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Confidence Intervals for Proportion

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.title[
# Confidence Intervals for Proportion
]
.subtitle[
## MA336 Statistics 
]
.author[
### Fei Ye Department of Mathematics and Computer Science 
]
.date[
### August 2022
]

---

## Learning Goals for Confidence Intervals

- Construct and interpret a confidence intervals for one population proportion.

- Describe how the following will affect the width of the confidence interval:

- increasing the sample size;

- increasing the confidence level.

---

## Confidence Interval for a Proportion (1/2)

- Recall that the standard error of sample proportions is `$\sigma_{\hat{P}}=\sqrt{\frac{p(1-p)}{n}}$`, where `$n$` is the sample size and `$p$` is the population proportion. As a consequence, when estimating the population proportion `$p$`, we only have a point estimate `$\hat{p}$` (phat) to use. For the standard error, we use the estimation
  `$$\sigma_{\hat{p}}\approx\hat{\sigma}_{\hat{p}}=\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}.$$`

- Based on the central limit theorem, when `$n$` is large enough, at the `$100(1-\alpha)\%$` level, the margin of error for `$p$` is defined as
  `$$E=z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$`
  In Excel, `$z_{\alpha/2}$`=`NORM.S.INV((1 + confidence level)/2)`. The marginal error can also be obtained by
  `CONFIDENCE.NORM(1-confidence level, SQRT(phat*(1-phat)/n, n)`.

---

## Confidence Interval for a Proportion (2/2)

- The confidence interval for `$p$` is defined by
 `$$[\hat{p}-E,\hat{p}+E]=\left[\hat{p}-z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right],$$`
 where [the critical value `$z_{\alpha/2}$` satisfies that `$P(Z< z_{\alpha/2})=1-\alpha/2$`](https://saylordotorg.github.io/text_introductory-statistics/s11-01-large-sample-estimation-of-a-p.html) for the standard normal variable `$Z$`.

- In practical, the sample size `$n$` is considered large enough if `$n\hat{p}\ge 10$` and `$n(1-\hat{p})\ge 10$`.

- The above defined confidence interval is known as the [normal approximation (or Wald's) confidence interval](https://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval).
  It is popular in introductory statistics books. However, it is unreliable when the sample size is small or the sample proportion is close to 0 or 1.
  Indeed, if the sample proportion is 0 or 1, the confidence interval defined here will have zero length.

???
By the central limit theorem, the random variable `$\hat{p}$` is normal distributed. The chance that `$p\in \left[\hat{p}-z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right]$` is the same as the chance that `$\hat{p}\in \left[p-z_{\alpha/2}\sqrt{\frac{p(1-p)}{n}}, p+z_{\alpha/2}\sqrt{\frac{p(1-p)}{n}}\right]$`. That shows `$z_{\alpha/2}$` satisfying
`$$P(-z_{\alpha/2}<\dfrac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}<z_{\alpha/2})=1-\alpha.$$`

---

## Example: Estimating the Proportion of Students Taking Busses (1/2)

In a random sample of 100 students in college, 65 said that they come to college by bus.

1. Give a point estimate of the proportion of all students who come to college by bus.

2. Construct a 99% confidence interval for that proportion.

**Solution:** A good point estimate would be a sample proportion. Here the sample proportion is `$\hat{p}=65/100=0.65$`.

As `$n\hat{p}=100\cdot 0.65=65>10$` and `$n(1-\hat{p})=100\cdot 0.35=35>10$`, which implies the sample is large enough, approximately the standard error is
`$$\hat{\sigma}_{\hat{P}}=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\sqrt{\frac{0.65(1-0.65)}{100}}\approx0.048.$$`

---

## Example: Estimating the Proportion of Students Taking Busses (2/2)

**Solution: (Continued)** At 99% level of confidence, the value `$1-\alpha/2=(1+\text{confidence level})/2=(1+0.99)/2=0.995$`. The critical value `$z_{\alpha/2}$` is determined by the equation `$P(Z<z_{\alpha/2})=0.9955$`. Using the Excel function `NORM.S.INV(0.995)`, we find the critical value \$z_{\alpha/2}\approx 2.576\$.

Thus the marginal error is
`$$E=z_{\alpha/2}\cdot \hat{\sigma}_{\hat{P}}=2.576\cdot 0.048=0.123,$$`
and the confidence interval at 99% level is
`$$[\hat{p}-E, \hat{p}+E]\approx [0.65-0.123, 0.65+0.123]=[0.527, 0.773].$$`

Conclusion: we are 99% confident that the proportion of all students at the college who take bus is in the interval `$[0.527, 0.773]$`.

.footmark[
  **Note:** The marginal error can also be obtained by the Excel function
  `CONFIDENCE.NORM(1-0.99, SQRT(65/100*(1-65/100)/100), 100)`.
]

<!-- 
Foothill College’s athletic department wants to calculate the proportion of students who have attended a women’s basketball game at the college. They use student email addresses, randomly choose 220 students, and email them. Of the 145 who responded, 22 had attended a women’s basketball game.

Calculate and interpret the approximate 90% confidence interval for the proportion of all Foothill College students who have attended a women’s basketball game. -->

<!-- 
**Solution:** Although 220 students were surveyed, only 145 responded. So the sample consists of those 145 students. The sample proportion is `$\hat{p}=\frac{22}{145}=0.152$`.

The estimated standard error is
`$$\hat{\sigma}_{\hat{P}}=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\sqrt{\frac{0.152\cdot(1-0.152)}{145}}\approx 0.03.$$` -->

<!-- **Solution: (Continued)**
At the 90% confidence level, `$\alpha=1-0.9=0.1$` and the critical value is `$z_{\alpha/2}$`=`NORM.S.INV(1-0.1/2)` `$\approx 1.645$`.

