Chapter 21, Problem 26E

(a)

To determine

To find: The method of simulating the proportion of an SRS of 25 adult Americans.

Answer to Problem 26E

Solution: The proportion of adults was simulated by considering the single digit lying between the values 0 and 9 in Table A, as the response of the adults.

Explanation of Solution

Calculation:

The possible responses for the number of Americans avoid drinking regular soda or pop will be designated as favorable (F) who avoid drinking and not favorable (NF) who do not avoid drinking.

By treating a single digit in the random number table as a response of an adult American, assign the response to the random digits from 0 to 9 as shown below:

Response of an adult	Digit
Favorable (F)	0, 1, 2, 3, 4, 5
Not favorable (NF)	6, 7, 8, 9

After this, simulate the proportion of randomly selected 25 adult American’s as shown below:

Digit	4	5	0	8	9	2	8	0	1	2	7	4	2
Response	F	F	F	NF	NF	F	NF	F	F	F	NF	F	F
Digit	7	5	6	2	8	0	5	9	1	3	7	1
Response	NF	F	NF	F	NF	F	F	NF	F	F	NF	F

Therefore, it can be concluded that the randomly selected 25 adult Americans were simulated by treating single digit as the response of an adult by using the random number table A.

(b)

To determine

To find: The sample proportion $\left(\widehat{p}\right)$ of simulated 10 randomly selected samples each of size 25 who avoid drinking regular soda or pop in the year 2015.

Answer to Problem 26E

Solution: The sample proportion $\left(\widehat{p}\right)$ of simulated 10 randomly selected samples each of size 25 who avoid drinking regular soda or pop in the year 2015 is shown below:

Samples	1	2	3	4	5	6	7	8	9	10
Number of favorable	16	14	14	16	11	15	12	14	17	12
Sample proportion $\left(\widehat{p}\right)$	0.64	0.56	0.56	0.64	0.44	0.60	0.48	0.56	0.68	0.48

Explanation of Solution

Calculation:

By treating a single digit in the random number table as a response of an adult American, assign the response to the random digits from 0 to 9 as shown below:

Response of an adult	Digit
Favorable (F)	0, 1, 2, 3, 4, 5
Not favorable (NF)	6, 7, 8, 9

Using the different lines from random number Table A, simulate the 10 randomly selected samples each of size 25. From the line 101, generate the random numbers for the sample 1. Similarly, from the line 102 to 110, generate the random numbers for the sample 2 to sample 10. The 10 samples are generated by using Table A and their corresponding responses are also mentioned as shown below:

Sample 1		Sample 2		Sample 3		Sample 4		Sample 5
Digit	Response	Digit	Response	Digit	Response	Digit	Response	Digit	Response
1	F	7	NF	4	F	5	F	9	NF
9	NF	3	F	5	F	2	F	5	F
2	F	6	NF	4	F	7	NF	5	F
2	F	7	NF	6	NF	1	F	9	NF
3	F	6	NF	7	NF	1	F	2	F
9	NF	4	F	7	NF	3	F	9	NF
5	F	7	NF	1	F	8	NF	4	F
0	F	1	F	7	NF	8	NF	0	F
3	F	5	F	0	F	8	NF	0	F
4	F	0	F	9	NF	9	NF	7	NF
0	F	9	NF	7	NF	9	NF	6	NF
5	F	9	NF	7	NF	3	F	9	NF
7	NF	4	F	5	F	0	F	9	NF
5	F	0	F	5	F	7	NF	7	NF
6	NF	0	F	8	NF	4	F	1	F
2	F	0	F	0	F	6	NF	9	NF
8	NF	1	F	0	F	0	F	1	F
7	NF	9	NF	0	F	2	F	4	F
1	F	2	F	9	NF	2	F	8	NF
3	F	7	NF	5	F	7	NF	1	F
9	NF	2	F	3	F	4	F	6	NF
6	NF	7	NF	2	F	0	F	0	F
4	F	7	NF	8	NF	0	F	7	NF
0	F	5	F	6	NF	1	F	7	NF
9	NF	4	F	3	F	1	F	9	NF

Sample 6		Sample 7		Sample 8		Sample 9		Sample 10
Digit	Response	Digit	Response	Digit	Response	Digit	Response	Digit	Response
6	NF	8	NF	6	NF	3	F	3	F
8	NF	2	F	0	F	6	NF	8	NF
4	F	7	NF	9	NF	0	F	4	F
1	F	3	F	4	F	0	F	4	F
7	NF	9	NF	0	F	9	NF	8	NF
3	F	5	F	7	NF	1	F	4	F
5	F	7	NF	2	F	9	NF	8	NF
0	F	8	NF	0	F	3	F	7	NF
1	F	9	NF	2	F	6	NF	8	NF
3	F	0	F	4	F	5	F	9	NF
1	F	2	F	1	F	1	F	1	F
5	F	0	F	7	NF	5	F	8	NF
5	F	8	NF	8	NF	4	F	3	F
2	F	0	F	6	NF	1	F	3	F
9	NF	7	NF	8	NF	2	F	8	NF
7	NF	4	F	2	F	3	F	2	F
2	F	7	NF	4	F	9	NF	4	F
7	NF	5	F	9	NF	6	NF	6	NF
6	NF	1	F	4	F	3	F	9	NF
5	F	1	F	3	F	8	NF	7	NF
8	NF	8	NF	6	NF	8	NF	3	F
5	F	1	F	1	F	5	F	9	NF
0	F	6	NF	7	NF	4	F	3	F
8	NF	7	NF	9	NF	5	F	6	NF
9	NF	6	NF	0	F	3	F	4	F

