20.3 Sampling distribution: Unknown proportion

In the die example (Sect. 20.1), an equation was given for computing the standard error for the sample proportion for samples of size $n$, when the value of $p$ was known.

However, usually the value of $p$ (the parameter) is unknown; after all, the reason for taking a sample is to estimate the unknown value of $p$. When $p$ is unknown, the best available estimate can be used, which is $\hat{p}$. When the value of $p$ is unknown, the standard error of the sample proportion (written $\text{s.e.}\left(\hat{p}\right)$) is approximately

$$\text{s.e.}\left(\hat{p}\right)=\sqrt{\frac{\hat{p}\times (1-\hat{p})}{n}}.$$

Definition 20.2 (Sampling distribution of a sample proportion when $p$ is unknown) When the value of $p$ is unknown, the sampling distribution of the sample proportion is described by

• an approximate normal distribution,
• centred around the (unknown) mean of $p$,
• with a standard deviation (called the standard error of $\hat{p}$) of

$$\begin{array}{}\text{s.e.}\left(\hat{p}\right)=\sqrt{\frac{\hat{p}\times (1-\hat{p})}{n}},& \text{(20.3)}\end{array}$$

when certain conditions are met, where $n$ is the size of the sample, and $\hat{p}$ is the sample proportion.

In general, the approximation gets better as the sample size gets larger.

Let’s pretend for the moment that the proportion of even rolls of a fair die is unknown (to demonstrate some points). In this case, an estimate of the proportion of even rolls can be found by rolling a die $n=25$ times and computing $\hat{p}$.

Suppose 11 of the $n=25$ rolls produced an even number, so that $\hat{p}=11/25=0.44$. Then (from Definition 20.2),

$$\text{s.e.}\left(\hat{p}\right)=\sqrt{\frac{0.44\times (1-0.44)}{25}}=\mathrm{0.099277.}$$ (This is very similar to the value of 0.1, the value of the standard error when the value of $p$ was known; see Eq. (20.2).)

Hence, the sample proportions would vary with an approximate normal distribution (Fig.20.3), centred around the unknown value of $p$ with a standard deviation of $\text{s.e.}\left(\hat{p}\right)=0.099277$.

FIGURE 20.3: The normal distribution, showing how the proportion of even rolls varies when a die is rolled 25 times

Using the 68–95–99.7 rule again:

About 95% of the values of $\hat{p}$ are expected to be between $p-0.199$ and $p+0.199$.

Though we are pretending the value of $p$ is unknown, the value of $\hat{p}$ is known however. What if the roles of $p$ and $\hat{p}$ were ‘reversed?’ Then,

About 95% of the values of $p$ are expected to be between $\hat{p}-0.199$ and $\hat{p}+0.199$.

Since $\hat{p}=0.44$, this is equivalent to:

About 95% of the values of $p$ are expected to be between $0.24$ and $0.64$.

This interpretation is not quite correct, but the idea seems reasonable. This is called a confidence interval (or CI), based on ideas from Sect. 20.2.

In summary, using $\hat{p}=0.44$ and $\text{s.e.}\left(\hat{p}\right)=0.0993$, the (approximate) 95% CI is

$$0.44\pm (2\times 0.0993),$$ or from 0.241 to 0.639. This CI straddles the population proportion of $p=0.5$, though we would not know this if $p$ truly was unknown.

In this case, we know the value of the population parameter: $p=0.5$.

Usually we do not know the value of the parameter: that’s why we are taking a sample.