Chapter 25, Problem 25.42E

a.

To determine

To construct: The 99% confidence interval for the proportion of Americans who are more than 50 years of age and also current smokers.

Answer to Problem 25.42E

The 99% confidence interval for the proportion of Americans who are more than 50 years of age and also current smokers is 0.0826 to 0.1057or 8.26% to 10.57%.

Explanation of Solution

Given info:

The data shows the results obtained by asking two questions to a random sample of 50 year old adults “How do your overall health, excellent, very good, good, fair and poor?” and the second question is “Do you smoke currently, yes or no?”

Calculation:

Let $\widehat{p}$ denote the proportion of Americans who are more than 50 years of age and currently smoking.

$\begin{array}{c}\widehat{p}=\frac{\text{Number of Americans who are currently smoking}}{\text{Total number of Americans}}\\ =\frac{404}{4,310}\\ =0.094\end{array}$

Thus, the proportion of Americans who are more than 50 years of age and currently smoking is 0.094.

Software procedure:

Step-by-step procedure for constructing 99% confidence interval for the given proportion is shown below:

• Click on Stat, Basic statistics and 1-proportion.
• Choose Summarized data, under Number of events enter 404, under Number of trials enter 4,310.
• Click on options, choose 99% confidence interval.
• Click ok.

Output using MINITAB is given below:

Thus, the 99% confidence interval for the proportion of Americans who are more than who are more than 50 years of age and also current smokers is 0.0826 to 0.1057.

b.

To determine

To compare: The two conditional distributions and also the graph of conditional distributions.

To find: The conditional distribution of age for the Americans who use social networking sitesand who don’t use social networking siteson their phones.

To construct: A suitable graph for the conditional distribution of age for the Americans who use social networking sitesand who don’t use social networking siteson their phones.

Answer to Problem 25.42E

• The bar graph and the conditional distributions show that Americans are currently smoking have lowest percentages in “excellent” and “very good” health evaluation whereas Americans who are not currently smoking have highest percentages in “excellent” and “very good” health evaluation.
• The percentages in “fair and poor” health evaluation is more in currently smoking group when compared to the percentages for the same categories under not currently smoking group.

The conditional distribution of health evaluation for the Americans who are currently smoking and not currently smoking is given below:

	Conditional distribution of health evaluation for the Americans who are currently smoking and not currently smoking
Health Evaluation	Yes	No
Excellent	6.20	12.4
Very good	28.50	39.9
Good	35.90	33.5
Fair	22.30	14.0
Poor	7.18	0.3

The bar graph showing the conditional distribution of health evaluation for the Americans who are currently smoking and not currently smoking is given below:

• Output obtained from MINITAB is given below:
• 

Explanation of Solution

Calculation:

The conditional distribution of health evaluation for the Americans who are current smokers is calculated as follows:

$\left(\begin{array}{l}\text{Conditional distribution of health}\\ \text{ evalution for currently}\\ \text{ smoking group}\end{array}\right)=\frac{\left(\begin{array}{l}\text{Number of Americans who have said yes }\\ \text{under each category of health evaluation }\end{array}\right)}{\text{Total number of Americans who have said yes}}$

The conditional distribution of health evaluation for the Americans who are not current smokers is calculated as follows:

$\left(\begin{array}{l}\text{Conditional distribution of health}\\ \text{ evalution for not currently}\\ \text{ smoking group}\end{array}\right)=\frac{\left(\begin{array}{l}\text{Number of Americans who have said no }\\ \text{under each category of health evaluation }\end{array}\right)}{\text{Total number of Americans who have said no}}$

The conditional distribution of health evaluation for the Americans who are currently smoking and not currently smoking is given below:

