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QUAN 2010 Notes: Introduction to Business Statistics

Omar Ortega

Section 8.3 The Sampling Distribution of the Proportion

So far, the focus in this chapter has been on the distribution of sample means. However, sometimes we deal with business scenarios where we are counting observations in a sample and in this case the sample proportion (or percentage), \(\overline{p}\text{,}\) is the statistic that is relevant rather than the sample mean, \(\overline{x}\text{.}\)

Activity 8.3.1.

We’re going to get in groups, and in each group, you’ll roll a die \(30\) times. (Use this Excel document to keep track of your rolls: external/sheets/DiceSamplingProportion.xlsx ) Count how many times you roll a “1” and use that to calculate the proportion of the 30 rolls for which the result was a “1”. After the whole class finishes finding this proportion, we’re going to draw a histogram of all the sample proportions. Before doing this, though, think about the following questions:

(a)

What should the distribution of your individual rolls look like (approximately)?

(b)

What do you think the approximate shape of the histogram of all sample proportions will be? (Will it be skewed left, uniform, normal/bell-shaped, or skewed right?)

(c)

What do you expect the approximate mean of the distribution of sample proportions to be? (Will it be \(1/100\text{,}\) \(1/6\text{,}\) \(6/100\text{,}\) or \(6\text{?}\))

Definition 8.3.1.

The sampling distribution of the proportion describes the pattern that the sample proportions tend to follow when randomly drawn from a population. This distribution has a mean, \(p\text{,}\) and a standard error (i.e. the standard deviation of the sample proportions), \(\sigma_p=\sqrt{\frac{p(1-p)}{n}}\text{.}\) (The sample proportion is \(\overline{p}=\frac{x}{n}\text{.}\))

Exercise 8.3.2.

(Donnelly 7.25)
For a population with a proportion equal to 0.32, calculate the standard error of the proportion for the following sample sizes.

(a)

\(n=35\)
Answer.
\begin{equation*} \sigma_p = \sqrt{\frac{0.32(1-0.32)}{35}}\approx 0.0788 \end{equation*}

(b)

\(n=70\)
Answer.
\begin{equation*} \sigma_p = \sqrt{\frac{0.32(1-0.32)}{70}}\approx 0.0558 \end{equation*}

(c)

\(n=105\)
Answer.
\begin{equation*} \sigma_p = \sqrt{\frac{0.32(1-0.32)}{105}}\approx 0.0455 \end{equation*}

Exercise 8.3.3.

    What can you conclude about the standard error as the sample size increases?
  • As the sample size increases, standard error decreases
  • As the sample size increases, standard error increases.
  • As the sample size increases, standard error stays the same.
  • As the sample size increases, the standard error might increase, decrease, or stay the same.

Exercise 8.3.4.

(Donnelly 7.32)
A social media survey found that \(69\%\) of parents “follow” their children on Instagram. A random sample of 140 parents was selected.

(a)

Calculate the standard error of the proportion.
Answer.
\begin{equation*} \sigma_p=\sqrt{\frac{0.69(1-0.69)}{140}}\approx 0.0391 \end{equation*}

(b)

What is the probability that 104 or more parents from this sample “follow” their children on Instagram?
Answer.
\begin{equation*} P\left(\overline{p}\geq \frac{104}{140}\right) = 1-P\left(\overline{p}\lt \frac{104}{140}\right)\approx 0.0869 \end{equation*}

(c)

What is the probability that between 97 and 104 parents from this sample “follow” their children on Instagram?
Answer.
\begin{equation*} P\left(\frac{97}{140}\lt \overline{p}\lt \frac{104}{140}\right) \approx 0.3812 \end{equation*}

(d)

If 81 parents responded that the “follow” their children on Instagram, does this result support the findings of the social media survey?
Answer.
\begin{equation*} P\left(\overline{p}\leq \frac{81}{140}\right) \approx 0.0023 \end{equation*}
Since \(0.0023\lt 0.05 \) (less than \(5\%\)), the sample with only 81 parents answering “yes” is unusual. Therefore, the survey results are questionable.
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