Us travel data center survey of 1500 adults found that 42% of respondents stated that they favor historical sites. Find the 95% confidence interval of the true proportion of all adults who favor visiting historical sites.
To find the 95% confidence interval of the true proportion of all adults who favor visiting historical sites, we can use the formula:
Confidence Interval = Sample Proportion ± (Z-score * Standard Error)
Where:
- Sample Proportion is the observed proportion in the survey (42% or 0.42)
- Z-score is the critical value corresponding to the desired confidence level (95% confidence level corresponds to a Z-score of approximately 1.96)
- Standard Error is calculated as the square root of [(Sample Proportion * (1 - Sample Proportion)) / Sample Size]
Given that the survey had 1500 respondents and 42% favored historical sites, we can calculate the confidence interval as follows:
Sample Proportion = 0.42
Sample Size = 1500
Standard Error = sqrt[(0.42 * (1 - 0.42)) / 1500] ≈ 0.014
Z-score for a 95% confidence level = 1.96
Now we can compute the confidence interval:
Confidence Interval = 0.42 ± (1.96 * 0.014)
Lower Bound = 0.42 - (1.96 * 0.014) ≈ 0.392
Upper Bound = 0.42 + (1.96 * 0.014) ≈ 0.448
Therefore, the 95% confidence interval for the true proportion of all adults who favor visiting historical sites is approximately 39.2% to 44.8%.
Use a proportional confidence interval formula:
CI95 = p + or - (1.96)(√pq/n)
Note: + or - 1.96 represents 95% confidence interval.
For p in your problem: .42
For q: 1 - p = 1 - .42 = .58
n = 1500
I let you take it from here to calculate the interval.