Therefore, the marginal error is
`$$E=z_{\alpha/2}\cdot\hat{\sigma}_{\hat{P}}\approx 1.645\cdot0.03=0.049,$$`
and the confidence interval is
`$$[\hat{p}-E, \hat{p}-E]\approx[0.152-0.049, 0.152+0.049]=[0.103, 0.049].$$`

Conclusion: we are 90% confident that the proportion of all Foothill College students who have attended a women’s basketball game is between 0.103 and 0.201. -->

---

## Factors Affect the Width of Confidence Intervals

- The width of a confidence interval, equals twice the standard error, gives a measure of precision of the estimation.

- Recall, for population proportion and mean,
  `$$\text{Marginal Error} = \text{Critical Value}\cdot \frac{\text{(estimated) Population SD}}{\sqrt{\text{Sample Size}}}$$`

- The formula tells us the precision of a confidence interval is affected by the confidence level, the variability, and the sample size.
  
  - Larger the confidence levels give larger critical values and errors.
  
  - Populations (and samples) with more variability gives larger errors.

- Larger sample sizes give smaller errors.

---

## Sample Size Determination

- In practice, we may desire a marginal error of `$E$`. With a fixed confidence level `$100(1-\alpha)\%$`, the larger the sample size the smaller the marginal error.

- When estimating population proportion, if we can produce a reasonable guess `$\hat{p}$` for population proportion, then an appropriate minimum sample size for the study is determined by
  `$$n=\left(\frac{z_{\alpha/2}}{{E}}\right)^2\cdot \hat{p}(1-\hat{p}).$$`

- When estimating population mean, if we can produce a reasonable guess `$\sigma$` for the population standard deviation, then an appropriate minimum sample size is given by
  `$$n=\left(\dfrac{z_{\alpha/2}\cdot \sigma}{{E}}\right)^2.$$`

---

## Example: Minimum Sample Size - Error in Proportion

Suppose you want to estimate the proportion of students at QCC who live in Queens. By surveying your classmates, you find around 70% live in Queens. Use this as a guess to determine how many students would need to be included in a random sample if you wanted the error of margin for a 95% confidence interval to be less than or equal to 2%.

**Solution:** We may use `$\hat{p}=0.7$` as a reasonable guess for the population proportion.

At the 95% level, the critical value is `$z_{\alpha/2}=$` `NORM.S.INV((1+0.95)/2)` `$\approx 1.96$`.

The marginal error is `$E=0.02$`.

Then the appropriate minimal sample size is determined by
`$$n=\left(\frac{z_{\alpha/2}}{{E}}\right)^2\cdot \hat{p}(1-\hat{p})=(1.96/0.02)^2\cdot 0.7\cdot(1-0.7)=2016.84.$$`

Since the sample size has to be an integer, to get a error no more than 2% at the level 95%,  the minimal sample size should be at least 2017.

---

## Example: Minimum Sample Size - Error in Mean

Find the minimum sample size necessary to construct a 99% confidence interval for the population mean with a margin of error `$E =0.2$`. Assume that the estimated population standard deviation is `$\sigma=1.3$`.

**Solution:** At the 99% level, the critical value `$z_{\alpha/2}=$` `NORM.S.INV((1+0.99)/2)` `$\approx 2.576$`.

The desired marginal error is `${E}=0.2$`.

The estimated population standard deviation is `$\sigma=1.3$`.

Then the minimal sample size is approximately
`$$n=\left(\dfrac{z_{\alpha/2}\cdot \sigma}{{E}}\right)^2\approx (2.576\cdot 1.3/0.2)^2 \approx 280.4.$$`

To get a error no more than 0.2 at the level 95%,  the minimal sample size should be at least 281.

---

## Practice: Confidence Interval of Proportion of Kids

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## Practice: Confidence Intervals for Product Quality

To understand the reason for returned goods, the manager of a store examines the records on 40 products that were returned in the last year. Reasons were coded by 1 for “defective,” 2 for “unsatisfactory,” and 0 for all other reasons, with the results shown in the table.
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<tbody>
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</tbody>
</table>

1. Give a point estimate of the proportion of all returns that are because of something wrong with the product, that is, either defective or performed unsatisfactorily.

2. Construct an 80% confidence interval for the proportion of all returns that are because of something wrong with the product.

---

## Practice: Sample Size for Mean with Given Error

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## Practice: Sample Size for Proportion with Given Error

<!-- 
A software engineer wishes to estimate, to within 5 seconds, the mean time that a new application takes to start up, with 95% confidence. Estimate the minimum size sample required if the standard deviation of start up times for similar software is 12 seconds.

.footmark[
 Source: [Exercise 7 in Section 7.4 in Introductory Statistics](https://saylordotorg.github.io/text_introductory-statistics/s11-04-sample-size-considerations.html)
] -->

<!-- 
The administration at a college wishes to estimate, to within two percentage points, the proportion of all its entering freshmen who graduate within four years, with 90% confidence. Estimate the minimum size sample required.

.footmark[
 Source: [Exercise 13 in Section 7.4 in Introductory Statistics](https://saylordotorg.github.io/text_introductory-statistics/s11-04-sample-size-considerations.html)
] -->

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# Lab Instructions in Excel

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## Excel Functions for Normal Distributions and Marginal Errors

- When a cumulative probability `$p=P(Z<z)$` of a standard normal random variable `$Z$` is given, we can find `$z$` using `NORM.S.INV(p)`.

- If a sample of size `$n$` has the proportion `$\hat{p}$` and the sampling distribution is approximately normal, the marginal error for the proportion can be obtained by the Excel function
`CONFIDENCE.NORM(1-confidence level, SQRT(phat*(1-phat)), n)`