2}}\cdot \sqrt{\frac{p\text{'}\cdot q\text{'}}{n}}=1.645\cdot \sqrt{\frac{\left(0.60\right)\cdot \left(0.40\right)}{500}}=0.036$

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

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

The confidence interval for the true binomial population proportion is $\left(p\text{'}-\text{EBP},p\text{'}+\text{EBP}\right)=$ $\left(0.564,0.636\right)$ .

## 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.

## Solution b

Using a function of the TI-83, 83+ or 84 calculators:

Press STAT and arrow over to TESTS .
Arrow down to A:1-PropZint . Press ENTER .
Arrow down to $x$ and enter 300.
Arrow down to $n$ and enter 500.
Arrow down to C-Level and enter .90.
Arrow down to Calculate and press ENTER .
The confidence interval is (0.564, 0.636).

## Calculating the sample size n

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 population proportion is

• $\mathrm{EBP}={z}_{\frac{\alpha }{2}}\cdot \sqrt{\frac{\mathrm{p\text{'}q\text{'}}}{n}}$
• Solving for $n$ gives you an equation for the sample size.
• $n=\frac{{{z}_{\frac{\alpha }{2}}}^{2}\cdot \mathrm{p\text{'}q\text{'}}}{{\mathrm{EBP}}^{2}}$

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.

## Solution

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 752 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.

**With contributions from Roberta Bloom.

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