	Conditional distribution of health evaluation for the Americans who are currently smoking and not currently smoking
Health Evaluation	Yes	No
Excellent	$\begin{array}{c}\frac{25}{404}\times 100=0.062\times 100\\ =6.2\end{array}$	$\begin{array}{c}\frac{484}{3,906}\times 100=0.124\times 100\\ =12.4\end{array}$
Very good	$\begin{array}{c}\frac{115}{404}\times 100=0.285\times 100\\ =28.5\end{array}$	$\begin{array}{c}\frac{1,557}{3,906}\times 100=0.399\times 100\\ =39.9\end{array}$
Good	$\begin{array}{c}\frac{145}{404}\times 100=0.359\times 100\\ =35.9\end{array}$	$\begin{array}{c}\frac{1,309}{3,906}\times 100=0.335\times 100\\ =33.5\end{array}$
Fair	$\begin{array}{c}\frac{90}{404}\times 100=0.223\times 100\\ =22.3\end{array}$	$\begin{array}{c}\frac{545}{3,906}\times 100=0.14\times 100\\ =14\end{array}$
Poor	$\begin{array}{c}\frac{29}{404}\times 100=0.0718\times 100\\ =7.18\end{array}$	$\begin{array}{c}\frac{11}{3,906}\times 100=0.003\times 100\\ =0.3\end{array}$

Software Procedure:

Step-by-step procedure to construct the Bar Chart using the MINITAB software:

• Choose Graph > Bar Chart.
• From Bars represent, choose Values from a table.
• Under Two Columns of values, choose Cluster. Click OK.
• In Graph variables, enter the columns of Yes and No.
• In Row labels, enter “Health Evaluation”.
• In Table arrangement, click “columns are outermost categories and columns are innermost”.
• Click OK.
• Interpretation:
• The bar chart for comparing two conditional distributions is constructed with bars to the extreme left side of the graph, which represents the currently smoking group.

• c.

To determine

• To test: Whether there is a significant difference between health evaluation and currently smoking and not currently smoking group.
• To find: The mean value of the test statistic given that the null hypothesis is true.
• To give: The P-value.

Answer to Problem 25.42E

• There is a significant difference between health evaluation and currently smoking and not currently smoking group.
• The mean value of the chi-square test statistic given that the null hypothesis is true is given is to be 4.
• The P-value is 0.000.

Explanation of Solution

• Calculation:
• The claim is to test whether there is any significant difference between health evaluation and currently smoking and not currently smoking group.

Cell frequency for using Chi-square test:

• When at most 20% of the cell frequencies are less than 5
• If all the individual frequencies are 1 or more than 1.
• All the expected frequencies must be 5 or greater than 5

The hypotheses used for testing are given below:

$H_0:$ There is no relationship between health evaluation and currently smoking.

$H_a:$ There is a relationship between health evaluation and currently smoking.

Software procedure:

Step-by-stepprocedure for calculating the chi-square test statistic is given below:

• Click on Stat, select Tables and then click on Chi-square Test (Two-way table in a worksheet).
• Under Columns containing the table: enter the columns of Yes and No.
• Click ok.

Output obtained from MINITAB is given below:

• 

Thus, the test statistic is 229.660, the degree of freedom is 4, and the P-value is 0.000.

Only one cell is having an expected frequency less than 5. The usage of chi-square test is appropriate.

Conclusion:

The P-value is 0.000 and the level of significance is 0.05.

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

Therefore, the null hypothesis is rejected.

Thus, there is sufficient evidence to support the claim that there is a significant difference between health evaluation and currently smoking and not currently smoking group.

Justification:

Fact:

If the null hypothesis is true, then the mean of any chi-square distribution is equal to its degrees of freedom.

From the MINITAB output, it can be observed that the degree of freedom is calculated as 3. So, the mean value of the chi-square statistic is 3.

Thus, the mean value of the chi-square statistic is 3.

Also, the observed chi-square value is ${\chi }^2=226.660$ and the mean value is 4. Thus, the observed value is larger than the mean value.

d.

To determine

To give: A comparison for the difference in the distribution of age across social media network usage.

Explanation of Solution

Comparison:

From the MINITAB output, it can be seen that the observed frequencies for good, fair and poor health evaluation for currently smoking group is higher than the expected frequency, given that the null hypothesis is true.

In non-smoking group the observed frequencies for good, fair and poor health evaluation is lesser than the expected frequency.

The people who are currently smoking have a poor health condition when compared to the non smoking group.

Hence, there is a significant difference in the health evaluation for currently smoking and not currently smoking group.