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?
ta/2 = a = 0.05 = 0.025
t = 2.02
a. 19.57
b. (253.47, 292.53)
a) a = 0.05 = z(0.025) = 1.96
Mean of error = Za/2* σ/√n
= 1.96 *65/√45
= 18.99
b) xbar-+ Za/2*σ/√n
= 273 -18.992, 273+ 18.992
= [254.008, 291.992]
Hello,
Thanks for your help the answers you got are the same as mine but the website where I do my homework says it is wrong.
Can you help me out please?
To determine the margin of error and the confidence interval estimate, we need to use the formula for confidence interval:
Margin of Error = (z-score) * (standard deviation / square root of sample size)
a. To find the margin of error, we need the critical value or z-score corresponding to a 95% confidence level. The z-score represents the number of standard deviations away from the mean to create a desired level of confidence. For a 95% confidence level, the z-score is 1.96.
Margin of Error = (1.96) * (65 / square root of 45)
Margin of Error ≈ 18.83 (rounded to 2 decimals)
Therefore, the margin of error is approximately $18.83.
b. The 95% confidence interval estimate is calculated by adding and subtracting the margin of error from the sample mean.
Confidence Interval = Sample Mean ± Margin of Error
Confidence Interval = $273 ± $18.83
Confidence Interval ≈ ($254.17, $291.83)
Therefore, using a 95% confidence level, the population mean of hotel room costs per night in New York City is estimated to be between $254.17 and $291.83.