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?

NOTE: This is what I got but it is wrong.
a. 18.99
b. [254.008, 291.992]

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To find the margin of error and the confidence interval estimate, you can use the formula:

Margin of Error = (Critical Value) * (Standard Deviation / √Sample Size)

where the Critical Value is determined based on the desired confidence level.

a. To find the margin of error with a 95% confidence level, you need to find the critical value for a 95% confidence interval. Since the sample size is large (n > 30), you can use the z-distribution. The critical value for a 95% confidence level is approximately 1.96.
Given:
Sample size (n) = 45
Standard deviation (σ) = $65

Margin of Error = 1.96 * (65 / √45)
Margin of Error ≈ 18.83 (rounded to 2 decimal places)

Therefore, the margin of error is approximately $18.83.

b. The confidence interval estimate is calculated by subtracting/adding the margin of error from/to the sample mean.
Given:
Sample mean (x̄) = $273
Margin of Error = $18.83

Confidence Interval = (Sample Mean - Margin of Error, Sample Mean + Margin of Error)
Confidence Interval = ($273 - $18.83, $273 + $18.83)
Confidence Interval ≈ ($254.17, $291.83) (rounded to 2 decimal places)

Therefore, the 95% confidence interval estimate of the population mean is approximately $254.17 to $291.83.

To solve this problem, we will use the formula for calculating the margin of error and the confidence interval estimate of the population mean.

a. Margin of Error:
The margin of error is calculated using the formula:
Margin of Error = (Z * Standard Deviation) / √n

Where:
Z = Z-value for the desired confidence level
Standard Deviation = Sample standard deviation
n = Sample size

Since the confidence level is 95%, we need to find the Z-value corresponding to that level. We can use the Z-table or a calculator to find the Z-value. For a 95% confidence level, the Z-value is approximately 1.96.

Now we can plug in the values into the formula:
Margin of Error = (1.96 * $65) / √45 ≈ $18.99

Therefore, the margin of error is approximately $18.99.

b. Confidence Interval Estimate:
The confidence interval estimate of the population mean is calculated using the formula:
Confidence Interval = Sample Mean ± Margin of Error

Since the sample mean is not given in the question, we cannot provide an exact confidence interval estimate. However, we can estimate it using the average cost per night of a hotel room in New York City ($273) as the sample mean.

Using the estimated sample mean and the margin of error calculated in part a, we can calculate the confidence interval:
Confidence Interval = $273 ± $18.99

The lower bound of the confidence interval:
Lower Bound = $273 - $18.99 = $254.01

The upper bound of the confidence interval:
Upper Bound = $273 + $18.99 = $291.99

Therefore, the 95% confidence interval estimate of the population mean is approximately [$254.01, $291.99].

Note: Your answer of a margin of error of 18.99 was correct, but you made a mistake in rounding the confidence interval. The correct confidence interval should be [254.01, 291.99], rounded to 2 decimal places.

a) I think the question requires the use of the t-distribution.

standard error=statistic/sqrt(45)
=65/sqrt(45)=9.689
critical value: df=45-1=44, area=.025 so t=2.015

margin of error=t*SE=19.52

b) Confidence interval: Use t-distr: sample mean +- margin of error
=273 +- 19.52
=(253.48, 292.52)