/*! 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 17 The International Air Transport ... [FREE SOLUTION] | 91Ó°ÊÓ

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The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airports. The maximum possible rating is \(10 .\) Suppose a simple random sample of 50 business travelers is selected and each traveler is asked to provide a rating for the Miami International Airport. The ratings obtained from the sample of 50 business travelers follow. $$\begin{array}{cccccccccccc}6 & 4 & 6 & 8 & 7 & 7 & 6 & 3 & 3 & 8 & 10 & 4 & 8 \\\7 & 8 & 7 & 5 & 9 & 5 & 8 & 4 & 3 & 8 & 5 & 5 & 4 \\\4 & 4 & 8 & 4 & 5 & 6 & 2 & 5 & 9 & 9 & 8 & 4 & 8 \\\9 & 9 & 5 & 9 & 7 & 8 & 3 & 10 & 8 & 9 & 6 & &\end{array}$$ Develop a \(95 \%\) confidence interval estimate of the population mean rating for Miami.

Short Answer

Expert verified
The 95% confidence interval for the mean rating is approximately (6.25, 7.47).

Step by step solution

01

Calculate the Sample Mean

First, add all the ratings together and then divide by the number of ratings to find the sample mean (\(\bar{x}\).Ratings: Adding up all ratings: \[ S = 6 + 4 + 6 + 8 + 7 + 7 + 6 + 3 + 3 + 8 + 10 + 4 + 8 + 7 + 8 + 7 + 5 + 9 + 5 + 8 + 4 + 3 + 8 + 5 + 5 + 4 + 4 + 4 + 8 + 4 + 5 + 6 + 2 + 5 + 9 + 9 + 8 + 4 + 8 + 9 + 9 + 5 + 9 + 7 + 8 + 3 + 10 + 8 + 9 + 6 = 343 \]Sample mean (\(\bar{x}\): \[\frac{343}{50} = 6.86\]
02

Calculate the Sample Standard Deviation

To calculate the sample standard deviation (\(s\), find the variance first by determining the squared difference of each score from the mean, summing them up, and then dividing by one less than the sample size, before taking the square root.Calculate each squared deviation: \[\text{Sum of } (x_i - \bar{x})^2 = 223.68\] Variance:\[\frac{223.68}{49} \approx 4.564 \] Sample standard deviation:\[s \approx \sqrt{4.564} \approx 2.136\]
03

Determine the Standard Error of the Mean

The standard error (SE) is found by dividing the sample standard deviation by the square root of the sample size.\[SE = \frac{s}{\sqrt{n}} = \frac{2.136}{\sqrt{50}} \approx 0.302\]
04

Use the T-Distribution to Find the Critical Value

For a 95% confidence interval and 49 degrees of freedom (\(n-1\), find the t-value from a t-distribution table. This value is approximately 2.009 for a two-tailed test.
05

Calculate the Confidence Interval

Now use the formula for the confidence interval:\[\bar{x} \pm t_{\text{critical}} \times SE \]\[6.86 \pm 2.009 \times 0.302 \]Calculate upper and lower bounds:Upper bound: \(6.86 + 0.606\,448\approx 7.466\) Lower bound: \(6.86 - 0.606\,448\approx 6.254\) The 95% confidence interval is approximately (6.25, 7.47).

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Key Concepts

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

Sample Mean Calculation
The sample mean, often represented by \( \bar{x} \), is a simple statistical metric to find the average of a data set. To calculate the sample mean, add together all the individual ratings given by the sample and then divide the sum by the total number of ratings in the sample. This produces an estimate of the central tendency of the data, giving us an easy way to understand where most data points lie on average. It serves as a crucial element in creating the confidence interval, as it acts as the midpoint of this range. In our example, the sum of all ratings was 343, and with 50 travelers providing these ratings, we obtained a sample mean of \( \bar{x} = \frac{343}{50} = 6.86 \). This means the average rating given by the travelers to Miami International Airport was approximately 6.86 out of 10.
Sample Standard Deviation
While the sample mean gives us an average, the sample standard deviation (represented by \( s \)) tells us how much the data deviates from this average. It's a measure of the dispersion or variability in the sample ratings. First, calculate the variance by determining the square of differences between each data point and the sample mean, adding those results, and then dividing by one less than the sample size. Afterward, take the square root of this variance to find the standard deviation. In our exercise, the squared differences summed up to 223.68, producing a variance of \( \frac{223.68}{49} \approx 4.564 \). Taking the square root of this variance, we find the sample standard deviation is \( s \approx 2.136 \). A lower value would indicate that the ratings are closer to the mean, while a higher value suggests more variability in traveler ratings.
T-Distribution
The t-distribution is pivotal when dealing with small sample sizes, especially when the population standard deviation is unknown. It helps in establishing a range within which we can be confident the true population parameter lies. Unlike the normal distribution, the t-distribution is slightly broader in shape, adjusting for the additional uncertainty introduced by small sample sizes. With 50 samples, we have 49 degrees of freedom (\( n-1 \)). For a 95% confidence interval, the critical t-value obtained from a t-table with 49 degrees of freedom is approximately 2.009. The t-distribution is then applied to the standard error to find the confidence interval around the sample mean. This interval provides a range where the true population mean is expected to fall 95% of the time, offering a balance of accuracy and reliability for our estimates.

