/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 A travel agent wants to gather i... [FREE SOLUTION] | 91Ó°ÊÓ

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A travel agent wants to gather information on the per-night cost at hotels in Caribbean countries. She took a random sample of 52 rooms from various hotels in those countries. The sample produced a mean cost for the 52 rooms to be \(\$ 208.35\) per night. If the population standard deviation of costs for a one- night stay in Caribbean hotels is \(\$ 47.45\), find a \(99 \%\) confidence interval for the average cost per night in Caribbean hotels.

Short Answer

Expert verified
Therefore, the 99% confidence interval for the average cost per night in Caribbean hotels is approximately between $192.91 and $223.79.

Step by step solution

01

Identify Given Information

We identify the following parameters from the problem: the sample mean (X) is $208.35, the population standard deviation (σ) is $47.45, the sample size (n) is 52. We aim to find a 99% confidence interval.
02

Calculate the Standard Error

The Standard Error (SE) is found using the formula \[SE = \frac{σ}{\sqrt{n}}\]. Substituting σ = $47.45 and n = 52, we calculate the value of SE.
03

Find the Z-Score

For a 99% confidence level, the Z-score (z) is approximately 2.576. This is a standard value from z-table corresponding to a desired confidence level.
04

Calculate the Lower and Upper Bound of Confidence Interval

The confidence interval (CI) is calculated using the formula \[CI = X ± Z * SE\]. We calculate the lower and upper bound of CI by subtracting and adding the product of Z and SE from X respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Mean
The sample mean is an essential concept in statistics. It's basically the average of a set of data points collected from a sample, which in this case, is the daily cost of hotel rooms in the Caribbean. To find the sample mean, add up all the prices collected and divide by the number of rooms. In our example, the travel agent found that the average nightly cost for the sample of 52 hotel rooms was $208.35. Calculating the sample mean helps in estimating the average cost in the entire population, which is all hotel rooms in Caribbean countries. It's important because it provides a point estimate that informs us about the general trend within a population. A key point to remember is that the sample mean is an estimate and can fluctuate based on different samples taken from the population.
Population Standard Deviation
The population standard deviation is a measure that quantifies how much individual data points in a population deviate from the mean. In simpler terms, it tells us how spread out the prices of all the hotel rooms are around the average price. This is valuable because it gives us an idea of volatility in prices — are they typically very similar, or is there a lot of variation? In our scenario, the population standard deviation for hotel room prices was known to be $47.45. This figure is predetermined for all rooms in Caribbean hotels, not just the sample. This helps in calculating the confidence interval because it portrays the natural dispersion of costs related to the mean cost. Understanding the standard deviation allows better predictions and calculations regarding the average costs.
Z-Score
The Z-score is a statistical measurement that describes a value's relationship to the mean in terms of standard deviations. When we talk about a Z-score in relation to confidence intervals, it represents the number of standard deviations a particular point is from the mean. For our confidence level of 99%, the Z-score used was approximately 2.576. This value is typically derived from standard Z-tables and tells us how extreme an observed result is, assuming the normal distribution. Choosing the correct Z-score is crucial for calculating precise confidence intervals. It fundamentally helps in determining how certain we can be about our estimate of the true mean. A larger Z-score corresponds to greater confidence in our interval estimate, meaning we're more certain our interval contains the true mean.
Standard Error
The standard error (SE) quantifies the variability or the standard deviation of the sample mean estimate of a population mean. In simple terms, it tells us how much the sample mean is expected to fluctuate around the true population mean.The formula for calculating the SE is \[SE = \frac{σ}{\sqrt{n}}\]where \(σ\) is the population standard deviation, and \(n\) is the sample size. For our case, the SE is \[SE = \frac{47.45}{\sqrt{52}}\].Calculating the standard error is a crucial step in forming confidence intervals. It essentially allows us to measure how far we expect the sample mean to be from the population mean. A smaller SE indicates that the sample mean is a more accurate estimate of the population mean, giving us greater confidence in our results.

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Most popular questions from this chapter

Because of inadequate public school budgets and lack of money available to teachers for classroom materials, many teachers often use their own money to buy materials used in the classrooms. A random sample of 100 public school teachers selected from an eastern state showed that they spent an average of \(\$ 290\) of their own money on such materials during the 2011-2012 school year. The population standard deviation was \(\$ 70\). a. What is the point estimate of the mean of such expenses incurred during the \(2011-2012\) school year by all public school teachers in this state? b. Make a \(95 \%\) confidence interval for the corresponding population mean.

Determine the most conservative sample size for the estimation of the population proportion for the following. a. \(E=.025\), confidence level \(=95 \%\) b. \(E=.05, \quad\) confidence level \(=90 \%\) c. \(E=.015\), confidence level \(=99 \%\)

What is the point estimator of the population proportion, \(p ?\)

a. How large a sample should be selected so that the margin of error of estimate for a \(98 \%\) confidence interval for \(p\) is \(.045\) when the value of the sample proportion obtained from a preliminary sample is \(.53\) ? b. Find the most conservative sample size that will produce the margin of error for a \(98 \%\) confidence interval for \(p\) equal to \(.045\)

A mail-order company promises its customers that the products ordered will be mailed within 72 hours after an order is placed. The quality control department at the company checks from time to time to see if this promise is fulfilled. Recently the quality control department took a sample of 50 orders and found that 35 of them were mailed within 72 hours of the placement of the orders. a. Construct a \(98 \%\) confidence interval for the percentage of all orders that are mailed within 72 hours of their placement. b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?

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