Confidence Intervals for Differences in Population Proportions (College Board AP® Statistics): Revision Note

The general formula for confidence intervals (including a table of standard errors) is given in the exam: $\mathrm{statistic}\pm \left(\mathrm{critical}\mathrm{value}\right)\left(\mathrm{standard}\mathrm{error}\mathrm{of}\mathrm{statistic}\right)$.

You will need to apply it appropriately using the difference between the sample proportions and the standard error of the difference of the sample proportions.

Some exam questions may change the four $\ge 10$ conditions into four $\ge 5$ conditions (changing the 10 into a 5), though this will be made clear in the question.

You need to know that the width of a confidence interval increases as the confidence level increases, whereas it decreases as the sample sizes increase!

Nova University and Terra University have over 10,000 students each. A random sample of 200 students at Nova University and a random sample of 150 students from Terra University were asked to complete a survey to measure their level of smartphone addiction. The results showed that 35% of the students sampled from Nova University were addicted to their smartphones, while 28% of the students sampled from Terra University were addicted to their smartphones.

Construct a 95% confidence interval for the difference in the proportion of students addicted to smartphones at Nova University and the proportion of students addicted to smartphones at Terra University.

Answer:

Define the population parameters, $p_1$ and $p_2$

Let $p_1$ be the proportion of all students at Nova University who are addicted to their smartphones

Let $p_2$ be the proportion of all students at Terra University who are addicted to their smartphones

State the type of interval being used and verify that the conditions for the interval are met

The correct inference procedure is a two-sample z-interval for the difference in population proportions at a 95% confidence level

• The independence condition is satisfied, as

  • both samples were selected randomly

  • the sample size from Nova University, 200, is less than 10% of the total number of students at Nova University (10% of 'over 10,000' is 'over 1000')

  • the sample size from Terra University, 150, is less than 10% of the total number of students at Terra University (10% of 'over 10,000' is 'over 1000')

    • These conditions are required as sampling was conducted without replacement

• The sample size is large enough for the sampling distribution of the difference in sample proportions to be approximately normally distributed, because the following conditions are satisfied

  • $n_1{\hat{p}}_1=200·0.35=70\ge 10$

  • $n_1\left(1-{\hat{p}}_1\right)=200·\left(1-0.35\right)=130\ge 10$

  • $n_2{\hat{p}}_2=150·0.28=42\ge 10$

  • $n_2\left(1-{\hat{p}}_2\right)=150·\left(1-0.28\right)=108\ge 10$

List the sample sizes, $n_1$ and $n_2$, the sample proportions, ${\hat{p}}_1$ and ${\hat{p}}_2$, and calculate the standard error of the difference in sample proportions, $\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$

$\begin{array}{rcl}{n}_1& =& 200\\ n_2& =& 150\\ {\hat{p}}_1& =& 0.35\\ {\hat{p}}_2& =& 0.28\\ \sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}& =& \sqrt{\frac{0.35\left(1-0.35\right)}{200}+\frac{0.28\left(1-0.28\right)}{150}}=0.0498146...\end{array}$

Find the z-score (critical value) for a confidence level of 95%, e.g. from the tables

Remember that a confidence level of 95% is 5% in both tails combined, so use 2.5% for a single tail in the table
(Alternatively, the row for $t_{\infty }$ in the t-tables are z-scores, together with the corresponding 'Confidence level C' shown below)

z-score = 1.960

Calculate the confidence interval using the formula given to you in the exam, $\mathrm{Confidence}\mathrm{interval}=\mathrm{statistic}\pm \left(\mathrm{critical}\mathrm{value}\right)\left(\mathrm{standard}\mathrm{error}\mathrm{of}\mathrm{statistic}\right)$

$\begin{array}{rcl}\mathrm{CI}& =& \left({\hat{p}}_1-{\hat{p}}_2\right)\pm z·\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}\\ & =& \left(0.35-0.28\right)\pm 1.960·0.0498146...\end{array}$

State the confidence interval

$\left(-0.0276,0.1676\right)$

Explain the confidence interval in the context of the question

We can be 95% confident that the interval from -0.0276 to 0.1676 captures the actual value of the difference in the proportion of students addicted to smartphones at Nova University and the proportion of students addicted to smartphones at Terra University