How to find p-hat (the sample proportion)

The formula for the sample proportion

The sample proportion is written $\hat{p}$ and read "p-hat". It is the fraction of your sample that has the characteristic you are counting:

Here $x$ is the number of successes in the sample and $n$ is the sample size. Both are counts of individuals, so $x$ is a whole number, $x$ is never larger than $n$, and $\hat{p}$ always lands between 0 and 1.

The hat is the whole point of the notation. It marks a number computed from one sample, as opposed to $p$ with no hat, which is the proportion in the entire population.

Because $\hat{p}$ is a fraction of two counts, it can only take the values $0/n$, $1/n$, $2/n$, and so on. With $n = 20$, $\hat{p}$ moves in steps of 0.05, so a reported sample proportion of 0.37 from 20 people is impossible.

Define what counts as a success first

A "success" is whichever category you are measuring the proportion of. The word carries no judgment: in a study of defective parts, a defective part is the success.

Write that definition down before you compute, because $\hat{p}$ flips when the definition flips. If 84 of 150 library patrons used the e-book service, then $\hat{p} = 84/150 = 0.56$ for users and $\hat{p} = 66/150 = 0.44$ for non-users. Both numbers are correct arithmetic, and only one answers the question that was asked.

Two habits keep this clean:

• Name the success in a sentence, such as "success = the patron used the e-book service".
• Confirm that your $x$ and your $n$ describe the same group of people, counted the same way.

The second habit matters more than it looks. Most wrong values of $\hat{p}$ come from a numerator and a denominator that were taken from different groups.

Finding p-hat in a word problem

A word problem hands you $x$ and $n$ in ordinary English, usually in that order. Your job is to spot them.

In a random sample of 250 households in a county, 90 said they compost food waste. Success is composting, so $x = 90$ and $n = 250$:

The phrases that signal $n$ are "a random sample of 250", "surveyed 250 households", or "250 were selected". The phrases that signal $x$ are "90 said yes", "90 of them", or "90 responded".

One trap is common enough to plan for. Some problems report the failures instead: "of the 250 households sampled, 160 do not compost". Subtract first, $x = 250 - 160 = 90$, and only then divide.

Finding p-hat in a two-way table

A two-way table gives you several different sample proportions, and the question tells you which denominator to use. Here are 400 library patrons classified by age group and by whether they used the e-book service.

Three questions, three denominators:

• Proportion of all patrons who used e-books: divide by the grand total, $\hat{p} = 186/400 = 0.465$.
• Proportion of the under-30 patrons who used e-books: divide by that row total, $\hat{p} = 84/150 = 0.56$.
• Proportion of the e-book users who are under 30: divide by that column total, $\hat{p} = 84/186 = 0.4516$.

The cell 84 appears in all three, and the answers differ because the group you are describing differs. Before dividing, finish this sentence out loud: "out of the ...". Whatever follows is your denominator. Working with the rows and columns of a table is covered in topic 2.2.

Finding p-hat from a reported percentage

A percentage is a sample proportion that has already been divided. Move the decimal two places left and you have $\hat{p}$ directly, so a reported 43% is $\hat{p} = 0.43$.

Never carry 43 into a formula. Every standard error, confidence interval, and test statistic for proportions expects a decimal between 0 and 1, and feeding in 43 produces answers that are wrong by a factor of 100 or that fail outright.

To recover the count, multiply the proportion by the sample size:

With 43% of $n = 400$ people, $x = 0.43 \times 400 = 172$. That count is what the Large Counts condition needs, since the condition is checked on counts of successes and failures rather than on proportions. See why you need 10 successes and 10 failures.

When the multiplication does not land on a whole number, the reported percentage was rounded. With 43% of 350, you get $0.43 \times 350 = 150.5$, so the true count was 150 or 151 and the percentage was rounded to the nearest whole percent. Say which one you chose and move on.

p-hat is a statistic, p is the parameter

These two symbols are not interchangeable, and the difference drives every inference procedure you will run.

• $\hat{p}$ is a statistic. You computed it from a sample, so you know its value, and a different sample would give a different value.
• $p$ is a parameter. It describes the whole population, its value is fixed, and you almost never know it.

A sample statistic is a point estimator of the matching population parameter, so $\hat{p}$ is the point estimate of $p$. It is also an unbiased estimator: across all possible samples of size $n$, the values of $\hat{p}$ center on $p$ rather than sitting systematically above or below it.

That is why hypotheses are always written about $p$ and never about $\hat{p}$. Writing $H_0: \hat{p} = 0.25$ claims something about a number you already computed, which is not a claim worth testing. More on the split in parameter vs statistic and null vs alternative hypothesis.

Why p-hat changes from sample to sample

Take a second random sample of 400 patrons and you will get a different $\hat{p}$. Nothing went wrong. Two samples contain different people, so their proportions differ, and that is called sampling variability.

Collect the values of $\hat{p}$ from every possible sample of size $n$ and you get the sampling distribution of the sample proportion, described in topic 3.2. When the sampled values are independent, that distribution has

The symbol $\mu_{\hat{p}}$ (mu sub p-hat) is the mean of all those sample proportions, and $\sigma_{\hat{p}}$ (sigma sub p-hat) is their standard deviation. Since $p$ is unknown in practice, you substitute your own $\hat{p}$ and call the result the standard error:

Two readable facts come straight out of the formula. Because $n$ sits in the denominator, larger samples give less variable estimates: at $p = 0.5$, moving from $n = 100$ to $n = 400$ drops the standard deviation from $\sqrt{0.25/100} = 0.05$ to $\sqrt{0.25/400} = 0.025$, so quadrupling $n$ halves it. Because $p(1-p)$ is largest at $p = 0.5$, proportions near a half are the noisiest to estimate.

That standard error is the engine of the confidence interval: the interval is $\hat{p}$ plus or minus a critical value times $SE_{\hat{p}}$. The difference between $\sigma_{\hat{p}}$, which uses the true $p$, and $SE_{\hat{p}}$, which uses your estimate, is worth keeping straight in every write-up.

Mistakes that cost points

• Using 43 instead of 0.43. Convert the percentage before it touches a formula.
• Swapping $x$ and $n$. Any $\hat{p}$ above 1 is that error, every time.
• Taking the numerator from one group and the denominator from another in a two-way table.
• Writing $\hat{p}$ in a hypothesis. Hypotheses are claims about $p$.
• Rounding $\hat{p}$ to two decimals and then building an interval from it. Carry the full value, or the fraction $x/n$, through the arithmetic and round only the final answer.
• Reporting $\hat{p}$ with no sample size attached. A proportion of 0.60 means something very different from 5 people than from 500, because the standard error changes by a factor of 10.