Chapter 13, Problem 5CRE

a.

To determine

Estimate the proportion of adults who view a landline phone as a necessity using 95% confidence interval.

Answer to Problem 5CRE

The 95% confidence interval of the proportion of adults who view a landline phone as a necessity is (0.6511, 0.7089).

Explanation of Solution

It is given that out of 1,003 adults, 68% thought a landline phone was a necessity.

Here, the sample size, $n$ is 1,003 and the proportion of adults thinks a landline phone is a necessity, $\widehat{p}$ is 0.68 (= 68%).

Conditions for 95% confidence interval:

1. Values of $n\widehat{p}$ and $n\left(1-\widehat{p}\right)$:

$\begin{array}{c}n\widehat{p}=1,003(0.68)\\ =682.04\\ n\left(1-\widehat{p}\right)=1,003(1-0.68)\\ =320.96\end{array}$

Since $n\widehat{p}$ and $n\left(1-\widehat{p}\right)$ are greater than 10, the sample size is large enough to proceed.

2. The sample size $n=1,003$ is much smaller than 10% of the number of adults (population size).

3. Since it is given that the survey is nationally representative, it is reasonable to assume the sample as a random sample from the population of adults.

Calculation:

The 95% confidence interval for p is given below.

$\widehat{p}\pm 1.96\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}$

Substitute the values of $n$ and $\widehat{p}$ in confidence interval formula as shown below.

$\begin{array}{c}\widehat{p}\pm 1.96\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}=0.68\pm 1.96\sqrt{\frac{0.68\left(1-0.68\right)}{1,003}}\\ =0.68\pm \left(1.96\times 0.0147\right)\\ =0.68\pm 0.0289\\ =\left(0.68-0.0289,0.68+0.0289\right)\\ =\left(0.6511,0.7089\right)\end{array}$

Interpretation:

The 95% confidence interval of the proportion of adults who view a landline phone as a necessity is (0.6511, 0.7089). Therefore, one can be 95% confident that the proportion of all adults who view a landline phone as a necessity is between 0.6511 and 0.7089.

b.

To determine

Check whether there is any convincing evidence that a majority of adults view a television set as a necessity at 0.05 significance level.

Answer to Problem 5CRE

There is no convincing evidence that a majority of adults consider a TV set as necessity.

Explanation of Solution

In order to test majority of adults view a television set as a necessity, population proportion test is appropriate.

The following nine steps carry out the test for proportions.

1. Population characteristic of interest:

Let $p$ denotes the proportion of all adults who consider a TV set as necessity.

2. Null hypothesis:

$H_0:p=0.5$

That is, the proportion of adults who view a television set as a necessity is 0.5.

3. Alternative hypothesis:

$H_a:p>0.5$

That is, the proportion of adults who view a television set as a necessity is greater than 0.5.

That is, majority of adults consider a TV set as necessity.

4. Significance level:

$\alpha =0.05$

5. Test statistic:

$\begin{array}{c}z=\frac{\widehat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}}\\ =\frac{\widehat{p}-0.5}{\sqrt{\frac{0.5\left(1-0.5\right)}{1,003}}}\end{array}$

6. Assumptions:

• Here, $n\widehat{p}=501.5$ (1,003 × 0.5) and $n\left(1-\widehat{p}\right)=501.5$ (1,003 × (1−0.5)). Since, both values are greater than 10, the sample is large enough to proceed.
• The sample is nationally representative. Hence, the sample can be treated as a random sample from the population.
• The sample size of 1,003 is large enough. Hence, all assumptions satisfy to conduct large sample test.

7. Calculation:

It is given that 52% (= 0.52) said they viewed a television set as a necessity. Therefore, $\widehat{p}=0.52$. Substitute the values in the test statistic formula as shown below.

$\begin{array}{c}z=\frac{0.52-0.5}{\sqrt{\frac{0.5\left(1-0.5\right)}{1,003}}}\\ =\frac{0.02}{0.01579}\\ =1.267\end{array}$

8. P-value:

Software procedure:

Step-by-step procedure to find P-value using MINITAB software:

• Select Graph > Probability Distribution Plots.
• Choose View Probability.
• In Distribution, select Normal under Distribution.
• In Shaded Area, choose Right tail and give X value as 1.267.
• Click OK.

The MINITAB output is as follows:

From the MINITAB output, the P-value is 0.1026.

