Chapter 10, Problem 43SE

Country Financial, a financial services company, uses surveys of adults age 18 and older to determine whether personal financial fitness is changing over time (USA Today, April 4, 2012). In February 2012, a sample of 1000 adults showed 410 indicating that their financial security was more than fair. In February 2010, a sample of 900 adults showed 315 indicating that their financial security was more than fair.

1. a. State the hypotheses that can be used to test for a significant difference between the population proportions for the two years.
2. b. What is the sample proportion indicating that their financial security was more than fair in 2012? In 2010?
3. c. Conduct the hypothesis test and compute the p-value. At a .05 level of significance, what is your conclusion?
4. d. What is the 95% confidence interval estimate of the difference between the two population proportions?

To determine

State the null and alternative hypotheses.

Answer to Problem 43SE

Null hypothesis:$H_0:p_1-p_2=0$

Alternative hypothesis:$H_a:p_1-p_2\ne 0$

Explanation of Solution

Calculation:

It is given that in February 2012, 410 out of 1,000 adults indicated that their financial security was more than fair and in February 2010, 315 out of 900 adults indicated that their financial security was more than fair.

Here, $p_1$ represents the population proportion of adults saying that their financial security was more than fair in 2012 and $p_2$ represents the population proportion of adults saying that their financial security was more than fair in 2010.

State the hypothesis:

Null hypothesis:

$H_0:p_1-p_2=0$

That is, there is no significant difference between the population proportions for the two years.

Alternative hypothesis:

$H_a:p_1-p_2\ne 0$

That is, there is a significant difference between the population proportions for the two years.

To determine

Find the sample proportion that indicates that their financial security was more than fair in 2012 and 2010.

Answer to Problem 43SE

The sample proportions that indicate that their financial security was more than fair in 2012 and 2010 are 0.41 and 0.35, respectively.

Explanation of Solution

Calculation:

Sample proportion for 2012:

The sample proportion for 2012 ${\overline{p}}_1$ is calculated as follows:

$\begin{array}{c}{\overline{p}}_1=\frac{x_1}{n_1}\\ =\frac{410}{1,000}\\ =0.41\end{array}$

Thus, the sample proportion for 2012 is 0.41.

Sample proportion for 2010:

The sample proportion for 2010 ${\overline{p}}_2$ is calculated as follows:

$\begin{array}{c}{\overline{p}}_2=\frac{x_2}{n_2}\\ =\frac{315}{900}\\ =0.35\end{array}$

Thus, the sample proportion for 2010 is 0.35.

To determine

Find the p-value and provide a conclusion at $\alpha =0.05$ level of significance.

Answer to Problem 43SE

The p-value is 0.007.

There is sufficient evidence to conclude that there is a significant difference between the population proportions for the two years.

Explanation of Solution

Calculation:

The test statistic for the hypothesis test about $p_1-p_2$ is calculated as follows:

$z=\frac{\left({\overline{p}}_1-{\overline{p}}_2\right)}{\sqrt{\overline{p}\left(1-\overline{p}\right)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$.

Here, ${\overline{p}}_1$ is the sample proportion for a simple random sample from population 1, ${\overline{p}}_2$ is the sample proportion for a simple random sample from population 2, $\overline{p}$ is the pooled estimator of p, $n_1$ is the sample size for population 1, $n_2$ is the sample size for population 2, and $z_{\frac{\alpha }{2}}$ is the critical value.

From Part (b), ${\overline{p}}_1=0.41$ and ${\overline{p}}_2=0.35$.

Pooled estimator:

$\begin{array}{c}\overline{p}=\frac{n_1{\overline{p}}_1+n_2{\overline{p}}_2}{n_1+n_2}\\ =\frac{1,000\left(0.41\right)+900\left(0.35\right)}{1,000+900}\\ =\frac{\text{410+315}}{\text{1,900}}\\ =0.3816\end{array}$

Substitute ${\overline{p}}_1$ for 0.41, ${\overline{p}}_2$ for 0.35, $n_1$ for 1,000, $n_2$ for 900, and $\overline{p}$ for 0.3816 in the z formula.

