/*! 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 111 Unoccupied seats on flights caus... [FREE SOLUTION] | 91Ó°ÊÓ

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Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its mean number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard deviation is 4.1 seats. a. i. \(\overline{x}=\) _____ ii. \(s_{x}=\) _____ iii. \(n=\) _____ iv. \(n-1=\) _____ b. Define the random variables \(X\) and \(\overline{X}\) in words. c. Which distribution should you use for this problem? Explain your choice. d. Construct a 92\(\%\) confidence interval for the population mean number of unoccupied seats per flight. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound.

Short Answer

Expert verified
i. 11.6; ii. 4.1; iii. 225; iv. 224; b. Define \(X\) and \(\overline{X}\) as the number and mean number of unoccupied seats, respectively; c. Use normal distribution due to large sample size; d. CI is [11.12, 12.08], error bound is 0.48.

Step by step solution

01

Identify Given Variables

We are given the sample mean, sample standard deviation, and sample size. These are: \(\overline{x} = 11.6\), \(s_{x} = 4.1\), and \(n = 225\).
02

Perform Calculations for Part a

Given that \(\overline{x} = 11.6\), \(s_{x} = 4.1\), \(n = 225\), calculate \(n-1\): \[n-1 = 225 - 1 = 224 \]
03

Define Random Variables

The random variable \(X\) represents the number of unoccupied seats on a single flight. The random variable \(\overline{X}\) represents the mean number of unoccupied seats over all sampled flights.
04

Choose the Distribution

Given that the sample size is large (225 flights) and the Central Limit Theorem applies, the sample mean \(\overline{X}\) follows a normal distribution, making the normal (Z) distribution appropriate for constructing the confidence interval.
05

Construct the Confidence Interval

For a 92% confidence interval, the Z-score corresponding to a 92% confidence level is approximately \(Z_{0.04} \approx 1.75\). Calculate the confidence interval using the formula:\[\overline{x} \pm Z \cdot \frac{s_{x}}{\sqrt{n}}\]Substitute the known values:\[11.6 \pm 1.75 \cdot \frac{4.1}{\sqrt{225}}\]\[11.6 \pm 1.75 \cdot \frac{4.1}{15}\]\[11.6 \pm 1.75 \cdot 0.2733\]\[11.6 \pm 0.4783\]Thus, the confidence interval is \([11.1217, 12.0783]\).
06

Error Bound Calculation

The error bound (margin of error) is calculated using the formula:\[Z \cdot \frac{s_{x}}{\sqrt{n}}\]Substitute the known values:\[1.75 \cdot \frac{4.1}{15}\]\[1.75 \cdot 0.2733 = 0.4783\]So, the error bound is \(0.4783\).
07

Graph the Confidence Interval

To graph the confidence interval, plot a normal distribution curve with the sample mean \(11.6\) at the center. Mark the interval \([11.1217, 12.0783]\) on the horizontal axis to indicate the range of values, representing the estimated interval for the population mean.

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics. It states that when you take a sufficiently large sample size from a population, the distribution of the sample mean will approximate a normal distribution, regardless of the original population's distribution. This is particularly useful because it allows us to make inferences about the population mean using the normal distribution.

Here's why CLT is important in our exercise: We have a sample size of 225 flights. According to CLT, the sample mean of the number of unoccupied seats should be normally distributed, allowing us to use normal distribution tools, like constructing confidence intervals.
  • This approximation works best with larger samples (usually n ≥ 30 is considered sufficient).
  • It applies to various original population distributions, making it versatile.
  • Helps in making statistically sound predictions about population parameters.
The beauty of CLT simplifies analysis, making it easier to work with data without knowing the exact population distribution. This is why we confidently used the normal distribution in this exercise.
Normal Distribution
The normal distribution is a continuous probability distribution defined by its bell-shaped curve. Most data around the average are close to the mean of the data set, which decreases symmetrically towards the tails. It's characterized by two parameters: the mean (\(\mu\)) and the standard deviation (\(\sigma\)).

In our scenario, normal distribution becomes relevant because of the Central Limit Theorem. Let's explore how it's applied:
  • We use the sample mean's (\(\overline{x}\)) normal distribution to determine confidence intervals.
  • The distribution allows us to estimate where the true population mean might lie.
  • Within the curve, intervals can be easily calculated, as the data is assumed to spread evenly around the sample mean.
With this understanding, the exercise leverages the symmetry and defined nature of the normal distribution to make predictions about the population parameter of unoccupied seats.
Sample Mean
The sample mean (\(\overline{x}\)) is the average of the data points in your sample. In our exercise, it represents the average number of unoccupied seats across the 225 sampled flights. It's calculated by summing all observations and dividing by the number of observations.

The significance of the sample mean in statistical analysis includes:
  • Being the estimator of the population mean, providing insight into the overall trend of the data.
  • A friendly point of comparison as a "center" of the sample distribution.
  • Vital for predictive modeling through confidence intervals, like in our exercise.
In summary, the sample mean is not just a number but a central figure in data interpretation, serving as a primary tool when making inferences about population parameters.
Sample Standard Deviation
The sample standard deviation (\(s_x\)) quantifies the amount of variation or dispersion in a set of data points. In this exercise, it reflects the variability in the number of unoccupied seats across the 225 flights.

Here's why understanding sample standard deviation is crucial:
  • It measures the average distance of each data point from the sample mean.
  • Helps determine how spread out the data is, which influences the width of confidence intervals.
  • Directly impacts the calculation of the error bound in inference tasks, as seen in the exercise.
In essence, knowing the sample standard deviation is pivotal when assessing the reliability of the mean as an estimator of the population mean. It is this measurement of spread that allows for precise confidence interval calculations, aiding in robust statistical inference.

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Most popular questions from this chapter

Public Policy polling recently conducted a survey asking adults across the U.S. about music preferences. When asked, 80 of the 571 participants admitted that they have illegally downloaded music. a. Create a 99% confidence interval for the true proportion of American adults who have illegally downloaded music. b. This survey was conducted through automated telephone interviews on May 6 and 7, 2013. The error bound of the survey compensates for sampling error, or natural variability among samples. List some factors that could affect the survey’s outcome that are not covered by the margin of error. c. Without performing any calculations, describe how the confidence interval would change if the confidence level changed from 99% to 90%.

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