# 5.3.1 - Construct and Interpret the CI

## Constructing a Confidence Interval for the Population Proportion Section

To construct a confidence interval we're going to use the following 3 steps:

1. CHECK CONDITIONS

Check all conditions before using the sampling distribution of the sample proportion.

We previously used $$np$$ and $$n(1-p)$$. But $$p$$ is not known. Therefore, for the confidence interval, we will use

• $$n\hat{p}>5$$ and
• $$n(1-\hat{p})>5$$
##### What can one do if the conditions are NOT satisfied?
For a confidence interval for a proportion, there is a technique called exact methods. These methods can be used if the software offers it. These exact methods are more complicated and are based on the relationship between the binomial and another distribution we will later learn called the F-distribution. The Z-method is much simpler and fairly easy to compute. In fact if you ever come across a published random survey (e.g. a Gallup poll) you can use the methods in this lesson to construct a reliable proportion confidence interval rather quickly.
2. CONSTRUCT THE GENERAL FORM

The general form of the confidence interval is '$$\text{point estimate }\pm M\times \hat{SE}(\text{estimate})$$.' The point estimate is the sample proportion, $$\hat{p}$$, and the estimated standard error is $$\hat{SE}(\hat{p})=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$. If the conditions are satisfied, then the sampling distribution is approximately normal. Therefore, the multiplier comes from the normal distribution. This interval is also known as the one-sample z-interval for $$p$$, or the Normal Approximation confidence interval for $$p$$.

$$\boldsymbol{\left(1-\alpha \right) 100\%}$$ confidence interval for the population proportion, $$\boldsymbol{p}$$
$$\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$$
where $$z_{\alpha/2}$$ represents a z-value with $$\alpha/2$$ area to the right of it.

General notes about the confidence interval...

• The $$\pm$$ in the formula above means "plus or minus". It is a shorthand way of writing

$$(\hat{p}-z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}})$$

• It is centered at the point estimate, $$\hat{p}$$.
• The width of the interval is determined by the margin of error.
• You must determine the multiplier.
3. INTERPRET THE CONFIDENCE INTERVAL

Applying the template from earlier in the lesson we can say we are $$(1-\alpha)100\%$$ confident that the population proportion is between $$\hat{p}-z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ and $$\hat{p}+z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$. The examples will go into more detail regarding the interpretation of the confidence interval.

Think about it! What terms in the margin of error would change the width of the confidence interval? Do the changes make it narrower or wider?

## Derivation of the Confidence Interval Section

To calculate the confidence interval, we need to know how to find the z-multiplier. So where does this $$z_{\alpha}$$ come from?

The confidence interval can be derived from the following fact:

\begin{align} P\left(\left|\frac{\hat{p}-p}{\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}}\right|\le z_{\alpha/2}\right)=1-\alpha \\ P\left(-z_{\alpha/2}\le \dfrac{\hat{p}-p}{\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}}\le z_{\alpha/2}\right)=1-\alpha \\ P\left(\hat{p}-z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\le p \le \hat{p}+z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\right)=1-\alpha  \end{align}

The figure shows the general confidence interval on the normal curve.

## How to find the multiplier using the Standard Normal Distribution Section

$$z_a$$ is the z-value having a tail area of $$a$$ to its right. With some calculation, one can use the Standard Normal Cumulative Probability Table to find the value.

## Example 5-1: Finding $$\boldsymbol{z_a}$$ Section

Find using the standard Normal table:  $$z_{0.15}$$

$$z_{0.15}$$ means $$P(Z>z_{0.15})=0.15$$. This implies that $$P(Z\le z_{0.15})=0.85$$. The value from the table is 1.04.

For more detailed directions on reading the z-table or using Minitab refer to the examples on this page: 3.3.2 The Standard Normal Distribution.

## Try it! Section

Use the Standard Normal Table to find the following:

$$z_{0.08}$$
$$z_{0.08}=1.40$$
$$z_{0.02}$$
$$z_{0.02}=2.05$$

## Commonly Used Alpha Levels Section

The table is a list of frequently used alphas and their  $$z_{\alpha/2}$$ multipliers.

Confidence level and corresponding multiplier.
Confidence Level $$\boldsymbol{\alpha}$$ $$\boldsymbol{z_{\alpha/2}}$$ $$\boldsymbol{z_{\alpha/2}}$$ Multiplier
90% .10 $$z_{0.05}$$ 1.645
95% .05 $$z_{0.025}$$ 1.960
98% .02 $$z_{0.01}$$ 2.326
99% .01 $$z_{0.005}$$ 2.576

The value of the multiplier increases as the confidence level increases. This leads to wider intervals for higher confidence levels. We are more confident of catching the population value when we use a wider interval.

## Example 5-2: Alpha Levels Section

For an 80% confidence interval find $$\alpha$$, $$\alpha/2$$, and $$z_{\alpha/2}$$.

Recall that $$\alpha$$ is used to find the confidence level by taking (1 - $$\alpha)*100%$$.

