According to a recent pole only 21% of Americans were satisfied with the direction the U.S. is heading. To investigate the reliability of the report an independent polling organization surveyed 500 Americans and inquired as to whether or not they were satisfied with the way things are progressing in the U.S. The response indicated 117 of 500 were satisfied.
A. Calculate a 95% confidence interval to estimate the true proportion of Americans who are currently satisfied with the direction the U.S. is heading .
B. If in the future the polling organization wanted to conduct a 99% confidence interval to estimate the true proportion of Americans who are currently satisfied with the direction the U.S. is heading and have the margin of error to be no larger than 3%, how many Americans should be included in the survey.
To calculate a confidence interval and determine the required sample size, we can use the formula for the proportion confidence interval. The formula is:
Confidence Interval = p̂ ± Z * √(p̂*(1-p̂)/n)
Where:
p̂ is the sample proportion (117/500 = 0.234)
Z is the z-score corresponding to the desired confidence level (e.g., 1.96 for a 95% confidence level or 2.58 for a 99% confidence level)
n is the sample size
A. Calculating a 95% confidence interval:
Using the given data, the sample proportion is 0.234, and the sample size is 500. The z-score for a 95% confidence level is 1.96.
Confidence Interval = 0.234 ± 1.96 * √(0.234*(1-0.234)/500)
Confidence Interval = 0.234 ± 1.96 * 0.0179
Confidence Interval = 0.234 ± 0.035
Therefore, the 95% confidence interval to estimate the true proportion of Americans satisfied with the direction the U.S. is heading is (0.199, 0.269).
B. Determining the required sample size for a 99% confidence interval:
Given that the desired margin of error is no larger than 3%, we can set up the following equation:
Z * √(p̂*(1-p̂)/n) ≤ margin of error
2.58 * √(0.234*(1-0.234)/n) ≤ 0.03
Squaring both sides of the equation:
(2.58 * √(0.234*(1-0.234)/n))^2 ≤ (0.03)^2
6.6564 * 0.234*(1-0.234)/n ≤ 0.0009
Simplifying:
0.3664/n ≤ 0.0009
n ≥ 0.3664/0.0009
Therefore, the required sample size to achieve a 99% confidence interval with a margin of error no larger than 3% is n ≥ 407.11. Since the sample size must be a whole number, we would need to round up. Therefore, a minimum sample size of 408 Americans should be included in the survey.