Final answer:
The 95% confidence interval estimate for the mean amount spent per day by a family of four vacationing at Niagara Falls, based on the sample provided, is between $233.83 and $271.07.
Step-by-step explanation:
The question is asking to calculate a 95% confidence interval for the mean amount spent per day by a family of four visiting Niagara Falls based on a sample of 64 families with a sample mean of $252.45 and a sample standard deviation of $74.50. To find the confidence interval, we will use the formula for the confidence interval of the mean when the population standard deviation is unknown, which is:
Sample Mean ± (t-value * (Sample Standard Deviation / sqrt(Sample Size)))
Since the sample size is 64, which is greater than 30, we can assume the distribution of the sample means is approximately normal. Therefore, we will use the t-distribution with 63 degrees of freedom (n-1) to find the critical t-value for a 95% confidence level.
Let's do the calculation:
- Sample Mean = $252.45
- Sample Standard Deviation = $74.50
- Sample Size (n) = 64
- Degrees of Freedom (df) = 63
Using a t-distribution table or calculator, we find that the critical t-value (two-tailed) for 63 degrees of freedom at the 95% confidence level is approximately 1.998. Now we can calculate the margin of error (ME):
ME = t-value * (Sample Standard Deviation / sqrt(Sample Size))
ME = 1.998 * ($74.50 / sqrt(64))
ME = 1.998 * ($74.50 / 8)
ME = 1.998 * 9.3125
ME ≈ $18.62
Finally, we calculate the confidence interval:
Lower Limit = Sample Mean - ME
Lower Limit = $252.45 - $18.62
Lower Limit ≈ $233.83
Upper Limit = Sample Mean + ME
Upper Limit = $252.45 + $18.62
Upper Limit ≈ $271.07
The 95% confidence interval estimate for the mean amount spent per day by a family of four vacationing at Niagara Falls is between $233.83 and $271.07.