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Where Is the Best Seat on the Plane? A survey of 1000 air travelers \(^{21}\) found that \(60 \%\) prefer a window seat. The sample size is large enough to use the normal distribution, and a bootstrap distribution shows that the standard error is \(S E=0.015 .\) Use a normal distribution to find and interpret a \(99 \%\) confidence interval for the proportion of air travelers who prefer a window seat.

Short Answer

Expert verified
The 99% confidence interval for the proportion of air travelers who prefer a window seat is approximately \((0.563, 0.637)\). This means that based on the given sample, we are 99% confident the actual proportion of all air travelers who prefer a window seat is between 56.3% and 63.7%.

Step by step solution

01

Identify Given Information

From the problem, we are given that 60% of the 1000 air travelers sampled prefer a window seat. Therefore, the sample proportion (\(p̂\)) is 0.60. We were also told that the standard error (\(SE\)) is 0.015.
02

Identify the Z-value Corresponding to the 99% Confidence Interval

The problem asks for a 99% confidence interval, which means that there is a total of 1% area in both tails (0.5% in each tail) of the normal distribution. Looking up on the z-table or standard normal distribution table, we see that the z-value that corresponds to 0.5% in the tail is roughly \(z = 2.58\).
03

Calculate Confidence Interval

The formula for a confidence interval is \(p̂ \pm z(SE)\). Plugging in the given values: \[ 0.60 \pm 2.58(0.015) \]. This gives us a confidence interval of approximately \((0.563, 0.637)\).

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Key Concepts

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

Normal Distribution
Understanding the normal distribution is essential when dealing with phenomena that have a natural variability around an average. Imagine a bell-shaped curve where the highest point represents the most common outcome, in this case, travelers' preference for a window seat. The normal distribution is symmetrical, with the mean, median, and mode being equal.

In the context of our exercise, a large sample of 1000 travelers follows this pattern. The distribution's symmetry implies that an equal proportion of values lie above and below the mean, which becomes particularly useful when we're calculating confidence intervals. To apply the normal distribution in practice, we use what's called the Z-score, which tells us how many standard deviations an element is from the mean.
Bootstrap Distribution
The bootstrap distribution is a modern statistical technique used to estimate the sampling distribution of a statistic. This is often done by repeatedly resampling with replacement from the sample data and aggregating the results. In simpler terms, think of it as creating a bunch of 'mini-samples' from your original sample, and then using those to see how much your statistic (like the sample proportion) varies.

For our air travelers, though the exact bootstrap distribution is not detailed in the problem, it has been used to determine the standard error of the sample proportion. This method allows us to embrace the variability and uncertainty inherent in the data, offering a robust way to understand the potential spread of the proportion estimate.
Standard Error
The term 'standard error' (SE) might sound complex, but it's really just a way to express the uncertainty in an estimate. Standard error tells us how far we can expect the sample statistic, like the sample proportion of window-seat lovers, to deviate from the true population parameter if we were to repeat the survey over and over. A smaller SE suggests our sample statistic is likely to be closer to the true population parameter.

In the given scenario, the standard error of 0.015 signifies that if multiple samples were taken, the sample proportions would typically vary by this amount from the true population proportion. It's a critical part of calculating the confidence interval, as it defines the precision of our estimate.
Sample Proportion
The sample proportion is the decimal value that represents the part of the sample meeting a particular criterion. For example, the proportion of air travelers preferring a window seat is 0.60, or 60%, in our survey. This percentage is not just a number; it's a snapshot of traveler preferences based on our sample.

Understanding the sample proportion is key since it is this value around which we build the confidence interval. Because samples can vary, we use the concept of a confidence interval to express the range within which we believe the true population proportion lies, with a certain level of confidence. Inaccuracies in estimating the sample proportion can lead to erroneous conclusions, making accurate sampling and understanding sample proportion crucial in statistics.

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Most popular questions from this chapter

Exercises 5.7 to 5.12 include a set of hypotheses, some information from one or more samples, and a standard error from a randomization distribution. Find the value of the standardized \(z\) -test statistic in each situation. Test \(H_{0}: p=0.25\) vs \(H_{a}: p<0.25\) when the sample has \(n=800\) and \(\hat{p}=0.235,\) with \(S E=0.018\).

Hearing Loss in Teenagers A recent study \(^{20}\) found that, of the 1771 participants aged 12 to 19 in the National Health and Nutrition Examination Survey, \(19.5 \%\) had some hearing loss (defined as a loss of 15 decibels in at least one ear). This is a dramatic increase from a decade ago. The sample size is large enough to use the normal distribution, and a bootstrap distribution shows that the standard error for the proportion is \(S E=0.009 .\) Find and interpret a \(90 \%\) confidence interval for the proportion of teenagers with some hearing loss.

Incentives to Exercise: How Well Do They Work? A study \(^{14}\) was designed to see what type of incentive might be most effective in encouraging people to exercise. In the study, 281 overweight or obese people were assigned the goal to walk 7000 steps a day, and their activity was tracked for 100 days. The participants were randomly assigned to one of four groups, with different incentives for each group. In this problem, we look at the overall success rate. For each participant, we record the number of days that the participant met the goal. For all 281 participants, the average number of days meeting the goal is \(36.5 .\) The standard error for this estimate is \(1.80 .\) Test to see if this provides evidence that the mean number of days meeting the goal, for people in a 100 -day program to encourage exercise, is greater than 35. Show all details of the test.

Find the \(z^{*}\) values based on a standard normal distribution for each of the following. (a) An \(86 \%\) confidence interval for a correlation. (b) A \(94 \%\) confidence interval for a difference in proportions. (c) A \(96 \%\) confidence interval for a proportion.

Exercises 5.7 to 5.12 include a set of hypotheses, some information from one or more samples, and a standard error from a randomization distribution. Find the value of the standardized \(z\) -test statistic in each situation. Test \(H_{0}: \mu=10\) vs \(H_{a}: \mu \neq 10\) when the sample has \(n=75, \bar{x}=11.3,\) and \(s=0.85,\) with \(S E=0.10\).

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