The sample proportion $\left(\widehat{p}\right)$ for each sample can be calculated by using the formula

$\widehat{p}=\frac{x}{n}$

where x is the number of successes and n is the sample size.

For sample 1, the number of adults who actively tried to avoid drinking regular soda or pop in the year 2015 is 16. So, the sample proportion is calculated as shown below:

$\begin{array}{c}\widehat{p}=\frac{x}{n}\\ =\frac{16}{25}\\ =0.64\end{array}$

For sample 2, the number of adults who actively tried to avoid drinking regular soda or pop in the year 2015 is 14. So, the sample proportion is calculated as shown below:

$\begin{array}{c}\widehat{p}=\frac{x}{n}\\ =\frac{14}{25}\\ =0.56\end{array}$

Similarly, the sample proportion who actively tried to avoid drinking regular soda or pop in the year 2015 for the remaining samples are shown below in the tabular manner:

Samples	1	2	3	4	5	6	7	8	9	10
Number of favorable	16	14	14	16	11	15	12	14	17	12
Sample proportion $\left(\widehat{p}\right)$	0.64	0.56	0.56	0.64	0.44	0.60	0.48	0.56	0.68	0.48

(c)

To determine

To find: The 68% confidence interval for the population proportion p from each of the 10 samples and then determine the number of confidence intervals in which the true parameter value 0.6 lies.

Answer to Problem 26E

Solution: The 68% confidence intervals for each of the 10 samples are shown below:

Samples	Sample proportion $\left(\widehat{p}\right)$	68% confidence interval
				Lower limit	Upper limit
1	0.64	0.544	0.736
2	0.56	0.461	0.659
3	0.56	0.461	0.659
4	0.64	0.544	0.736
5	0.44	0.341	0.539
6	0.60	0.502	0.698
7	0.48	0.381	0.579
8	0.56	0.461	0.659
9	0.68	0.587	0.773
10	0.48	0.381	0.579

From the obtained confidence intervals of population proportion p from each of the 10 samples, it can be said that true parameter value 0.6 will lie in the intervals of sample 1, sample 2, sample 3, sample 4, sample 6, sample 7, sample 8, and sample 9.

Explanation of Solution

Calculation:

Compute the confidence interval for true proportion p using formula as shown below:

$\text{Confidence interval}=\left[\widehat{p}\pm z\left(\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\right)\right]$

where

$\begin{array}{l}\widehat{p}\text{ is the sample proportion}\text{.}\\ z\text{ is the tabulated value for desired confidence interval}\text{.}\\ n\text{ is the sample size}\text{.}\end{array}$

According to empirical rule, 68% of the sample proportion lies within the one standard deviation about mean proportion, so the tabulated value of $z$ statistic is 1.0 for a two-tailed test.

For the calculation of confidence interval of the population proportion p from each sample, use the sample proportion obtained in part (b).

The 68% confidence interval of p is calculated from first sample as shown below:

$\begin{array}{c}\text{Confidence interval}=\left[\widehat{p}\pm z\times \left(\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\right)\right]\\ =0.64\pm 1\times \left(\sqrt{\frac{0.64\left(1-0.64\right)}{25}}\right)\\ =\left(0.64\pm 0.096\right)\\ =\left(0.544,0.736\right)\end{array}$

The 68% confidence interval of p is calculated from first sample as shown below:

$\begin{array}{c}\text{Confidence interval}=\left[\widehat{p}\pm z\times \left(\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\right)\right]\\ =0.56\pm 1\times \left(\sqrt{\frac{0.56\left(1-0.56\right)}{25}}\right)\\ =\left(0.56\pm 0.099\right)\\ =\left(0.461,0.659\right)\end{array}$

Similarly, the 68% confidence interval for each of the remaining samples is obtained, which is shown below in the tabular manner:

Samples	Sample proportion $\left(\widehat{p}\right)$	68% confidence interval
				Lower limit	Upper limit
1	0.64	0.544	0.736
2	0.56	0.461	0.659
3	0.56	0.461	0.659
4	0.64	0.544	0.736
5	0.44	0.341	0.539
6	0.60	0.502	0.698
7	0.48	0.381	0.579
8	0.56	0.461	0.659
9	0.68	0.587	0.773
10	0.48	0.381	0.579

Interpretation: Therefore, it can be concluded all the intervals capture the true parameter value 0.6 except the sample 5 and sample 10.