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Most popular questions from this chapter

A Phoenix Wealth Management/Harris Interactive survey of 1500 individuals with net worth of \(\$ 1\) million or more provided a variety of statistics on wealthy people (BusinessWeek, September 22,2003 ). The previous three-year period had been bad for the stock market, which motivated some of the questions asked. a. The survey reported that \(53 \%\) of the respondents lost \(25 \%\) or more of their portfolio value over the past three years. Develop a \(95 \%\) confidence interval for the proportion of wealthy people who lost \(25 \%\) or more of their portfolio value over the past three years. b. The survey reported that \(31 \%\) of the respondents feel they have to save more for retirement to make up for what they lost. Develop a \(95 \%\) confidence interval for the population proportion. c. Five percent of the respondents gave \(\$ 25,000\) or more to charity over the previous year. Develop a \(95 \%\) confidence interval for the proportion who gave \(\$ 25,000\) or more to charity. d. Compare the margin of error for the interval estimates in parts (a), (b), and (c). How is the margin of error related to \(\bar{p} ?\) When the same sample is being used to estimate a variety of proportions, which of the proportions should be used to choose the planning value \(p^{*} ?\) Why do you think \(p^{*}=.50\) is often used in these cases?

The travel-to-work time for residents of the 15 largest cities in the United States is reported in the 2003 Information Please Almanac. Suppose that a preliminary simple random sample of residents of San Francisco is used to develop a planning value of 6.25 minutes for the population standard deviation. a. If we want to estimate the population mean travel-to-work time for San Francisco residents with a margin of error of 2 minutes, what sample size should be used? Assume \(95 \%\) confidence. b. If we want to estimate the population mean travel-to-work time for San Francisco residents with a margin of error of 1 minute, what sample size should be used? Assume \(95 \%\) confidence.

The motion picture Harry Potter and the Sorcerer's Stone shattered the box office debut record previously held by The Lost World: Jurassic Park (The Wall Street Journal, November 19,2001 ). A sample of 100 movie theaters showed that the mean three-day weekend gross was \(\$ 25,467\) per theater. The sample standard deviation was \(\$ 4980\) a. What is the margin of error for this study? Use \(95 \%\) confidence. b. What is the \(95 \%\) confidence interval estimate for the population mean weekend gross per theater? c. The Lost World took in \(\$ 72.1\) million in its first three-day weekend. Harry Potter and the Sorcerer's Stone was shown in 3672 theaters. What is an estimate of the total Harry Potter and the Sorcerer's Stone took in during its first three-day weekend? d. An Associated Press article claimed Harry Potter "shattered" the box office debut record held by The Lost World. Do your results agree with this claim?

A simple random sample of 40 items resulted in a sample mean of \(25 .\) The population standard deviation is \(\sigma=5\) a. What is the standard error of the mean, \(\sigma_{z} ?\) b. \(\quad\) At \(95 \%\) confidence, what is the margin of error?

Find the \(t\) value(s) for each of the following cases. a. Upper tail area of .025 with 12 degrees of freedom b. Lower tail area of .05 with 50 degrees of freedom c. Upper tail area of .01 with 30 degrees of freedom d. Where \(90 \%\) of the area falls between these two \(t\) values with 25 degrees of freedom e. Where \(95 \%\) of the area falls between these two \(t\) values with 45 degrees of freedom

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