# Ch 8.3 Confidence Interval for Population Proportion

## Ch 8.3  Confidence Interval for population proportion:

Terms: Population proportion: p

Sample proportion: $$\hat{p}$$ = (x/n)

Number of success: x;  sample size: n

Confidence Level: C-level

Significant Level: α = 1 – C-level  (probability of unlikely)

EBP: Error Bound for proportion, margin of error

### A) To Estimate p:

$$\fbox {1) point estimate: \( \hat{p} = x/n$$ }  \)

$$\fbox{ 2) Interval estimate: \( \hat{p} - E \text{ to } \hat{p} + E$$ } \)

$$E(EBP) = z_ {\alpha /2} \sqrt{\frac{\hat{p}\hat{q}}{n}}$$

Note: E is the approximated value of sampling error of sampling distribution of $$\hat{p}$$, which has a normal distribution when x and n-x  ≥ 5.

Use Statdisk online calculator to find the confidence interval.

- identify C-level, sample size n and success = x.

- https://www.statdisk.com/# , Analysis/ Confidence Intervals/Proportion one sample/

output: E (EBP) and   lower < p < upper

$$z_ {\alpha /2}$$ is the critical value with C-level in the middle.

To find  $$z_ {\alpha /2}$$

Use online Inverse Normal calculator, set area = $$\alpha /2$$, mean = 0, sd =1,

click above, recalculate.

Explanation:

X = number of success is a binomial distribution with mean = np and sd = $$\sqrt{npq}$$.

When np and nq ≥ 5, distribution of X is normal with mean = np and sd = $$\sqrt{npq}$$ .

So  distribution of  $$\hat{p} = \frac{x}{n}$$ is normal with mean = p and SD =  $$\sqrt{\frac{\hat{p}\hat{q}}{n}}$$ .

At a given C-level, the maximum error of $$\hat{p}$$ and p is E where  $$E(EBP) = z_ {\alpha /2} \sqrt{\frac{\hat{p}\hat{q}}{n}}$$

Note: the requirement for this confidence interval is not n > 30 but np and nq ≥  5.

Interpret a 95% Confidence Interval:

We are 95% confidence that the interval from ____ to ____   actual contain the true value of the population proportion of the category of interest.

The Confidence interval also shows that 95% of all confidence Intervals contain the true value of p.

Make conclusion from confidence interval:

1) Any value in the confidence Interval can be p.

2) If the whole interval > a, we can conclude p > a

3) If the whole interval < a, we can conclude p < a.

4) When two confidence interval overlap, we can conclude that the two p may be the same. We cannot conclude one of the p is higher.

Ex1. A research is conducted to determine how many household use Netflix to stream videos. A random sample of 500 households show that 442 households use Netflix.

a) Use a 90% confidence level to compute a confidence interval estimate of true proportion of households using Netflix.

use Statdisk Analysis/Confidence Intervals/Proportion one sample : Clevel = 0.9, n = 500, x = 442,

b) Find the critical value.

Use Inverse Normal calculator, since clevel = 90%,  $$\alpha = 0.1, \alpha /2 = 0.05$$, set area = 0.05, mean = 0, sd = 1,

click above, recalculate.

$$z_ {\alpha /2}$$ = 1.645

c) Interpret the confidence interval in non-technical term.

We estimate with 90% confidence that the true proportion of all households that use Netflix is between 86.0% to 90.8%

d) Can we conclude with 90% confidence that more than 80% of households use Netflix?

Since the interval contains 86.0% to 90.8%, all values are more than 80%, so we can conclude that.

Ex2: A Gallup poll of 1487 adults showed that 43% of the respondents have Facebook pages.

a) Find the number in the sample who have Facebook pages.   $$x = n(\hat{p})$$ = 1487 (0.43) = 639

c) Find the interval estimate of p at 95% confidence level  and margin of error E.

Use statdisk with Clevel = 0.95, n = 1487, x = 639

Interval estimate is  0.405 < p < 0.455, E = 0.025

d) Write a non-technical interpretation of the above

We are 95% confidence that the true percent of adults who have Facebook pages are between 40.5% to 45.5%.

e) Can we claim that less than 60% of all adults have Facebook pages?

Since the whole interval is less than 0.6, yes, we can conclude less than 60% of all adults have Facebook pages.

### B) Determine sample size for a desired E

Since   $$E(EBP) = z_ {\alpha /2} \sqrt{\frac{\hat{p}\hat{q}}{n}}$$

$$n = \frac{(z_ {\alpha/2})^2\hat{p}\hat{q}}{E^2}$$ when an estimate $$\hat{p}$$ is known.

$$n = \frac{(z_ {\alpha/2})^2\cdot 0.25}{E^2}$$     when no estimate $$\hat{p}$$  is known.

Use statdisk / Analysis/Sample size Determination/ Estimate proportion to find sample size for a given error E.

Input C-Level,  Desired E,  Estimate of p = $$\hat{p} = \frac{x}{n}$$, evaluate.

Ex3. What sample size should be used if we want the keep the margin of error within 2.1% when estimating a proportion at a 90% confidence interval. use $$\hat{p} = 0.43$$ as an old estimation of p.

C-level = 0.9,  E = 0.021,  estimate of p = 0.43, Evaluate

Sample size p = 1504.

b) If no previous study has been done , what sample size will be needed. (do not use p-hat = 0.43)

Use statdisk / Analysis/Sample size Determination/ Estimate proportion

C-level = 0.9,  E = 0.021,  estimate of p = blank, evaluate

sample size p = 1534