# 6.2: The Sampling Distribution for Proportions

- Page ID
- 25665

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

- To recognize that the sample proportion \(\hat{p}\) is a random variable.
- To understand the meaning of the formulas for the mean and standard deviation of the sample proportion.
- To learn what the sampling distribution of \(\hat{p}\) is when the sample size is large.

Often sampling is done in order to estimate the proportion of a population that has a specific characteristic, such as the proportion of all items coming off an assembly line that are defective or the proportion of all people entering a retail store who make a purchase before leaving. The population proportion is denoted \(p\) and the sample proportion is denoted \(\hat{p}\). Thus if in reality \(43\%\) of people entering a store make a purchase before leaving,

\[p = 0.43 \nonumber\]

if in a sample of \(200\) people entering the store, \(78\) make a purchase,

\[\hat{p}=\dfrac{78}{200}=0.39. \nonumber\]

The sample proportion is a random variable: it varies from sample to sample in a way that cannot be predicted with certainty. Viewed as a random variable it will be written \(\hat{P}\). It has a mean \(μ_{\hat{P}}\) and a standard deviation \(σ_{\hat{P}}\). Here are formulas for their values.

Suppose random samples of size \(n\) are drawn from a population in which the proportion with a characteristic of interest is \(p\). The mean \(μ_{\hat{P}}\) and standard deviation \(σ_{\hat{P}}\) of the sample proportion \(\hat{P}\) satisfy

\[μ_{\hat{P}}=p\]

and

\[σ_{\hat{P}}= \sqrt{\dfrac{pq}{n}}\]

where \(q=1−p\).

The Central Limit Theorem has an analogue for the population proportion \(\hat{p}\). To see how, imagine that every element of the population that has the characteristic of interest is labeled with a \(1\), and that every element that does not is labeled with a \(0\). This gives a numerical population consisting entirely of zeros and ones. Clearly the proportion of the population with the special characteristic is the proportion of the numerical population that are ones; in symbols,

\[p=\dfrac{\text{number of 1s}}{N}\]

But of course the sum of all the zeros and ones is simply the number of ones, so the mean \(μ\) of the numerical population is

\[μ=\dfrac{ \sum x}{N}= \dfrac{\text{number of 1s}}{N}\]

Thus the population proportion \(p\) is the same as the mean \(μ\) of the corresponding population of zeros and ones. In the same way the sample proportion \(\hat{p}\) is the same as the sample mean \(\bar{x}\). Thus the Central Limit Theorem applies to \(\hat{p}\). However, the condition that the sample be large is a little more complicated than just being of size at least \(30\).

## The Sampling Distribution of the Sample Proportion

For large samples, the sample proportion is approximately normally distributed, with mean \(μ_{\hat{P}}=p\) and standard deviation \(\sigma _{\hat{P}}=\sqrt{\frac{pq}{n}}\).

A sample is large if the interval \(\left [ p-3\sigma _{\hat{p}},\, p+3\sigma _{\hat{p}} \right ]\) lies wholly within the interval \([0,1]\).

In actual practice \(p\) is not known, hence neither is \(σ_{\hat{P}}\). In that case in order to check that the sample is sufficiently large we substitute the known quantity \(\hat{p}\) for \(p\). This means checking that the interval

\[\left [ \hat{p}-3\sqrt{\frac{\hat{p}(1-\hat{p})}{n}},\, \hat{p}+3\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \right ]\]

lies wholly within the interval \([0,1]\). This is illustrated in the examples.

Figure 6.2.1 shows that when \(p = 0.1\), a sample of size \(15\) is too small but a sample of size \(100\) is acceptable.

Figure 6.2.2 shows that when \(p=0.5\) a sample of size \(15\) is acceptable.

Suppose that in a population of voters in a certain region \(38\%\) are in favor of particular bond issue. Nine hundred randomly selected voters are asked if they favor the bond issue.

- Verify that the sample proportion \(\hat{p}\) computed from samples of size \(900\) meets the condition that its sampling distribution be approximately normal.
- Find the probability that the sample proportion computed from a sample of size \(900\) will be within \(5\) percentage points of the true population proportion.

**Solution**:

- The information given is that \(p=0.38\), hence \(q=1-p=0.62\). First we use the formulas to compute the mean and standard deviation of \(\hat{p}\):

\[\mu _{\hat{p}}=p=0.38\; \text{and}\; \sigma _{\hat{P}}=\sqrt{\frac{pq}{n}}=\sqrt{\frac{(0.38)(0.62)}{900}}=0.01618 \nonumber\]

Then \(3\sigma _{\hat{P}}=3(0.01618)=0.04854\approx 0.05\) so

\[\left [ \hat{p} - 3\sqrt{\frac{\hat{p}(1-\hat{p})}{n}},\, \hat{p}+3\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \right ]=[0.38-0.05,0.38+0.05]=[0.33,0.43] \nonumber\]

which lies wholly within the interval \([0,1]\), so it is safe to assume that \(\hat{p}\) is approximately normally distributed.

- To be within \(5\) percentage points of the true population proportion \(0.38\) means to be between \(0.38-0.05=0.33\) and \(0.38+0.05=0.43\). Thus

\[\begin{align*} P(0.33<\hat{P}<0.43) &= P\left ( \frac{0.33-\mu _{\hat{P}}}{\sigma _{\hat{P}}} <Z< \frac{0.43-\mu _{\hat{P}}}{\sigma _{\hat{P}}} \right )\\[4pt] &= P\left ( \frac{0.33-0.38}{0.01618} <Z< \frac{0.43-0.38}{0.01618}\right )\\[4pt] &= P(-3.09<Z<3.09)\\[4pt] &= P(3.09)-P(-3.09)\\[4pt] &= 0.9990-0.0010\\[4pt] &= 0.9980 \end{align*}\]

An online retailer claims that \(90\%\) of all orders are shipped within \(12\) hours of being received. A consumer group placed \(121\) orders of different sizes and at different times of day; \(102\) orders were shipped within \(12\) hours.

- Compute the sample proportion of items shipped within \(12\) hours.
- Confirm that the sample is large enough to assume that the sample proportion is normally distributed. Use \(p=0.90\), corresponding to the assumption that the retailer’s claim is valid.
- Assuming the retailer’s claim is true, find the probability that a sample of size \(121\) would produce a sample proportion so low as was observed in this sample.
- Based on the answer to part (c), draw a conclusion about the retailer’s claim.

**Solution**:

- The sample proportion is the number \(x\) of orders that are shipped within \(12\) hours divided by the number \(n\) of orders in the sample:

\[\hat{p} =\frac{x}{n}=\frac{102}{121}=0.84\nonumber\]

- Since \(p=0.90\), \(q=1-p=0.10\), and \(n=121\),

\[\sigma _{\hat{P}}=\sqrt{\frac{(0.90)(0.10)}{121}}=0.0\overline{27}\nonumber\]

hence

\[\left [ p-3\sigma _{\hat{P}},\, p+3\sigma _{\hat{P}} \right ]=[0.90-0.08,0.90+0.08]=[0.82,0.98]\nonumber\]

Because\[[0.82,0.98]⊂[0,1]\nonumber\]

it is appropriate to use the normal distribution to compute probabilities related to the sample proportion \(\hat{P}\).

- Using the value of \(\hat{P}\) from part (a) and the computation in part (b),

\[\begin{align*} P(\hat{P}\leq 0.84) &= P\left ( Z\leq \frac{0.84-\mu _{\hat{P}}}{\sigma _{\hat{P}}} \right )\\[4pt] &= P\left ( Z\leq \frac{0.84-0.90}{0.0\overline{27}} \right )\\[4pt] &= P(Z\leq -2.20)\\[4pt] &= 0.0139 \end{align*}\]

- The computation shows that a random sample of size \(121\) has only about a \(1.4\%\) chance of producing a sample proportion as the one that was observed, \(\hat{p} =0.84\), when taken from a population in which the actual proportion is \(0.90\). This is so unlikely that it is reasonable to conclude that the actual value of \(p\) is less than the \(90\%\) claimed.

## Key Takeaway

- The sample proportion is a random variable \(\hat{P}\).
- There are formulas for the mean \(μ_{\hat{P}}\), and standard deviation \(σ_{\hat{P}}\) of the sample proportion.
- When the sample size is large the sample proportion is normally distributed.