So for an 80% confidence we would take...

$$(1 - \alpha)*100 = 80$$ or...

$$(1 - \alpha) = .8$$

$$\alpha = .2$$

Therefore, $$\alpha/2 = .2/2 = .1$$

We would have $$z_{0.10}$$ which means $$P(Z>z_{0.10})=0.10$$.

This implies that $$P(Z\le z_{0.10})=0.90$$. The value from the table is 1.28.

Visually, you can see how these numbers relate to the normal distribution in the graph below.

## Example 5-3: Approval Ratings Section

A random sample of 1500 U.S. adults is taken. They are asked whether they approve or disapprove of the current president's performance so far (i.e. an approval rating). Of the 1500 surveyed, 660 respond with "approve". Calculate a 95% confidence interval for the overall approval rating of the the president.

Since we're dealing with a single proportion, we will examine the number of "successes" and the number of "failures". In this example there were 660 successes and 840 failures. With both successes and failures being at least 5, the condition to use the z-method to calculate the interval is acceptable.

For 95% confidence, the alpha value is 5% or 0.05 The multiplier would be a z-value with $$\alpha/2$$, or 0.025 area to the right of it. Examining the standard normal table, we find that this corresponds to a z-value of 1.96.

Important Note: Many students tend to use the multiplier of 2 instead of 1.96 due to the empirical rule. As a general rule, it is always best to use the exact values rather than the rounded value.

In this example, we have a sample proportion, $$\hat{p}$$, of 660/1500 = 0.44 and a sample size, $$n$$, of 1500.

\begin{align} \hat{p}& \pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p}}{n}} &&\text{(General Form)} \\0.44 &\pm 1.96 \sqrt{\dfrac{0.44(1-0.44)}{1500}} &&\text{(Plug in the numbers)} \\ 0.44 &\pm 0.025 &&\text{(Simplify)}\end{align}

"We are 95% confident that the overall U.S. adult approval rating for the current president is from 41.5% to 46.5%." You could also see this written as, "The current U.S. approval rating for the president is 44% with a 95% margin of error of 2.5%." Commonly, the standard level of confidence is 95% so that reference is often left out as that is the assumed level of confidence unless otherwise stated. Also, the method calculates a proportion but often the reported values are converted to percentages. If you use the decimal formal (e.g. 0.415 and 0.465) then reference these as proportion and not percentage.

#### View the video explanation from Dr. Bulathsinhala

To construct a 1-proportion confidence interval...
1. In Minitab choose Stat > Basic Statistics > 1 proportion .
2. From the drop down box select the Summarized data option button. (If you have the raw data you would use the default drop down of One or more samples, each in a column.)
3. Enter the number of successes in the Number of Events text box, and the sample size in the Number of Trials text box.
4. Choose the Options button. The default confidence level is 95. If your desire another confidence level edit appropriately.
5. To use the z- interval method choose Normal Approximation from the Method text box. The exact interval is always appropriate and is the default. Under the conditions that: $n \hat{p} \ge 5, n(1− \hat{p}) \ge 5$, one can also use the z-interval to approximate the answers. The exact interval and the z-interval should be very similar when the conditions are satisfied.
6. Choose OK and OK again.

#### Using Minitab: Approval Ratings Example

We will now use Minitab to verify our by-hand results. Recall in that example a random sample of 1500 was taken from the population of U.S. adults, with 660 responding with a positive approval.

In Minitab and following the steps above, we would enter 660 for the Number of Events and 1500 for the Number of Trials. The confidence level was 95% and we satisfied the necessary conditions to use the Normal Approximation (or z-interval) method. The results are:

Test and CI for One Proportion

Sample X N Sample p 95% CI
1 660 1500 0.440000 (0.414880, 0.465120)

Using the normal approximation.

These results closely match our by-hand interval of 0.415 to 0.465

What if we had calculated the exact confidence interval (i.e. did not choose Normal Approximation as the method)? With the exact method the interval is (0.414685, 0.465550). Consistent to three decimal places in this case. You will notice that in the output Minitab does provide a notification that the normal approximation was used.

We want to know the proportion of graduate students at Penn State who are Democrats. To answer the question, we give out the following survey:

• Yes
• No

Suppose that we get 10 people that circled Yes and 20 people that circled No (that includes the case when people don't know whether they are Democrats!!)

• Let X = the number of successes (number of students who chose Yes) = 10
• n = number of trials = 30

Find a 90% confidence interval for the proportion of graduate students who are democrats.

You should first check the conditions. We know $$\hat{p}=\frac{10}{30}=0.333$$ and $$n=30$$ Therefore, $$n\hat{p}=30(0.333)=10$$ and $$n(1-\hat{p})=20$$. Since both values are greater than 5, we can use the Normal distribution.

The z multiplier will be $$z_{0.1/2}=1.645$$

$$\hat{p}\pm 1.645\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}=0.333\pm 1.645\sqrt{\dfrac{0.333(1-0.333)}{30}}=0.333\pm0.1415=(0.1915, 0.4745)$$.

We are 90% confident that the population proportion of graduate students at Penn State who are democrats is between 19.15% and 47.45%.

The video demonstrates this same example using Minitab.