9. Conclusion:

Decision Rule:

If P-value is less than significance level, reject H­₀.

Here the significance level is 0.05.

Hence, $0.103>0.05$.

Since P-value is greater than 0.05, fail to reject the null hypothesis. Therefore, there is no convincing evidence that a majority of adults consider a TV set as necessity.

c.

To determine

Check whether there is any convincing evidence that the proportion of adults who regard a microwave oven as a necessity decreased between 2003 and 2009 at 0.01 significance level.

Answer to Problem 5CRE

There is a convincing evidence that the proportion of adults who regarded a microwave oven as a necessity decreased between 2003 and 2009.

Explanation of Solution

In order to test the proportion of adults who regard a microwave oven as a necessity decreased between 2003 and 2009, population proportion test is appropriate.

The following nine steps carry out the test for proportions.

1. Population characteristic of interest:

Let $p_1$ denotes the proportion of adults who regarded a microwave oven as a necessity in 2003 and $p_2$ denotes the proportion of adults who regarded a microwave oven as a necessity in 2009.

2. Null hypothesis:

$H_0:p_1-p_2=0$

That is, the proportions of adults who regarded a microwave oven as a necessity in 2003 and 2009 are equal.

3. Alternative hypothesis:

$H_a:p_1-p_2>0$

That is, the proportion of adults who regarded a microwave oven as a necessity in 2003 is greater than the proportion of adults who regarded a microwave oven as a necessity in 2009.

4. Significance level:

$\alpha =0.01$

5. Test statistic:

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

Where, ${\widehat{p}}_c$ is the combined estimate of common population proportion.

6. Assumptions:

Here, $n_1$ is 1,003, $n_2$ is 1,003, ${\widehat{p}}_1$ is 0.68, and ${\widehat{p}}_2$ is 0.47

 $\begin{array}{c}{n}_1{\widehat{p}}_1=1,003(0.68)\\ =682.04\\ n_1\left(1-{\widehat{p}}_1\right)=1,003(1-0.68)\\ =320.96\end{array}$

$\begin{array}{c}{n}_2{\widehat{p}}_2=1,003(0.47)\\ =471.41\\ n_2\left(1-{\widehat{p}}_2\right)=1,003(1-0.47)\\ =531.59\end{array}$

• Since the values of $n_1{\widehat{p}}_1$, $n_1\left(1-{\widehat{p}}_1\right)$, $n_2{\widehat{p}}_2$, and $n_2\left(1-{\widehat{p}}_2\right)$ are greater than 10, the samples are large enough to proceed.
• Also, the sample is nationally representative. Hence, the samples can be treated as a random sample from the population.
• The sample size of 1,003 is large enough. Hence, all assumptions satisfy to conduct large sample test.

7. Calculation:

It is given that 68% of the 2003 sample regarded a microwave oven as a necessity and 47% of the 2009 sample regarded a microwave oven as a necessity.

$\begin{array}{c}{\widehat{p}}_c=\frac{n_1{\widehat{p}}_1+n_2{\widehat{p}}_2}{n_1+n_2}\\ =\frac{1,003\left(0.68\right)+1,003\left(0.47\right)}{1,003+1,003}\\ =\frac{1,153.45}{2,006}\\ =0.575\\ z=\frac{0.68-0.47}{\sqrt{\frac{0.575\left(1-0.575\right)}{1,003}+\frac{0.575\left(1-0.575\right)}{1,003}}}\\ =\frac{0.21}{\sqrt{0.000244+0.000244}}\\ =\frac{0.21}{\sqrt{0.000487}}\\ =\frac{0.21}{0.022075}\\ =9.5132\end{array}$

8. P-value:

Software procedure:

Step-by-step procedure to find P-value using MINITAB software:

• Select Graph > Probability Distribution Plots.
• Choose View Probability.
• In Distribution, select Normal under Distribution.
• In Shaded Area, choose left tail and give X value as 9.5132.
• Click OK.

The MINITAB output is as follows:

From the MINITAB output, the P-value is 0.000.

9. Conclusion:

Decision Rule:

If P-value is less than significance level, reject H­₀. Otherwise, fail to reject $H_0$.

Here, the significance level is 0.01.

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

Hence, $0.000<0.01$.

Since P-value is less than 0.01, reject the null hypothesis. Therefore, there is convincing evidence that the proportion of adults who regarded a microwave oven as a necessity decreased between 2003 and 2009.