1. Perform and interpret $Z$ tests for $p_1-p_2$.
2. Compute and interpret $Z$ intervals for $p_1-p_2$.
3. Use $Z$ intervals for $p_1-p_2$ to perform two-tailed $Z$ tests.

Developing Your Statistical Sense

$Z=\frac{({\hat{p}}_1-{\hat{p}}_2)-(p_1-p_2)}{\sqrt{\frac{p_1{q}_1}{n_1}+\frac{p_2{q}_2}{n_2}}}$

$\begin{array}{cccc}{x}_1\ge 5,& (n_1-x_1)\ge 5,& x_2\ge 5,& (n_2-x_2)\ge 5\end{array}$

Note: As a check on your arithmetic, ${\hat{p}}_{\text{pooled}}$ must also lie between ${\hat{p}}_1$ and ${\hat{p}}_2$.

• Step 1 State the hypotheses.

  Use one of the forms from Table 12 (page 609). State the meaning of $p_1$ and $p_2$.

• Step 2 Find ${\mathit{Z}}_{\mathbf{\text{crit}}}$ and state the rejection rule.

  Use Table 12 on page 609.

• Step 3 Calculate ${\mathit{Z}}_{\mathbf{\text{data}}}$

  $Z_{\text{data}}=\frac{{\hat{p}}_1-{\hat{p}}_2}{\sqrt{{\hat{p}}_{\text{pooled}}\cdot (1-{\hat{p}}_{\text{pooled}})\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$

  where

  ${\hat{p}}_{\text{pooled}}=\frac{x_1+x_2}{n_1+n_2}$

  $Z_{\text{data}}$ follows an approximately standard normal distribution if the required conditions are satisfied.

• Step 4 State the conclusion and the interpretation.

  Compare $Z_{\text{data}}$ with $Z_{\text{crit}}$.

1. Find the point estimate of the difference in the population proportions of males and females, ${\hat{p}}_1-{\hat{p}}_2$.
2. Compute the pooled estimate of the common proportion, ${\hat{p}}_{\text{pooled}}$.
3. Calculate the value of the test statistic $Z_{\text{data}}$.
4. Check whether the conditions for performing the $Z$ test for $p_1-p_2$ are met.
5. Test whether the population proportion of males who agree that technology will lead to a better future is greater than the population proportion of females who agree. Use the critical-value method at level of significance $\alpha =0.01$.

1. The point estimate is ${\hat{p}}_1-{\hat{p}}_2=0.67-0.51=0.16$
2. ${\hat{p}}_{\text{pooled}}=\frac{x_1+x_2}{n_1+n_2}=\frac{335+255}{500+500}=0.59$
   

   FIGURE 16 TI-83/84 results.
3. $Z_{\text{data}}=\frac{{\hat{p}}_1-{\hat{p}}_2}{\sqrt{{\hat{p}}_{\text{pooled}}\cdot \left(1-{\hat{p}}_{\text{pooled}}\right)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}=\frac{0.67-0.51}{\sqrt{\left(0.59\right)\left(0.41\right)\left(\frac{1}{500}+\frac{1}{500}\right)}}\approx 5.1$
4. We check the conditions for performing the $Z$ test for $p_1-p_2$. We have: $x_1=335\ge 5$, $x_2=255\ge 5$, $n_1-x_1=500-335=165\ge 5$, and $n_2-x_2=500-255=245\ge 5$. We may thus proceed with the hypothesis test.
5. The $Z$ test for $p_1\le p_2$ follows the steps below.

• Step 1 State the hypotheses.

  The key words “greater than,” together with the fact that sample 1 represents the males, indicate that we have a right-tailed test:

  $\begin{array}{ccc}{H}_0:p_1=p_2& \text{versus}& H_a:p_1>p_2\end{array}$

  where $p_1$ and $p_2$ represent the population proportion of males and females, respectively, who agree that technology will lead to a better future.

  

  FIGURE 17 $Z_{\text{data}}=5.1$ is extreme, leading to rejection of $H_0$.

• Step 2 Find $Z_{\text{crit}}$ and state the rejection rule.

  For a right-tailed test with level of significance $\alpha =0.01$, Table 12 gives us $Z_{\text{crit}}=2.33$ and our rejection rule: Reject $H_0$ if $Z_{\text{data}}\ge 2.33$.

• Step 3 Calculate ${\mathit{Z}}_{\mathbf{\text{data}}}$.

  From (c), we have $Z_{\text{data}}\approx 5.1$ (also see Figure 16).

• Step 4 State the conclusion and the interpretation.

  $Z_{\text{data}}\approx 5.1\ge 2.33$; therefore, reject $H_0$ (see Figure 17). There is evidence at level of significance $\alpha =0.01$ that the population proportion of males who agree that technology will lead to a better future is greater than the population proportion of females who agree.

• Step 1 State the hypotheses and the rejection rule.

  Use one of the forms from Table 12. State the meaning of $p_1$ and $p_2$. The rejection rule is: Reject $H_0$ if the $p$-value $\le \alpha$.

• Step 2 calculate ${\mathit{Z}}_{\mathbf{\text{data}}}$.

