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.

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.

\(z_a\) 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.

In the graph below, we show 10 replications (for each replication, we sample 30 students and ask them whether they are Democrats) and compute an 80% Confidence Interval each time. We are lucky in this set of 10 replications and get exactly 8 out of 10 intervals that contain the parameter. Due to the small number of replications (only 10), it is quite possible that we get 9 out of 10 or 7 out of 10 that contain the true parameter. On the other hand, if we try it 10,000 (a large number of) times, the percentage that contains the true proportions will be very close to 80%.

If we repeatedly draw random samples of size n from the population where the proportion of success in the population is $p$ and calculate the confidence interval each time, we would expect that $100(1 - \alpha)$% of the intervals would contain the true parameter, $p$.

Sample Size Computation for the Population Proportion Confidence Interval

An important part of obtaining desired results is to get a large enough sample size. We can use what we know about the margin of error and the desired level of confidence to determine an appropriate sample size.

Recall that the margin of error, E, is half of the width of the confidence interval. Therefore for a one sample proportion,

\(E=z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\)

Since the confidence level reflects the success rate of the method we use to get the confidence interval, we like to have a narrower interval while keeping the confidence level at a reasonably higher level.

For most newspapers and magazine polls, it is understood that the margin of error is calculated for a 95% confidence interval (if not stated otherwise). A 3% margin of error is a popular choice also. For instance, you might see a television poll state that the "approval rating of the president is 72%; the margin of error of the poll is plus or minus 3%."

If we want the margin of error smaller (i.e., narrower intervals), we can increase the sample size. Or, if you calculate a 90% confidence interval instead of a 95% confidence interval, the margin of error will also be smaller. However, when one reports it, remember to state that the confidence interval is only 90% because otherwise, people will assume 95% confidence.

If the desired margin of error E is specified and the desired confidence level is specified, the required sample size to meet the requirements can be calculated by two methods:

The sample size obtained from using the educated guess is usually smaller than the one obtained using the conservative method. This smaller sample size means there is some risk that the resulting confidence interval may be wider than desired. Using the sample size by the conservative method has no such risk.

To construct a confidence interval for a population mean, we're going to apply the same three steps as with the population proportion, but first, let's look at the two possible cases.

1. CHECK THE CONDITIONS

   One of the following conditions need to be satisfied:

   1. If the sample comes from a Normal distribution, then the sample mean will also be normal. In this case, \(\dfrac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\) will follow a \(t\)-distribution with \(n-1\) degrees of freedom.
   2. If the sample does not come from a normal distribution but the sample size is large (\(n\ge 30\)), we can apply the Central Limit Theorem and state that \(\bar{X}\) is approximately normal. Therefore, \(\dfrac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\) will follow a \(t\)-distribution with \(n-1\) degrees of freedom.

2. CONSTRUCT THE GENERAL FORM

   \((1-\alpha)\)100% Confidence Interval for the Population Mean, \(\mu\)

   \(\bar{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}\)

   where the t-distribution has \(df = n - 1\). This interval is also known as the one-sample t-interval for the population mean.

3. INTERPRET THE CONFIDENCE INTERVAL

   We are \((1-\alpha)100\%\) confident that the population mean, \(\mu\), is between \(\bar{x}-t_{\alpha/2}\frac{s}{\sqrt{n}}\) and \(\bar{x}+t_{\alpha/2}\frac{s}{\sqrt{n}}\).

In 1908, William Sealy Gosset from Guinness Breweries discovered the t-distribution. His pen-name was Student and thus it is called the "Student's t-distribution."

The t-distribution is different for different sample size, n. Thus, tables, as detailed as the standard normal table, are not provided in the usual statistics books. The graph below shows the t-distribution for degrees of freedom of 10 (blue) and 30 (red dashed).

If the sample size is less than 30, one needs to use a Normal Probability Plot to check whether the assumption that the data come from a normal distribution is valid.

Recall that a \((1-\alpha)\)100% confidence interval for \(\mu\) is \(\bar{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}\) where the multiplier \(t\) has a t-distribution with \(df = n - 1\). Thus, the margin of error, E, is equal to:

\(E=t_{\alpha/2}\dfrac{s}{\sqrt{n}}\)

To determine the sample size, one first decides the confidence level and the half width of the interval one wants. Then we can find the sample size to yield an interval with that confidence level and with a half width not more than the specified one. The crude method to find the sample size: \(n=\left(\dfrac{z_{\alpha/2}\sigma}{E}\right)^2\) Then round up to the next whole integer.

The Iterative Method

A more accurate method to estimate the sample size: iteratively evaluate the formula since the t value also depends on n.

\(n=\left(\dfrac{t_{\alpha/2}s}{E}\right)^2\)

Use the example above for illustration. Start with an initial guess for $n$, plug in the formula, and iteratively solve for \(n\).

If the initial guess for \(n\) is 20, \(t_{0.05} = 1.729\) and degrees of freedom = 19,

\(n=\left(\dfrac{t_{\alpha/2}s}{E}\right)^2=n=\left(\dfrac{1.729(400)}{120}\right)^2=33.21\)

For \(n = 34\), degree of freedom = 33, and \(t_{0.05} = 1.697\)

\(n=\left(\dfrac{t_{\alpha/2}s}{E}\right)^2=n=\left(\dfrac{1.697(400)}{120}\right)^2=31.99\)

If we use \(n = 32\), the result is the same. Thus, the more accurate answer to the example is to sample 32 students.