$\begin{array}{c}z=\frac{\left(0.41-0.35\right)}{\sqrt{0.3816\left(1-0.3816\right)\left(\frac{1}{1,000}+\frac{1}{900}\right)}}\\ =\frac{\text{0}\text{.06}}{\sqrt{\left(\text{0}\text{.2360}\right)\left(\text{0}\text{.0021}\right)}}\\ =\frac{\text{0}\text{.06}}{\text{0}\text{.0223}}\\ =2.69\end{array}$

Thus, the test statistic z-value is 2.69.

Software procedure:

Step-by-step procedure to obtain the probability value using Excel.

• Open an EXCEL sheet and select the cell A1.
• Enter the formula =NORM.S.DIST(2.69,TRUE) in the cell A1.
• Press Enter.

The output obtained using EXCEL software is given below:

From the output, the p-value for the left tail is 0.9964.

For the upper-tail test, the p-value is $\left(1-\text{ area to the left of 2}\text{.69}\right)$, and for two-tail test, the p-value is $2\times \left(1-\text{ area to the left of 2}\text{.69}\right)$.

From the output, the result is calculated as follows:

$\begin{array}{c}p\text{-value}=2\left(1-0.9964\right)\\ =\text{0}\text{.007}\end{array}$

Thus, the p-value is 0.007.

Rejection rule:

If the $p\text{-value}\le \alpha$, reject the null hypothesis

If the $p\text{-value}>\alpha$, do not reject the null hypothesis.

Conclusion:

Here, the p-value is less than the level of significance.

That is, $p\text{-value}\left(=0.007\right)\le \alpha \left(=0.05\right)$.

From the rejection rule, the null hypothesis is rejected.

Therefore, there is sufficient evidence to conclude that there is a significant difference between the population proportions for the two years.

To determine

Obtain a 95% confidence interval for the difference between the proportions of the two populations.

Answer to Problem 43SE

The 95% confidence interval is $\underset{\_}{\left(\text{0}\text{.017},\text{0}\text{.103}\right)}$.

Explanation of Solution

Calculation:

The formula for the confidence interval estimate of the difference between two populations is as follows:

${\overline{p}}_1-{\overline{p}}_2\pm z_{\frac{\alpha }{2}}\sqrt{\frac{{\overline{p}}_1\left(1-{\overline{p}}_1\right)}{n_1}+\frac{{\overline{p}}_2\left(1-{\overline{p}}_2\right)}{n_2}}$.

Here, ${\overline{p}}_1$ is the sample proportion for a simple random sample from population 1, ${\overline{p}}_2$ is the sample proportion for a simple random sample from population 2, $n_1$ is the sample size for population 1, $n_2$ is the sample size for population 2, and $z_{\frac{\alpha }{2}}$ is the critical value.

Critical value:

For the 95% confidence level:

$\begin{array}{c}\alpha =1-0.95\\ =0.05\\ \frac{\alpha }{2}=\frac{0.05}{2}\\ =0.025\end{array}$

Hence, cumulative area to the left is calculated as follows:

$\begin{array}{c}\text{Area to the left}=1-\text{Area to the right}\\ =1-0.025\\ =0.975\end{array}$

From Table II of the standard normal distribution in Appendix B, the critical value is 1.96.

Confidence interval:

The 95% confidence interval for the difference between two population proportions is obtained as shown below:

$\begin{array}{c}{\overline{p}}_1-{\overline{p}}_2\pm z_{\frac{\alpha }{2}}\sqrt{\frac{{\overline{p}}_1\left(1-{\overline{p}}_1\right)}{n_1}+\frac{{\overline{p}}_2\left(1-{\overline{p}}_2\right)}{n_2}}=\left(0.41-0.35\right)\pm 1.96\sqrt{\frac{0.41\left(1-0.41\right)}{1,000}+\frac{0.35\left(1-0.35\right)}{900}}\\ =0.06\pm 1.96\sqrt{\text{0}\text{.00024}+0.00025}\\ =0.06\pm 1.96\left(\text{0}\text{.0221}\right)\\ =0.06\pm 0.043\end{array}$                                                            $\begin{array}{c}=\left(0.06-0.043,0.06+0.043\right)\\ =\left(\text{0}\text{.017},\text{0}\text{.103}\right)\end{array}$

Thus, the 95% confidence interval is $\underset{\_}{\left(\text{0}\text{.017},\text{0}\text{.103}\right)}$.

Interpretation:

There is 95% confidence that the population proportion of adults who say that their financial security is more than fair between 0.017 and 0.103.