10.3: The One Proportion Z Procedure

In One Mean Z and T Procedures, we estimate the unknown population mean by constructing a confidence interval around the sample mean. But we can estimate other unknown parameters as well. To construct the confidence interval for the population mean we used the Central Limit Theorem for Sample Means. Similarly, we can use the Central Limit Theorem for Sample Proportions to construct a confidence interval for an unknown population proportion.

By the Central Limit Theorem for proportions

$\hat{p}\sim N\left(\mu_{\hat{p}}=p, \sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}\right)$

thus, we can apply the Generalized Empirical Rule for $\hat{p}\sim N\left(\mu_{\hat{p}}, \sigma_{\hat{p}}\right)$:

$P\left(\mu_{\hat{p}}-z_{\alpha/2}\sigma_{\hat{p}}<X<\mu_{\hat{p}}+z_{\alpha/2}\sigma_{\hat{p}}\right)=1-\alpha$

After plugging in $\mu_{\hat{p}}=p$ and $\sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}$:

$P\left(p-z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}<X<p+z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}\right)=1-\alpha$

Therefore, we have the following two interpretations:

We are $(1-\alpha)\cdot100\%$ confident that

(1) $\hat{p}_i$ is within $z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}$ from $p$

OR

(2) $p$ is within $z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}$ from $\hat{p}_i$.

Hence, we are $(1-\alpha)\cdot100\%$ confident that $p$ is between $\hat{p}_i-z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}$ and $\hat{p}_i+z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}$.

Try It Yourself!

Similarly to the other procedures, it appears that it is reasonable to build the confidence interval around the sample proportion $\hat{p}_i$ at the center with the margin of error $z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}$. However, the estimate for the unknown value $p$ cannot depend on itself, therefore we cannot use $p$ in $ME=z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}$ to calculate the margin of error. As in One Mean T Procedure, where we replaced the unknown $\sigma$ with $s$, we are going to replace the unknown $p$ with $\hat{p}_i$. Thus, the formula to approximate the margin of error is

$ME=z_{\alpha/2}\sqrt{\dfrac{\hat{p}_i(1-\hat{p}_i)}{n}}$

Try It Yourself!

As a result, the following template will be used to construct and interpret the confidence intervals:

Table $\PageIndex{1.1}$: Template of the procedure to construct a confidence interval for unknown population proportion.

Unknown Parameter	$p$, population proportion
Point Estimate	$\hat{p}_i=\dfrac{x}{n}$, the sample proportion where $n$ is the sample size and $x$ is the number of observations that satisfy certain condition.
Confidence Level (of the point estimate)	0%
Confidence Level (for a new interval estimate)	$(1-\alpha)100\%$ $\alpha$ $\alpha/2$
Critical Value(s)	$z_{\alpha/2}$
Margin of Error	$ME=z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}$
Lower Bound	$LB=\hat{p}_i-ME$
Upper Bound	$UB=\hat{p}_i+ME$
Interpretation	We are $(1-\alpha)100\%$ confident that the unknown population proportion $p$ is between LB and UB.

For the procedure to work the following assumptions must be made:

• The sample is simple random.
• The CLT for sample proportions must be applicable that is $np\geq10$ and $n(1-p)\geq10$.

Try It Yourself!

Let’s say we want to construct a $(1-\alpha)100\%$ confidence interval with the margin of error no more than $E_0$, how large we should choose the sample size?

By solving the following inequality:

$ME=z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}<E_0$

We get:

$n>\left(z_{\alpha/2}\cdot\dfrac{\sqrt{p(1-p)}}{E_0}\right)^2$

Note that the right-hand side contains the unknown $p$ which makes the formula useless! The good news is that while $p$ is unknown the following inequality is true for ANY $p$!

$p(1-p)\leq\frac{1}{4}$

So that

$\left(\dfrac{z_{\alpha/2}}{2E_0}\right)^2>\left(z_{\alpha/2}\cdot\dfrac{\sqrt{p(1-p)}}{E_0}\right)^2$

Thus, we can guarantee the desired margin of error by choosing $n$ to satisfy the following inequality:

$n>\left(\dfrac{z_{\alpha/2}}{2E_0}\right)^2$

Try It Yourself!

Consider the following example.

Example $\PageIndex{1}$

The ABC News/Ipsos poll was conducted by Ipsos Public Affairs’ Knowledge Panel® on March 18-19, 2020, in English and Spanish, among a random national sample of 512 adults. In the poll, 282 Americans approve of the president's management of the crisis.

1. Construct and interpret a 90% confidence interval for the proportion of the Americans that approve of the president's management of the crisis.

First, let’s check if all necessary assumptions are satisfied:

• The sample is simple random.
• $np=282\geq10$ and $n(1-p)=230\geq10$ thus the CLT for sample proportions is applicable.

Also, note that the sample proportion is

$\hat{p}_i=\dfrac{282}{512}=0.5508=55.08\%$

Once the assumptions are verified, we may apply the procedure by using the template.

Table 10.3.1.2: The summary of the procedure to construct a 90% confidence interval for the population of Americans that approve of the president's management of the crisis.

Table $\PageIndex{1.2}$: The summary of the procedure to construct a 90% confidence interval for the population of Americans that approve of the president's management of the crisis.

	Example	Template
Unknown Parameter	$p$, the approval rating	$p$, population proportion
Point Estimate	$\hat{p}_i=0.5508$, the proportion of the sample that approve the president's handling of the crisis.	$\hat{p}_i=\dfrac{x}{n}$, the sample proportion where $n$ is the sample size and $x$ is the number of observations that satisfy certain condition.
Confidence Level (of the point estimate)	0%	0%
Confidence Level (for a new interval estimate)	$(1-\alpha)100\%=90\%$ $\alpha=0.10$ $\alpha/2=0.05$	$(1-\alpha)100\%$ $\alpha$ $\alpha/2$
Critical Value(s)	$z_{\alpha/2}=z_{0.05}=1.65$	$z_{\alpha/2}$
Margin of Error	$ME=1.65\sqrt{\dfrac{0.5508(1-0.5508)}{512}}=0.0363$	$ME=z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}$
Lower Bound	$LB=0.5508-0.0363=0.5145$	$LB=\hat{p}_i-ME$
Upper Bound	$UB=0.5508+0.0363=0.5871$	$UB=\hat{p}_i+ME$
Interpretation	We are $90\%$ confident that the approval rating is between 51.45% and 58.71%.	We are $(1-\alpha)100\%$ confident that the unknown population proportion $p$ is between LB and UB.

Now we can interpret the results: We are $90\%$ confident that the approval rating is between 51.45% and 58.71%.

In addition, we may also answer the following question:

2. How large must the sample be so that the margin of error is less than 2.5%?

To answer this question, we will use the following formula:

$n>\left(\dfrac{z_{\alpha/2}}{2E_0}\right)^2$

in which we can plug in all the values:

$n>\left(\dfrac{z_{\alpha/2}}{2E_0}\right)^2=\left(\dfrac{1.65}{2\cdot0.025}\right)^2=1089.0\uparrow1090$

We conclude that 1090 is the smallest sample size for which the margin of error will be less than 2.5% for a 90% confidence level.

Try It Yourself! $\PageIndex{1}$

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