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

Collaborative statistics: custom Page 2 / 7

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

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

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

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

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. $X$ ~ $B (500, \frac{421}{500})$ .

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

$n = 500 x$ = the number of successes $= 421$

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

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

$q' = 1 - p' = 1 - 0.842 = 0.158$

Since $CL = 0.95$ , then $α = 1 - CL = 1 - 0.95 = 0.05 \frac{α}{2} = 0.025$ .

$z_{\frac{α}{2}} = z_{.025} = 1.96$

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.

$EBP = z_{\frac{α}{2}} \cdot \sqrt{\frac{p' \cdot q'}{n}} = 1.96 \cdot \sqrt{[\frac{(.842) \cdot (.158)}{500}]} = 0.032$

$p' - EBP = 0.842 - 0.032 = 0.81$

$p' + EBP = 0.842 + 0.032 = 0.874$

The confidence interval for the true binomial population proportion is $(p' - EBP, p' + EBP) =$ $(0.810, 0.874)$ .

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).

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

$q' = 1 - p' = 1 - 0.600 = 0.400$

Since $CL = 0.90$ , then $α = 1 - CL = 1 - 0.90 = 0.10 \frac{α}{2} = 0.05$ .

$z_{\frac{α}{2}} = z_{.05} = 1.645$

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.

$EBP = z_{\frac{α}{2}} \cdot \sqrt{\frac{p' \cdot q'}{n}} = 1.645 \cdot \sqrt{[\frac{(.60) \cdot (.40)}{500}]} = 0.036$

$p' - EBP = 0.60 - 0.036 = 0.564$

$p' + EBP = 0.60 + 0.036 = 0.636$

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 $EBP = z_{\frac{α}{2}} \cdot \sqrt{\frac{p' \cdot q'}{n}}$ . Solving for $n$ gives you an equation for the sample size:

$n z 2 p' q' EBP 2$ , where $z = z_{\frac{α}{2}}$

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 $z_{\frac{α}{2}} = z_{.05} = 1.645$ 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.

$n z 2 p' q' EBP 2$ gives $n 1.645 2 (.5) (.5) .03 2$ =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.

Questions & Answers

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Answer

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Source: OpenStax, Collaborative statistics: custom version modified by r. bloom. OpenStax CNX. Nov 15, 2010 Download for free at http://legacy.cnx.org/content/col10617/1.4

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