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Chapter 8: Problem 19

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? b. What is the \(95 \%\) confidence interval estimate of the population mean? c. Two years ago the average cost of a hotel room in New York City was \(\$ 229 .\) Discuss the change in cost over the two-year period.

Short Answer

Expert verified
The margin of error for a 95% confidence level is approximately $19.58. The 95% confidence interval estimate of the population mean is approximately (\(253.42, \)292.58). Over the two-year period, the average cost per night of a hotel room in New York City has increased by at least \(24.42, suggesting a significant increase in cost.

Step by step solution

01

Identify the given information

We know the following: - Sample mean (\(\bar{X}\)) = $273 - Sample size (n) = 45 - Sample standard deviation (s) = $65 - Confidence level = 0.95
02

Calculate the t-score for the given confidence level and sample size

The t-distribution table provides us with the t-score for the given confidence level and degrees of freedom (n - 1). In this case, we have (45 - 1) = 44 degrees of freedom. For a 95% confidence level, we find the t-score to be approximately 2.015.
03

Calculate the margin of error

The margin of error (E) is calculated using the following formula: \(E = t * \frac{s}{\sqrt{n}}\) Plugging in the values, we get: \(E = 2.015 * \frac{65}{\sqrt{45}} ≈ 19.58\) The margin of error is approximately $19.58.
04

Calculate the 95% confidence interval estimate of the population mean

The confidence interval is calculated using the following formula: \((\bar{X} - E, \bar{X} + E)\) Plugging in the values, we get: \((273 - 19.58, 273 + 19.58) ≈ (253.42, 292.58)\) The 95% confidence interval estimate of the population mean is approximately (\(253.42, \)292.58).
05

Discuss the change in cost over the two-year period

Two years ago, the average cost of a hotel room in New York City was \(229. The current average cost is \)273. The difference between the two average costs is: \(273 - \)229 = $44 Over the two-year period, the average cost per night of a hotel room in New York City has increased by \(44. However, since the actual population mean could be anywhere within the confidence interval, we cannot definitively say that the average cost has increased by exactly \)44. It could have increased by as little as \(24.42 or as much as \)63.58. Nonetheless, the data suggests that there has been a significant increase in the cost over the two-year period.

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Most popular questions from this chapter

A \(95 \%\) confidence interval for a population mean was reported to be 152 to \(160 .\) If \(\sigma=15\) what sample size was used in this study?

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