Confidence Interval for a Single Proportion

# Confidence Interval for a Single Proportion
### Dr. Dogucu

---

<div class="my-footer"> 
 Copyright &copy; <a href="https://mdogucu.ics.uci.edu">Dr. Mine Dogucu</a>. <a href="https://creativecommons.org/licenses/by-nc-sa/4.0/">CC BY-NC-SA 4.0</a></div>

---

## Remembering CLT

Let `$\pi$` represent the proportion of bike owners on campus then `$\pi =$` 0.15.

---

## Getting to sampling distribution of single proportion

`$p_1$` - Proportion of first sample (n = 100)

```
## [1] 0.17
```

`$p_2$` -Proportion of second sample (n = 100)

```
## [1] 0.12
```

`$p_3$` -Proportion of third sample (n = 100)

```
## [1] 0.14
```

....

---

### Sampling Distribution of Single Proportion

---

If certain conditions are met then

`$$p \sim \text{approximately } N(\pi, \frac{\pi(1-\pi)}{n})$$`

---

## In Reality

- We only have one sample and thus one point estimate of the population parameter. How can make use of it?

- First we will assume the sample proportion is the best thing we have at hand and use it as a point estimate of the population proportion.

- Second, even though we embrace the sample proportion as a point estimate of the population proportion, we will need to acknowledge that it has some error.

---

## Standard Error

`$p \sim  \text{approximately } N(\text{mean} = \pi, \text{sd} = \sqrt{\frac{\pi(1-\pi)}{n}})$`

We call the standard deviation of the sampling distribution __standard error__ of the estimate.

Standard error of single proportion is `$\sqrt{\frac{p(1-p)}{n}}$`.

---

## Confidence Interval

CI = `$\text{point estimate} \pm \text { margin of error}$`

CI = `$\text{point estimate} \pm \text { critical value} \times \text{standard error}$`

CI for single proportion = `$p \pm \text {critical value} \times \text{standard error}$`

CI for single proportion = `$p \pm \text {critical value} \times \sqrt{\frac{p(1-p)}{n}}$`

95% CI for single proportion = `$p \pm 1.96 \times \sqrt{\frac{p(1-p)}{n}}$` because ...

---

95% of the data falls within 1.96 standard deviations in the normal distribution.
      
<img src="slide-4-prop-ci_files/figure-html/unnamed-chunk-7-1.png" style="display: block; margin: auto;" />
      
---

## How do we know that?

```r
qnorm(0.025, mean = 0 , sd = 1)
```

```
## [1] -1.959964
```

```r
qnorm(0.975, mean = 0 , sd = 1)
```

```
## [1] 1.959964
```

---

## 95% CI for the first sample

Recall `$p = 0.17$` and `$n = 100$`

95% CI for single proportion = `$p \pm 1.96 \times \sqrt{\frac{p(1-p)}{n}}$`

95% CI = `$0.17 \pm 1.96 \times \sqrt{\frac{0.17(1-0.17)}{100}}$`

95% CI = `$0.17 \pm 1.96 \times 0.03756328$`

95%CI = `$0.17 \pm 0.07362403$`

95%CI = (0.09637597, 0.243624)

---

## 95% CI for the first sample

95%CI = (0.09637597, 0.243624)

We are 95% confident that the true population proportion of bike owners is in this confidence interval.

---
class: middle center

95%CI = (0.09637597, 0.243624)

---

## Understanding Confidence Intervals

I have taken 100 samples with `$n = 100$`, calculated the sample proportion, standard error, and 95% CI interval for each sample

```
## # A tibble: 100 x 4
##        p     SE lower_bound upper_bound
##    <dbl>  <dbl>       <dbl>       <dbl>
##  1  0.19 0.0392      0.113        0.267
##  2  0.21 0.0407      0.130        0.290
##  3  0.15 0.0357      0.0800       0.220
##  4  0.15 0.0357      0.0800       0.220
##  5  0.13 0.0336      0.0641       0.196
##  6  0.11 0.0313      0.0487       0.171
##  7  0.16 0.0367      0.0881       0.232
##  8  0.11 0.0313      0.0487       0.171
##  9  0.19 0.0392      0.113        0.267
## 10  0.16 0.0367      0.0881       0.232
## # ... with 90 more rows
```

---

## Understanding Confidence Intervals

---

## Understanding Confidence Intervals

---

## Confidence Interval Width

Which of the following confidence intervals would be the widest? Why?

