The following is taken from a State of the Park Report for Banff National Park.
Visitors Numbers
Total Visitors: 2004= 3,135,727. 2005=3,164,906. 2006=3,281,435
Total visitor days: 2004= 7,453,465. 2005= 7,518,997. 2006= 7,784,044
Total visitors have increased by 4.6% and total visitor days by 4.4% in this period. Group tour visitors have increased slightly from 11.6% to 12.6% of the total. These figures are not absolutes, as the margin of error for total visitors is 7.5% and for total visitor days, 7.3%.
1.) determine the confidence interval for total visitors in 2006.
Explanation:
Total visitors for 2006 = 3,281,435
Margin of error = 7.5% of 3,281,435
= 0.075 * 3,281,435
= 246, 107.625
Confidence interval:
Lower bound = Total visitors for 2006 - margin of error
= 3,281,435 - 246, 107.625
= 3035327. 375
= 3035328 ( round up)
Upper bound = Total visitors for 2006 + margin of error
= 3,281,435 + 246, 107.625
= 3527542.625
= 3527543 ( round up)
To determine the confidence interval for total visitors in 2006, we need to use the information given, including the total visitors for each year (2004, 2005, and 2006) and the margin of error for total visitors.
First, let's calculate the mean total visitors for the three years:
Mean Total Visitors = (Total Visitors in 2004 + Total Visitors in 2005 + Total Visitors in 2006) / 3
Mean Total Visitors = (3,135,727 + 3,164,906 + 3,281,435) / 3
Mean Total Visitors = 9,581,068 / 3
Mean Total Visitors = 3,193,689.33 (approximately)
Next, we need to calculate the standard deviation. To do this, we'll use the margin of error for total visitors (7.5%) as an estimate for the standard deviation:
Standard Deviation ≈ Margin of Error × √Number of Observations
Standard Deviation ≈ 0.075 × √3
Standard Deviation ≈ 0.075 × 1.732 (approximately)
Standard Deviation ≈ 0.1299 (approximately)
Now, we can calculate the confidence interval using the formula:
Confidence Interval = Mean Total Visitors ± (Z-Score × Standard Deviation)
The Z-Score is a value that corresponds to the desired level of confidence. Let's assume we want a 95% confidence level, which corresponds to a Z-Score of 1.96 (for a large sample size).
Confidence Interval = 3,193,689.33 ± (1.96 × 0.1299)
Confidence Interval = 3,193,689.33 ± 0.2548 (approximately)
Therefore, the confidence interval for total visitors in 2006 is approximately 3,193,689.33 ± 0.2548, which can be rounded to:
Confidence Interval ≈ 3,193,689 ± 0.255 (approximately)