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(a)
To find: The method of simulating the proportion of an SRS of 25 adult Americans.
(a)
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 find: The sample proportion
(b)
Answer to Problem 26E
Solution: The sample proportion
| 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 |
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
where x is the number of successes and n is the
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:
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:
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 |
0.64 | 0.56 | 0.56 | 0.64 | 0.44 | 0.60 | 0.48 | 0.56 | 0.68 | 0.48 |
(c)
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.
(c)
Answer to Problem 26E
Solution: The 68% confidence intervals for each of the 10 samples are shown below:
| Samples | Sample proportion |
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:
where
According to
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:
The 68% confidence interval of p is calculated from first sample as shown below:
Similarly, the 68% confidence interval for each of the remaining samples is obtained, which is shown below in the tabular manner:
| Samples | Sample proportion |
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.
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Chapter 21 Solutions
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