- 90% CI
- 95% CI
- 99% CI

---

CI = `$\text{point estimate} \pm \text { critical value} \times \text{standard error}$`

```r
qnorm(0.05, mean = 0, sd = 1) # critical value for 90%CI
```

```
## [1] -1.644854
```

---

CI = `$\text{point estimate} \pm \text { critical value} \times \text{standard error}$`

```r
qnorm(0.025, mean = 0, sd = 1) # critical value for 95%CI
```

```
## [1] -1.959964
```

---

CI = `$\text{point estimate} \pm \text { critical value} \times \text{standard error}$`

```r
qnorm(0.005, mean = 0, sd = 1) # critical value for 99%CI
```

```
## [1] -2.575829
```

<img src="slide-4-prop-ci_files/figure-html/unnamed-chunk-20-1.png" style="display: block; margin: auto;" />
---

CI = `$\text{point estimate} \pm \text { critical value} \times \text{standard error}$`

- 99% CI has the highest critical value.
- Higher critical value means higher margin of error.
- Higher margin of error means wider CI.

Thus 99% CI would be the widest.

---

---

## Effect of Sample Size on Confidence Interval

Researchers A, B, and C are interested in proportion of bike ownership took samples. They each take separate samples of size 100, 500, and 1000 respectively. They each have a sample proportion of 0.18. What a surprise! Which of the researchers will find the narrowest 95% CI?

---

**Researcher A:** 95% CI for single proportion = `$0.18 \pm 1.96 \times \sqrt{\frac{0.18(1-0.18)}{100}}$`

**Researcher B:** 95% CI for single proportion = `$0.18 \pm 1.96 \times \sqrt{\frac{0.18(1-0.18)}{500}}$`

**Researcher C:** 95% CI for single proportion = `$0.18 \pm 1.96 \times \sqrt{\frac{0.18(1-0.18)}{1000}}$`

As sample size increases, the standard error decreases and the margin of error also decreases, thus the confidence interval interval gets narrower. The Researcher C would have the narrowest CI.

---

## CLT

If these conditions are met then

`$p \sim \text{approximately } N(\pi, \frac{\pi(1-\pi)}{n})$`

---

## Conditions

1. The sample data are independent. 
2. There needs to be at least 10 successes and 10 failures in the sample.
3. The sample size is smaller than 10% of the population.

---

__Example__

According to a Gallup Survey  of 1017 adults living in US 66% of Americans favor legalizing marijuana.

Compute 95% confidence interval for the population proportion of those who favor legalizing marijuana.

.footnote[Information on the survey can be found [here](https://news.gallup.com/poll/267698/support-legal-marijuana-steady-past-year.aspx)]

---

## Checking Conditions

`$n = 1017$` and `$p = 0.66$`

1) The survey link indicates that the respondents were chosen from a random sample. We would expect such sample to be independent.

2) We need at least 10 people favoring legalizing marijuana and 10 people opposing this.

`$np = 1017 \cdot 0.66 = 671.22$`. There are more than 10 people favoring legalizing marijuana.
`$n(1-p) = 1017 \cdot (1-0.66) = 345.78$`. There are more than 10 people opposing legalizing marijuana.

3) 1017 is less than 10% of US population.

---

## Confidence Interval

CI = `$\text{point estimate} \pm \text { margin of error}$`

CI = `$\text{point estimate} \pm \text { critical value} \times \text{standard error}$`

95% CI for single proportion = `$p \pm 1.96 \times \sqrt{\frac{p(1-p)}{n}}$`

95% CI = `$0.66 \pm 1.96 \times \sqrt{\frac{0.66(1-0.66)}{1017}}$`

95% CI = (0.6308857, 0.6891143)

We are 95% confident that the true proportion of Americans who support legalizing marijuana falls between 0.6308857 and 0.6891143.

---

## Confidence Interval Using R

```r
p <- 0.66 #sample proportion
n <- 1017 #sample size

se <- sqrt(p*(1-p)/n) #standard error
cv <- qnorm(0.975) #critical value

p - cv*se #lower bound of the CI
```

```
## [1] 0.6308862
```

```r
p + cv*se #upper bound of the CI
```

```
## [1] 0.6891138
```