What happens when the sample comes from a population that is not normally distributed? This is where the Central Limit Theorem (CLT) comes in.

We should stop here to break down what this theorem is saying because the Central Limit Theorem is very powerful!

The Central Limit Theorem applies to a sample mean from any distribution. We could have a left-skewed or a right-skewed distribution. As long as the sample size is large, the distribution of the sample means will follow an approximate Normal distribution.

For the purposes of this course, a sample size of \(n>30\) is considered a large sample.

For many people just learning statistics there is a "so what" thought about the CLT. Why is this important and why do I care? If you recall, when we introduced the idea of Z scores we did so with the caveat that the distribution was normal. We take the observed data, that is normally distributed, and convert the data to z scores creating a standard normal distribution. We then leveraged this distribution to find percentiles (and will in future units leverage this to find probabilities.  

The CLT allows us to assume a distribution IS normal as long as the sample size is greater than 30 observations. With this, we can apply most of our inferential statistics without having to compensate for non-normal distributions. This will take on greater relevance as we move through the course. 

With the Central Limit Theorem, we can finally define the sampling distribution of the sample mean.

If the population is normally distributed with mean \(\mu\) and standard deviation \(\sigma\), then the sampling distribution of the sample mean is also normally distributed no matter what the sample size is. When the sampling is done with replacement or if the population size is large compared to the sample size, then \(\bar{x}\) has mean \(\mu\) and standard deviation \(\dfrac{\sigma}{\sqrt{n}}\). We use the term standard error for the standard deviation of a statistic, and since sample average, \(\bar{x}\) is a statistic, standard deviation of \(\bar{x}\) is also called standard error of \(\bar{x}\). However, in some books you may find the term standard error for the estimated standard deviation of \(\bar{x}\). In this class we use the former definition, that is, standard error of \(\bar{x}\) is the same as standard deviation of \(\bar{x}\).

For the sampling distribution of the sample mean, we learned how to apply the Central Limit Theorem when the underlying distribution is not normal. In this section, we will present how we can apply the Central Limit Theorem to find the sampling distribution of the sample proportion. Remember when we introduced quantitative and categorical data? In this example, we are working with a special type of categorical variable called Bernoulli random variable, \(Y\).  

A side note for those who are curious:  A Bernoulli random variable is a very simple kind of variable. It only has two possible values, 0 and 1 and there is only one trial. This is different from a binomial random variable in that there are repeated independent trails. We will not focus too much on these differences in this course but if you are curious this might be information to have!

The Bernoulli random variable is a special case of the Binomial random variable, where the number of trials is equal to one.

Suppose we have, say \(n\), independent trials of this same experiment. Then we would have \(n\) values of \(Y\), namely \(Y_1, Y_2, ...Y_n\).

If we define \(X\) to be the sum of those values, we get...

\(X=\sum_{i=1}^n Y_i\)

\(X\) is then a Binomial random variable with parameters \(n\) and \(p\).

You are probably wondering what this has to do with the sampling distribution of the sample proportion. Well, suppose we have a random sample of size \(n\) from a population and are interested in a particular “success”. Let the probability of success be \(p\). We can label the successes as 1 and the failures as 0. The sample proportion, \(\hat{p}\) would be the sum of all the successes divided by the number in our sample. Therefore,

\(\hat{p}=\dfrac{\sum_{i=1}^n Y_i}{n}=\dfrac{X}{n}\)

In other words, \(\hat{p}\) could be thought of as a mean! If this is the case, we can apply the Central Limit Theorem for large samples!

Therefore, for large samples, the shape of the sampling distribution for $\hat{p}$ will be approximately normal. What about the mean and the standard deviation?

The distribution of the sample proportion approximates a normal distribution under the following 2 conditions.

Over the years the values of the conditions have changed. The examples that follow in the remaining lessons will use the first set of conditions at 5, however, you may come across other books or software that may use 10 or 15 for this value.

In determining what makes a good estimator, there are two key features:

1. The center of the sampling distribution for the estimate is the same as that of the population. When this property is true, the estimate is said to be unbiased. The most often-used measure of the center is the mean.
2. The estimate has the smallest standard error when compared to other estimators. For example, in the normal distribution, the mean and median are essentially the same. However, the standard error of the median is about 1.25 times that of the standard error of the mean. We know the standard error of the mean is \(\frac{\sigma}{\sqrt{n}}\). Therefore in a normal distribution, the SE(median) is about 1.25 times \(\frac{\sigma}{\sqrt{n}}\). This is why the mean is a better estimator than the median when the data is normal (or approximately normal).

In putting the two properties above together, the center of our interval should be the point estimate for the parameter of interest. With the estimated standard error of the point estimate, we can include a measure of confidence to our estimate by forming a margin of error.

This you may have readily seen whenever you have heard or read a sample survey result (e.g. a survey of the current approval rating of the President, or attitude citizens have on some new policy). In such surveys, you may hear reference to the "44% of those surveyed approved of the President's reaction" (this is the sample proportion), and "the survey had a 3.5% margin or error, or ± 3.5%." This latter number is the margin of error.

With the point estimate and the margin of error, we have an interval for which the group conducting the survey is confident the parameter value falls (i.e. the proportion of U.S. citizens who approve of the President's reaction). In this example, that interval would be from 40.5% to 47.5%.

This example provides the general construction of a confidence interval:

The margin of error will consist of two pieces. One is the standard error of the sample statistic. The other is some multiplier, \(M\), of this standard error, based on how confident we want to be in our estimate. This multiplier will come from the same distribution as the sampling distribution of the point estimate; for example, as we will see with the sample proportion this multiplier will come from the standard normal distribution. The general form of the margin of error is shown below.

The interpretation of a confidence interval has the basic template of: "We are 'some level of percent confident' that the 'population of interest' is from 'lower bound to upper bound'. After Ellie calculates at 95% confidence interval, she could say she is 95% confident that the true population average number of miles run by marathon runners is between “the values of the confidence interval”. The phrases in single quotes are replaced with the specific language of the problem. We will discuss more about the interpretation of a confidence interval after we provide a few more examples.

Moving forward...

We're going to begin exploring confidence intervals for one population proportions. The important issue of determining the required sample size to estimate a population proportion will also be discussed in detail in this lesson.

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

1. Step 1: Check Condition

   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
   • \(n(1-\hat{p})>5\)

2. 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.

   What can one do if the conditions are NOT satisfied?

3. Step 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.
4. Step 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.

How to find the multiplier using the Standard Normal Distribution

\(z_a\) 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.

The table is a list of frequently used alphas andtheir  \(z_{\alpha/2}\) multipliers.

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.

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, \(p\).

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.

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).

1. t is symmetric about 0
2. t-distribution is more variable than the Standard Normal distribution
3. t-distributions are different for different degrees of freedom (d.f.).
4. The larger \(n\) gets (or as \(n\) goes to infinity), the closer the \(t\)-distribution is to the \(z\).
5. The meaning of \(t_\alpha\) is the \(t\)-value having the area "\(\alpha\)" to the right of it.

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.

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.