Introduction to confidence intervals

Confidence intervals

A confidence interval gives a range of plausible values for a population parameter (such as a true proportion or a true mean), based on sample data.

Example:

Suppose there are $100$ products being tested. In this sample, the manufacturer decides to move forward with $6$ of them. If a $90\%$ confidence interval is $(3,9)$, this means we can be $90\%$ confident that the true percentage of products the manufacturer would move forward with (in the overall population or long run) is between $3\%$ and $9\%$. This does not mean there is a $90\%$ probability that the true percentage is in the interval. The true percentage is a single fixed value. It is either inside the interval (between $3\%$ and $9\%$) or it is not.

As the confidence level increases, the interval must get wider to capture the true value more often. As the confidence level decreases, the interval can be narrower because we’re willing to be less certain. In practice, $90\%$, $95\%$, and $99\%$ confidence intervals are the most common, but you can construct a confidence interval for any confidence level.

Conditions for inference

Before constructing a confidence interval, we need to verify that certain conditions are met so that our inference methods are valid.

Definitions

Independence

The trials in the sample or experiment must be independent of one another. For samples, this means using random sampling methods and having a relatively large sample size. Also, the sample size $n$ should typically be no larger than $10\%$ of the overall population (because sampling without replacement can create dependence when the sample is too large). For experiments, subjects should be randomly assigned to treatments, and there should be a control group.

Normality

Inference for proportions uses the normal model. When you run multiple independent trials with two outcomes, the underlying model is binomial, but the binomial distribution is approximately normal as long as $n\hat{p}$ and $n(1-\hat{p})$ are both at least $10$ (where $\hat{p}$ is the sample proportion and $n$ is the number of trials).

Inference for means also uses the normal model. This is exact when the population is normal. Even when the population is not normal, the central limit theorem says that if the sample size is large enough (typically at least $30$), the distribution of sample means is approximately normal.

Confidence intervals for a proportion

If we want to estimate a population proportion $p$ using a sample proportion $\hat{p}$, it helps to remember the key facts about the sampling distribution of $\hat{p}$:

Conditions for using the normal model

• To use the normal approximation methods, we need:

$$\begin{array}{rl}n\hat{p}& \ge 10\\ n(1-\hat{p})& \ge 10\end{array}$$

Otherwise, the sampling distribution may not be approximately normal.

Shape and center

• The set of all sample proportions is approximately normally distributed (by the central limit theorem).
• The mean $\widehat{p_{\mu }}$ of the sampling distribution of sample proportions is equal to $\hat{p}$, the population proportion.

Spread

The standard deviation $\widehat{p_{\sigma }}$ of the sampling distribution is approximately:

$$\begin{array}{rl}\widehat{p_{\sigma }}& =\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\end{array}$$

Once those conditions are satisfied, we can use the sampling distribution to build an interval around our sample estimate.

A confidence interval is equal to the estimate $\pm$ the margin of error.

$$\begin{array}{rl}\text{Margin of Error (ME)}& =z^{\ast }\times \text{SE}\end{array}$$

Practice problem

The example below walks through constructing a confidence interval for a proportion, using it to evaluate a claim, and thinking about how the confidence level affects the width of the interval.

Example 1:

Suppose a sample of $1000$ students from the University of Michigan is surveyed. Suppose that $330$ of them have at least some student loans.

Part a.

Determine a $90\%$ confidence interval for the true proportion of students from this university who have student loans.

From the problem:

• Sample size: $n=1000$
• Sample proportion: $\hat{p}=\frac{330}{1000}=0.33$
• Confidence level: $90\%$

Check the large-counts condition:

$$\begin{array}{rl}n\hat{p}& =1000\times \frac{330}{1000}\\ & =330\end{array}$$

$$\begin{array}{rl}n(1-\hat{p})& =1000\times \frac{670}{1000}\\ & =670\end{array}$$

Both are at least $10$, so using the normal model is reasonable.

For a $90\%$ confidence interval, the two-tailed critical z-value is $1.645$.

Now compute the standard error and margin of error:

$$\begin{array}{rl}\text{SE}& =\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\\ & =\sqrt{\frac{0.33\times 0.67}{1000}}\\ & =\sqrt{0.0002211}\\ & =0.01487\end{array}$$

$$\begin{array}{rl}\text{ME}& =z^{\ast }\times \text{SE}\\ & =1.645\times 0.01487\\ & =0.0245\end{array}$$

Construct the interval:

$$\begin{array}{rl}\hat{p}\pm \text{ME}& =0.33\pm 0.0245\end{array}$$

Solution:

With the interval in hand, we can now use it to evaluate specific claims about the population proportion.

Part b.

Does your answer from Part (a.) support the claim that at least $50\%$ of all students at the University of Michigan have student loans?

Solution:

Finally, let’s think about how changing the confidence level would change the width of the interval itself.

Part c.

Suppose you were to calculate a $95\%$ confidence interval next. Would you expect this interval to be larger or smaller than the $90\%$ one calculated in Part (a.)? Why?

Solution: