> For the complete documentation index, see [llms.txt](https://kmis.gitbook.io/statistics/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://kmis.gitbook.io/statistics/chapter-7.-estimation/7-3.-large-sample-estimation-of-a-population-proportion.md).

# 7-3. Large Sample Estimation of a Population Proportion

Since from [Section 6.3 "The Sample Proportion"](https://saylordotorg.github.io/text_introductory-statistics/fwk-shafer-ch06_s03#fwk-shafer-ch06_s03) in [Chapter 6 "Sampling Distributions"](https://saylordotorg.github.io/text_introductory-statistics/s10-sampling-distributions.html) we know the mean, standard deviation, and sampling distribution of the sample proportion $$\hat{p}$$ , the ideas of the previous two sections can be applied to produce a confidence interval for a population proportion. Here is the formula.

> **Large Sample** $$100(1−α)%$$ **Confidence Interval for a Population Proportion** <br>
>
> &#x20;                                                           $$\hat{p}±z\_{α∕2} \sqrt{  \frac{\hat{p}(1−\hat{p})}{n}}$$&#x20;
>
> A sample is large if the interval \
> &#x20;                                                        $$\[p−3 σ\_\hat{P}, p+3 σ\_\hat{P}]$$ \
> lies wholly within the interval $$\[0,1]$$ .

In actual practice the value of $$p$$ is not known, hence neither is $$σ\_\hat{P}$$ . In that case we substitute the known quantity $$\hat{p}$$ for $$p$$ in making the check; this means checking that the interval

&#x20;                                                 $$\[ \hat{p}-3 \sqrt{  \frac{\hat{p}(1−\hat{p})}{n}}, \hat{p}+3 \sqrt{  \frac{\hat{p}(1−\hat{p})}{n}}]$$&#x20;

lies wholly within the interval $$\[0,1]$$ .

**EXAMPLE 7.** To estimate the proportion of students at a large college who are female, a random sample of 120 students is selected. There are 69 female students in the sample. Construct a 90% confidence interval for the proportion of all students at the college who are female.

**\[ Solution ]**

The proportion of students in the sample who are female is $$\hat {p}=69∕120=0.575$$ .

Confidence level 90% means that $$α=1−0.90=0.10$$, so $$α∕2=0.05$$ . From the last line of [Figure 12.3 "Critical Values of "](https://saylordotorg.github.io/text_introductory-statistics/s16-appendix.html) we obtain $$z\_{0.05}=1.645$$ .

Thus $$\hat{p}±z\_{α∕2} \sqrt{ \hat p (1−\hat p) / n }=0.575±1.645 \sqrt{ (0.575)(0.425) / 120 }=0.575±0.074$$ .

One may be 90% confident that the true proportion of all students at the college who are female is contained in the interval $$(0.575−0.074,0.575+0.074)=(0.501,0.649)$$ .<br>

{% tabs %}
{% tab title="R Source" %}

```
n <- 120                  # number of samples
p <- 69/120               # sample proportion
alpha <- 0.10             # 

se <- sqrt(p*(1-p)/n)     # std of sample proportion
z <- qnorm(1-alpha/2); z

ll <- p - z * se
ul <- p + z * se
ll   # lower limit
ul   # upper limit
```

{% endtab %}

{% tab title="Result" %}

```
> z <- qnorm(1-alpha/2); z
## [1] 1.644854
> 
> ll <- p - z * se
> ul <- p + z * se
> ll   # lower limit
## [1] 0.5007725
> ul   # upper limit
## [1] 0.6492275
```

{% endtab %}
{% endtabs %}

* Using **Rstat** Package : `prob.ci()`

{% tabs %}
{% tab title="R Source" %}

```
library(Rstat)

prob.ci(n=120, x=69, alp=0.10, dig=3)
```

{% endtab %}

{% tab title="Result" %}

```
> prob.ci(n=120, x=69, alp=0.10, dig=3)
## [0.575 ± 1.645×√(0.575×0.425/120)] = [0.575 ± 0.074] = [0.501, 0.649] 
```

{% endtab %}
{% endtabs %}