Sample proportions

Sampling distribution of sample proportions

When you work with sample proportions, you’re measuring the proportion of successes in a sample.

Definitions

Success

An outcome where the individual has the characteristic being studied.

Failure

An outcome where the individual does not have the characteristic being studied.

For example, if the question is “What proportion of high school students have at least one cat?”, then a success is a student who has at least one cat, and a failure is a student who has no cats. The population is split into people who have the characteristic and people who don’t.

If you want to estimate a population proportion using a single sample proportion, it helps to understand how sample proportions typically vary from sample to sample.

Recall the binomial setting:

• $np$ is the mean number of successes.
• $\sqrt{np(1-p)}$ is the standard deviation of the number of successes.
• Large values of $n$ are needed for the binomial distribution to be well-approximated by a normal distribution.

Proportion rather than count

Here, we care about the proportion of successes rather than the count. The idea is:

1. Start with a population where the true success proportion is $p$.
2. Consider all possible samples of size $n$.
3. Compute the sample proportion $\hat{p}$ for each sample.

Sidenote

Important to know!

1. The set of all sample proportions is approximately normally distributed.
2. The mean of the sampling distribution, ${\mu }_{\hat{p}}$, equals the population proportion $p$. That is, ${\mu }_{\hat{p}}=p$.
3. The standard deviation of the sampling distribution is approximately:

$${\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$$

Overall, the sampling distribution of $\hat{p}$ is approximately normal with mean $p$ and standard deviation $\sqrt{\frac{p(1-p)}{n}}$, as long as:

• $np\ge 10$
• $n(1-p)\ge 10$

Where $n$ is the sample size and $p$ is the probability of one success.

Practice problem

Example 1:

It is estimated that $30\%$ of people will experience cancer in their lifetime. In a random sample of $150$ people, what is the probability that less than $50$ of them will experience cancer in their lifetime?

First, check the normal-approximation conditions $np\ge 10$ and $n(1-p)\ge 10$, where $n=150$ and $p=0.3$. In this context, a “success” means a person experiences cancer in their lifetime.

$$\begin{array}{rl}np& =150\times 0.3\\ & =45\\ & \ge 10✓\\ & \\ n(1-p)& =150\times 0.7\\ & =105\\ & \ge 10✓\end{array}$$

The sampling distribution of $\hat{p}$ is approximately normal. Next, compute the standard deviation:

$$\begin{array}{rl}{\sigma }_{\hat{p}}& =\sqrt{\frac{p(1-p)}{n}}\\ & =\sqrt{\frac{0.3(0.7)}{150}}\\ & =\sqrt{0.0014}\\ & \approx 0.0374\end{array}$$

The event “less than $50$ out of $150$” converts to a sample proportion:

$$\hat{p}<\frac{50}{150}=0.33$$

Now compute the probability using a z-score:

$$\begin{array}{rl}P(\hat{p}<0.33)& =P\left(z<\frac{0.33-0.3}{0.0374}\right)\\ & =P(z<0.802)\\ & =0.7887\end{array}$$

Solution:

Comparing two proportions

There are many real-world situations where you compare two population proportions.

For example:

• Is the proportion of parents satisfied with public school education greater than the proportion satisfied with private school education?
• How does the proportion of dentists who recommend toothbrush brand “A” compare to the proportion who recommend brand “B”?

In these settings, individuals might fall into different categories — a dentist might recommend only “A”, only “B”, both, or neither. The key question is whether an observed difference in sample proportions is large enough to be meaningful, or whether it can be explained by natural sample-to-sample variation.

The set of all differences of sample proportions is approximately normally distributed.

To compare sampling distributions of sample proportions:

1. Start with two populations with proportions $p_1$ and $p_2$. Take samples of sizes $n_1$ and $n_2$.
2. Compute the difference ${\hat{p}}_1-{\hat{p}}_2$ for each pair of samples.
3. The mean of the sampling distribution of differences is $p_1-p_2$.
4. The standard deviation (also the standard error) is:

$${\sigma }_{{\hat{p}}_1-{\hat{p}}_2}=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}$$

Practice problem

Example 2:

A researcher is testing whether a new study method improves test performance among students. He randomly assigns students to one of two groups: Group A uses the new study method and Group B uses the old study method. At the end of the term, $33$ out of $80$ students in Group A passed the final exam, while $52$ out of $130$ students in Group B passed. What is the probability that the observed difference in passing rates is at least $10\%$, assuming there is no actual difference between the two methods?

First, define and compute the sample proportions:

$$\begin{array}{rl}{\hat{p}}_1& =\frac{33}{80}=0.4125\text{(Group A, new method)}\\ {\hat{p}}_2& =\frac{52}{130}=0.4\text{(Group B, old method)}\end{array}$$

Assume the null hypothesis is true — that there is no real difference in population passing rates, so any observed difference is due to sampling variation:

$$H_0:p_1=p_2$$

Compute the standard error:

$$\begin{array}{rl}\text{SE}& =\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}\\ & =\sqrt{\frac{0.4125(0.5875)}{80}+\frac{0.4(0.6)}{130}}\\ & =\sqrt{\frac{0.242}{80}+\frac{0.24}{130}}\\ & \approx 0.0698\end{array}$$

Now find the probability that the difference is at least $10\%$:

$$\begin{array}{rl}P({\hat{p}}_1-{\hat{p}}_2\ge 0.10)& =P\left(Z\ge \frac{0.10-0.0125}{0.0698}\right)\\ & =P(Z\ge 1.2536)\\ & \approx 0.1056\end{array}$$

Solution:

Continuity correction

When using the normal distribution to approximate binomial probabilities, it’s important to remember that the binomial distribution is discrete (counts like $0,1,2,\dots$) while the normal distribution is continuous. The continuity correction adjusts for this mismatch by shifting boundaries by $0.5$ units.

Definitions

Continuity correction

An adjustment applied when using the normal distribution to approximate a discrete binomial probability, shifting each boundary by $0.5$.

• For $P(X\le k)$, use $P(X\le k+0.5)$
• For $P(X<k)$, use $P(X\le k-0.5)$
• For $P(X\ge k)$, use $P(X\ge k-0.5)$
• For $P(X>k)$, use $P(X\ge k+0.5)$
• For $P(X=k)$, use $P(k-0.5\le X\le k+0.5)$

Practice problem

Example 3:

The census confirms that $13\%$ of people are left-handed. In a sample of $100$ people, what is the probability that at least $17$ are left-handed?

With the continuity correction (more accurate):

Since we want $P(X\ge 17)$, apply the correction and use $X\ge 16.5$, which converts to $\hat{p}=0.165$:

$$\begin{array}{rl}z& =\frac{0.165-0.13}{\sqrt{\frac{0.13(0.87)}{100}}}\approx 1.04\\ P& \approx 0.149\end{array}$$

Without the continuity correction (less accurate):

$$\begin{array}{rl}z& =\frac{0.17-0.13}{\sqrt{\frac{0.13(0.87)}{100}}}\approx 1.19\\ P& \approx 0.117\end{array}$$

Solution: