Section 7.4: The Distribution of the Sample Proportion

Chapter 7 is written to summarize continuous random variables and their sampling distributions. So far, we have only discussed quantitative variables: Wait times for trains, lifespan of cell phones, weights of chocolate bars, etc. Our next step is to develop a method to view qualitative variables with the same types of methods. We will do this through random sampling and counting the numbers in the sample that have a qualitative characteristic, then computing the sample proportions with that characteristic. This process creates a new sampling variable that is technically not continuous, but can be approximated using the normal distribution, which is continuous.

Sample Proportions

Solution:This could have been a question in chapter 5. The sample proportion is simply the probability that a randomly selected person selected “Has a positive impact on the country’s culture”.

So,  $\widehat{p}=\frac{2121}{3314}\approx 0.6400$. From this sample of n = 3314, about 64% of them believe that the ethnic and racial makeup of our country has a positive impact on our culture.

Now, if we took another random sample of a different group of 3314 people, the sample proportion would likely be different. If we took a second sample and found 2185 people who selected “Has a positive impact”, then the sample proportion would have been $\widehat{p}=\frac{2185}{3314}\approx 0.6593$. Yet a third sample might have had 2115 select “Has a positive impact”, with a resulting sample proportion $\widehat{p}=\frac{2115}{3314}\approx 0.6382$. Each sample we take will have a different sample proportion. If we create all possible sample proportions and their probability, we have created the distribution of sample proportions.

Table 2 contains 100 simulated samples for surveys of 3314 individuals. Each simulated survey was given to 3314 individuals and reports the number of “Has a positive impact on the country’s culture” responses and the sample proportion. You can open the Table 2 data (opens in a new tab) in STATBELT (download CSV).

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Solution:Open STATBELT Utility. Open the data using the link, copy the values in the “Positive Impact” and “Sample Proportion” columns (not the column names), and paste them into the first two columns of the spreadsheet. Type “Positive Impact” and “Sample Proportion” in the column headings. You can ignore the “sample number” column.

To make the histogram, select Graphs → Histogram, and choose “Sample Proportion” for the column. You can type the start and bin width or let STATBELT choose them for you. Click “Open graph in a new window.” The histogram is shown in figure 7.29.

Figure 7.29: Histogram for simulated sample proportions.

You may not be surprised to see that the distribution of sample proportions looks bell-shaped. The distribution will be normal under the right conditions, which we will present shortly.

To get the mean and standard deviation, you can select Descriptive Statistics → Basic Statistics, check “Sample Proportion” under “Columns to include,” and click “Show results in a new window,” as we did in chapter 3. The mean is given as 0.6388, and the standard deviation is given as 0.0089.

Another option is to mark the mean on the histogram. When you choose Graphs → Histogram and choose “Sample Proportion”, check “Show mean (x̄) in green.” The histogram then shows a green triangle at the mean, labeled with its value. STATBELT does not draw a normal curve over a histogram, so we judge the bell shape by eye. See figure 7.30.

Figure 7.30: Histogram for simulated sample proportions with the mean marked.

Once again, we see that the mean is 0.6388, and Basic Statistics gave the standard deviation as 0.0089.

So, it appears that the distribution of sample proportions has the following description:

•Center- The mean is approximately $0.6388$.

•Variation- The standard deviation is approximately 0.0089.

•Shape- The distribution appears to be normally distributed.

What we have seen so far is that if we consider the distribution of sample proportions, it appears to follow a bell shaped distribution. But is that always the case? Sadly, no.

Solution:In the first sample, there were X = 2 twos, so the sample proportion is  $\widehat{p}=\frac{2}{10}=0.2$.

The second sample had only X = 1 two, so the sample proportion for that sample is $\widehat{p}=\frac{1}{10}=0.1$

The third sample had X = 2 twos, so the sample proportion for that sample is $\widehat{p}=\frac{2}{10}=0.2$.

The theoretical probability of rolling a two is $\frac{1}{6}\approx 0.1667$. The sample proportions we have are 0.2, 0.1, and 0.2. These sample proportions cannot be equal to 1/6 because the sample size is 10, so all sample proportions will be tenths. But the mean of the first three sample proportions is

$$\begin{array}{cc}\mu & =\frac{0.2+0.1+0.2}{3}=\frac{0.5}{3}=\frac{1}{6}\approx 0.1667.\end{array}$$

So, it appears that the mean of the sample proportions is close to the probability of rolling a two. This makes sense. Rolling dice and counting the number of twos is a binomial experiment with expected value $\mathrm{np}=10\times \frac{1}{6}\approx 1.667$ twos per roll of 10 dice. Some samples will have more twos, and some will have fewer. But they will average out to about 1.667 twos per sample of 10, giving a proportion of $\frac{1.667}{10}=0.1667$, which is also the probability of rolling a two in one roll. (The mean of the distribution of sample proportions equals the probability of success in the binomial experiment!)

Table 3 contains the sample proportions for all 100 samples of size n = 10. You can open the Table 3 data (opens in a new tab) in STATBELT (download CSV). If you make a histogram of the sample proportions, we can see that it is skewed to the right. See figure 7.31.

Figure 7.31: Histogram for sample proportions for the number of Twos in 10 rolled dice.

The histogram looks like it is a little bell-shaped, but not close enough for us to call it a normal distribution. This shows that the distribution of sample proportions is not normal in this case.

So, why is the distribution of sample proportions normal in example 2, but not in example 3? The answer has to do with the binomial random variable X. We created $\widehat{p}=\frac{X}{n}$ by dividing the binomial variable X by n. Dividing by n merely changes the vertical scale on the distribution but does not affect the shape. So, if X is approximately normally distributed, then $\widehat{p}=\frac{X}{n}$ is too. We saw in section 6.3 that a binomial random variable X will have a bell-shaped distribution if $n\cdot p\cdot \left(1-p\right)\ge 10$.

In example 1, we had n = 3314 and p = 0.64 (approximated), so

$$\begin{array}{cc}n\cdot p\cdot \left(1-p\right)& =\left(3314\right)\times \left(0.64\right)\times \left(1-0.64\right)=763.5\ge 10.\end{array}$$

This confirms that X is approximately normally distributed, and so $\widehat{p}=\frac{X}{n}$ is as well.

In example 3, we had n = 10 and p = 1/6, so

$$\begin{array}{cc}n\cdot p\cdot \left(1-p\right)& =\left(10\right)\times \left(\frac{1}{6}\right)\times \left(1-\frac{1}{6}\right)=\frac{50}{36}\approx 1.39<10.\end{array}$$

This shows that X is not bell-shaped, so neither X nor $\widehat{p}=\frac{X}{n}$ are approximately normally distributed.

Solution:We will leave it to the reader to check that the variable X is binomial, with n = 10 and p = 1/6.

•According to the central limit theorem, the mean of the $\widehat{p}$ distribution is ${\mu }_{\widehat{p}}=p=\frac{1}{6}$. (When we found the mean of the first three sample proportions in example 3, the result was already exactly 1/6.) The mean of the 100 sample proportions in Table 3 is 0.169, which is very close to $p=\frac{1}{6}\approx 0.167$. See figure 7.32. The mean of the sample proportions appears to be equal to p = 1/6 based on the experiment.

Figure 7.32: Histogram for sample proportions for the number of Twos in 10 rolled dice, with the mean marked.

•The standard deviation for the distribution of sample proportions is ${\sigma }_{\widehat{p}}=\sqrt{\left(\frac{p\left(1-p\right)}{n}\right)}$. If we substitute n = 10 and p = 1/6, we get

$$\begin{array}{cc}{\sigma }_{\widehat{p}}& =\sqrt{\left(\frac{p\left(1-p\right)}{n}\right)}=\sqrt{\left(\frac{\left(\frac{1}{6}\right)\left(1-\frac{1}{6}\right)}{10}\right)}\end{array}$$

$$\begin{array}{cc}& =\sqrt{\left(\frac{\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)}{10}\right)}\end{array}$$

$$\begin{array}{cc}& =\sqrt{\frac{5}{360}}\end{array}$$

$$\begin{array}{cc}& \approx \sqrt{0.013888889}\end{array}$$

$$\begin{array}{cc}& \approx 0.11785\end{array}$$

The theoretical standard deviation for the distribution of sample proportions is approximately 0.1179. The standard deviation from the 100 samples was 0.1237. These are close as they both round off to 0.12 at two decimal places. This suggests that the formula for the standard deviation given in the central limit theorem for proportions is reasonable.

We have already noted that the distribution in this case is not normal, so we have described the center using the mean, the variation using the standard deviation, and the shape as not normal.

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Solution:To describe the sampling distribution for sample proportions $\widehat{p}=\frac{X}{n}$, we need to know the population proportion p and the sample size n. In this example, it is reported that 65% of college students skip purchasing textbooks due to the cost, so we will use p = 0.65. Random samples of size n = 160 are taken.

•The mean is ${\mu }_{\widehat{p}}=p=0.65$.

•The standard deviation must be calculated.

$$\begin{array}{cc}{\sigma }_{\widehat{p}}& =\sqrt{\left(\frac{p\left(1-p\right)}{n}\right)}=\sqrt{\left(\frac{\left(0.65\right)\left(1-0.65\right)}{160}\right)}\end{array}$$

$$\begin{array}{cc}& =\sqrt{\left(\frac{\left(0.65\right)\left(0.35\right)}{160}\right)}\end{array}$$

$$\begin{array}{cc}& =\sqrt{0.001421875}\end{array}$$

$$\begin{array}{cc}& \approx 0.037708\end{array}$$

•To determine the shape, we check if $n\cdot p\cdot \left(1-p\right)\ge 10$.

$$\begin{array}{cc}& n\cdot p\cdot \left(1-p\right)=160\times \left(0.65\right)\times \left(1-0.65\right)=160\times \left(0.65\right)\times \left(0.35\right)=36.4\ge 10\mathrm{.}\end{array}$$

Since $n\cdot p\cdot \left(1-p\right)\ge 10$, the distribution of sampling proportions will have an approximately normal distribution.

Tip!

Solution:a)To describe the sampling distribution for sample proportions $\widehat{p}=\frac{X}{n}$, we need to know the population proportion p and the sample size n. In this example, it is reported that 82% of the population has been infected with COVID-19 at least once, so p = 0.82. Random samples of size n = 200 are taken.

◦The mean is ${\mu }_{\widehat{p}}=p=0.82$.

◦The standard deviation is calculated as follows.

$$\begin{array}{cc}{\sigma }_{\widehat{p}}& =\sqrt{\left(\frac{p\left(1-p\right)}{n}\right)}=\sqrt{\left(\frac{\left(0.82\right)\left(1-0.82\right)}{200}\right)}\end{array}$$

$$\begin{array}{cc}& =\sqrt{\left(\frac{\left(0.82\right)\left(0.18\right)}{200}\right)}\end{array}$$

$$\begin{array}{cc}& =\sqrt{0.000738}\end{array}$$

$$\begin{array}{cc}& \approx 0.027166\end{array}$$

◦To determine the shape, we check if $n\cdot p\cdot \left(1-p\right)\ge 10$.

$$\begin{array}{cc}n\cdot p\cdot \left(1-p\right)& =200\times \left(0.82\right)\times \left(1-0.82\right)\\ & =200\times \left(0.82\right)\times \left(0.18\right)\\ & =29.52\ge 10\mathrm{.}\end{array}$$

Since $n\cdot p\cdot \left(1-p\right)\ge 10$, the distribution of sampling proportions will have an approximately normal distribution.

b)Since we know the distribution of sample proportions has a normal distribution with mean $0.82$ and standard deviation $0.027166$, we can use a normal calculator. Open STATBELT Utility and select Calculators → Normal Calculator. Type in the mean as $0.82$ and standard deviation as $0.027166$. Symbolically, the probability that a sample proportion is less than 75% is

$P(\widehat{p}<0.75)$. Choose “P(X < a)” and type 0.75 for the value a. The result is $P\left(\widehat{p}<0.75\right)\approx 0.0049867$, or 0.0050 rounded to 4 decimal places.

Note on part b): If we treat X as binomial, then we can answer part (b) using the binomial calculator and get the exact answer rather than the approximation. Since 75% of 200 is 150, having a sample proportion less than 75% means the same as having X < 150. The exact answer using n = 200 and p = 0.82 is $P\left(X<150\right)\approx 0.0051$, which is very close to the answer we found before. Don’t forget that the binomial calculator can give the exact probability where the normal approximation to the distribution of sample proportions gives us a close approximation!

Tip!

c)In part c) we are asked to use the distribution of sample proportions to estimate the probability that a sample would have 175 or more COVID-19 infections. We start with $X\ge 175$ and convert this statement to a statement about the sample proportion. To do this, we divide each side of the inequality by the sample size n = 200.

$$\begin{array}{cc}X& \ge 175\end{array}$$

$\frac{X}{200}\ge \frac{175}{200}$ divide both sides by 200

$\widehat{p}=\frac{X}{200}\ge 0.875$ since  $\widehat{p}=\frac{X}{n}$ and $\frac{175}{200}=0.875$

So $P\left(X\ge 175\right)=P(\widehat{p}\ge 0.875)$. Using the normal calculator with mean 0.82 and standard deviation 0.027166 we choose “P(X > a)” and type 0.875 for the value a. The result is $P\left(\widehat{p}\ge 0.875\right)\approx 0.0215$ rounded to 4 decimal places.

Note that if we had used the binomial calculator instead, we could have found $P\left(X\ge 175\right)$ directly, and it would have been the exact answer rather than the approximation given by the normal curve. The exact answer is $P\left(X\ge 175\right)\approx 0.0230$.

Once again, using the distribution of the sample proportions gave us a very close approximation to the answer we could find directly using the binomial calculator, 0.0215 compared to 0.0230.

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