  $Z_{\text{data}}=\frac{{\hat{p}}_1-{\hat{p}}_2}{\sqrt{{\hat{p}}_{\text{pooled}}\cdot \left(1-{\hat{p}}_{\text{pooled}}\right)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$

  where ${\hat{p}}_{\text{pooled}}=\frac{x_1+x_2}{n_1+n_2}$. If the required conditions are satisfied, $Z_{\text{data}}$ follows an approximately standard normal distribution.

• Step 3 Find the $\mathit{p}$-value.

  Either use technology or calculate the $p$-value using one of the forms in Table 14.

• Step 4 State the conclusion and the interpretation.

  Compare the $p$-value with $\alpha$.

• Step 1 State the hypotheses and the rejection rule.

  We are interested in whether the proportion of females who report being very happily married is smaller than that of males and because the females represent sample 1, the hypotheses are

  $\begin{array}{cc}{H}_0:p_1=p_2& H_a:p_1<p_2\end{array}$

  Page 612

  where $p_1$ and $p_2$ represent the population proportions of all females and males, respectively, who report being very happily married. We will reject $H_0$ if the $p–\text{value}\le \alpha =0.05$.

  

  FIGURE 18 $p$-Value for left-tailed $Z$ test.

• Step 2 Find ${\mathit{Z}}_{\mathbf{\text{data}}}$.

  First, use the data from Table 15 to find the value of ${\hat{\mathit{p}}}_{\text{pooled}}$.

  ${\hat{p}}_{\text{pooled}}=\frac{x_1+x_2}{n_1+n_2}=\frac{257+242}{423+366}\approx 0.63245$

  Then

  $Z_{\text{data}}=\frac{(0.6076-0.6612)}{\sqrt{0.63245}\cdot (1-0.63245)\left(\frac{1}{423}+\frac{1}{366}\right)}\approx -1.56$

• Step 3 Find the $p$-value.

  Because it is a left-tailed test, the $p$-value is given by Table 14 as $P(Z<Z_{\text{data}})=P(Z<-1.56)$, as shown in Figure 18. This amounts to a Case 1 problem from Table 8 in Chapter 6 on page 357:

  $P(Z<-1.56)=0.0594$

• Step 4 State the conclusion and the interpretation.

  The $p\text{-value}=0.0594$ is not less than or equal to $\alpha =0.05$, so we do not reject $H_0$. There is insufficient evidence that the proportion of females who report being very happily married is smaller than the proportion of males who do so.

${\hat{p}}_1-{\hat{p}}_2\pm Z_{\alpha /2}\sqrt{\frac{{\hat{p}}_1\cdot {\hat{q}}_1}{n_1}+\frac{{\hat{p}}_2\cdot {\hat{q}}_2}{n_2}}$

$E=Z_{\alpha /2}\cdot \sqrt{\frac{{\hat{p}}_1\cdot {\hat{q}}_1}{n_1}+\frac{{\hat{p}}_2\cdot {\hat{q}}_2}{n_2}}$

1. Calculate and interpret the margin of error $E$ for confidence level 99%.
2. Construct and interpret a 99% confidence interval for $p_1-p_2$.

1. $\begin{array}{cc}{\hat{q}}_1=1-{\hat{p}}_1=1-0.67=0.33& {\hat{q}}_2=1-{\hat{p}}_2=1-0.51=\mathrm{0.49.}\end{array}$

   From Table 1 in Chapter 8 on page 432, the $Z_{\alpha /2}$ value for a 99% confidence level is 2.576. Therefore, the margin of error is

   $E=Z_{\alpha /2}\cdot \sqrt{\frac{{\hat{p}}_1\cdot {\hat{q}}_1}{n_1}+\frac{{\hat{p}}_2\cdot {\hat{q}}_2}{n_1}}=\left(2.576\right)\sqrt{\frac{\left(0.67\right)\left(0.33\right)}{500}+\frac{\left(0.51\right)\left(0.49\right)}{500}}\approx 0.079$

   The margin of error is 0.079, so we may estimate $p_1-p_2$ to within 0.079 with 99% confidence.

2. The point estimate is ${\hat{p}}_1-{\hat{p}}_2=0.67-0.51=0.16$. The 99% confidence interval is therefore

   ${\hat{p}}_1-{\hat{p}}_2\pm E=0.16\pm 0.079=(0.081,0.239)$

   We are 99% confident that the difference in population proportions of males and females who agree that technology will lead to a better future lies between 0.081 and 0.239.

1. $H_0:p_1-p_2=0.1$ versus $H_a:p_1-p_2\ne 0.1$.

   The hypothesized value 0.1 lies outside the interval (0.081, 0.239), so we reject $H_0$.

2. $H_0:p_1-p_2=0.2$ versus $H_a:p_1-p_2\ne 0.2$.

   The hypothesized value 0.2 lies inside the interval, so we do not reject $H_0$.

3. $H_0:p_1-p_2=0.3$ versus $H_a:p_1-p_2\ne 0.3$.

   The hypothesized value 0.3 lies outside the interval, so we reject $H_0$.