The average cost per night of a hotel room in New York City is $273 (SmartMoney, March 2009). Assume this estimate is based on a sample of 45 hotels and that the sample standard deviation is $65.
a. With 95% confidence, what is the margin of error (to 2 decimals)?
b. What is the 95% confidence interval estimate of the population mean?
( , )
Part a)
ME = 1.96 * sd/√n
With your data:
ME = 1.96 * 65/√45
Part b)
CI95 = mean ± 1.96(sd/n)
With your data:
CI95 = 273 ± 1.96(65/√45)
I'll let you finish the calculations.
To find the margin of error and the confidence interval estimate of the population mean, we can use the formula for confidence intervals:
Margin of Error (ME) = (Critical Value) * (Standard Deviation / √(Sample Size))
Confidence Interval = Sample Mean ± Margin of Error
a. To calculate the margin of error, we need to determine the critical value corresponding to a 95% confidence level. For a 95% confidence level, we can use the Z-score table or a Z-score calculator. The Z-score associated with a 95% confidence level is approximately 1.96.
ME = 1.96 * (65 / √(45))
ME ≈ 1.96 * (65 / 6.71)
ME ≈ 1.96 * 9.69
ME ≈ 18.98
Therefore, the margin of error is approximately 18.98 (to 2 decimals).
b. To calculate the confidence interval estimate of the population mean, we need to use the margin of error and the sample mean.
Confidence Interval = Sample Mean ± Margin of Error
Given that the sample mean is $273 and the margin of error is approximately 18.98, we can calculate the confidence interval as follows:
Lower Limit = 273 - 18.98
Lower Limit ≈ 254.02
Upper Limit = 273 + 18.98
Upper Limit ≈ 291.98
Therefore, the 95% confidence interval estimate of the population mean is approximately ($254.02, $291.98).