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Chapter 9: Problem 48

Example 9.3 gave the following airbome times (in minutes) for 10 randomly selected flights from San Francisco to Washington Dulles airport: \(\begin{array}{llllllllll}270 & 256 & 267 & 285 & 274 & 275 & 266 & 258 & 271 & 281\end{array}\) a. Compute and interpret a \(90 \%\) confidence interval for the mean airborne time for flights from San Francisco to Washington Dulles. b. Give an interpretation of the \(90 \%\) confidence level associated with the interval estimate in Part (a). c. If a flight from San Francisco to Washington Dulles is scheduled to depart at \(10 \mathrm{A.M}\), what would you recommend for the published arrival time? Explain.

Short Answer

Expert verified
a) The 90% confidence interval for the mean airborne time is computed using the sample mean and standard deviation. b) The 90% confidence level means that we are confident that 90% of the confidence intervals calculated from the random samples of flights would contain the true mean flight time. c) The recommended published arrival time can be estimated by adding the mean airborne time (plus possibly a buffer time, e.g. standard deviation) to the departure time of 10 A.M.

Step by step solution

01

Compute the Mean Airborne Time

Add all the individual flight times together and divide by the total number of flights to get the mean. Remember that the mean is the sum of all data divided by the number of data points. In this case, we add up all flight times and divide by the total number of flights, which is 10
02

Compute the Standard Deviation

Subtract each flight time from the mean, square the result, sum up these squared values and then divide by the number of flights, finally, take the square root of the outcome. A standard deviation is a measure of how spread out numbers are from the mean.
03

Compute the 90% Confidence Interval

Subtract and add the product of standard deviation, standard score for the 90% confidence level (which is 1.645 for a one-tailed test), and the square root of the number of flights divided by the square root of the number of flights to the mean flight time respectively. The formula for the confidence interval is mean 卤 (standard score 脳 (standard deviation 梅 鈭歯)) where n is the number of data points.
04

Interpret the Confidence Interval

A 90% confidence interval means that we are 90% confident that the true mean flight time lies within the calculated interval. It gives an indication of the 'reliability' of the estimated mean.
05

Recommend a Published Arrival Time

Add the mean airborne time to the departure time (10 A.M.). An additional buffer time (e.g., the standard deviation) can also be added to accommodate any variations in airborne time.

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Key Concepts

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

Mean Airborne Time
To compute the mean airborne time, we perform a simple calculation involving the sum of flight times. Imagine you have a list of durations collected from different flights. The mean is simply the average time the plane spends in the air for these flights. You take all the flight durations, add them together, and then divide by the number of flights you have. This gives you a single number that represents the central tendency of those times.

For instance, with our data:
  • Add up all the airborne times: 270 + 256 + 267 + 285 + 274 + 275 + 266 + 258 + 271 + 281.
  • The total sum is 2703 minutes.
  • Divide this sum by the number of flights: 2703 梅 10 = 270.3 minutes.
So, the mean airborne time is 270.3 minutes. This tells us that on average, these flights spend about 270.3 minutes between San Francisco and Washington Dulles.
Standard Deviation
Standard deviation is a vital statistic in understanding the spread of flight times around the mean. It tells us how much variability exists in our dataset. If the standard deviation is high, the flight times vary a lot from the mean; if it's low, they are closer to the mean.

To compute it, follow these steps:
  • First, subtract the mean time from each flight time to see how far each one is from the average.
  • Then, square these differences to eliminate any negative signs.
  • Sum up all these squared numbers.
  • Divide by the number of flights to get the variance.
  • The standard deviation is the square root of this variance.
This process gives us an overview of how much flights deviate from the average time of 270.3 minutes. A clear picture of the spread helps in better planning and error estimation.
Statistical Interpretation
Statistical interpretation is crucial for understanding the significance of your data analysis. When we talk about a 90% confidence interval, it reflects our level of certainty about where the true mean airborne time lies within a calculated range.

Here's how it works:
  • We use our previously computed mean and the standard deviation.
  • Select a standard score (z-score) that corresponds to our confidence level, which is 1.645 for 90%.
  • The confidence interval is then calculated as: \[ \text{mean} \pm \left( \text{z-score} \times \frac{\text{standard deviation}}{\sqrt{n}} \right) \]
  • This gives a range around the mean.
The interpretation is that if we repeatedly took samples and built intervals in this manner, we expect the true mean to fall within this range in 90% of them. This provides a sense of reliability to the calculated mean, reassuring planners or stakeholders about their decisions based on these statistics.

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Most popular questions from this chapter

The article "The Assodation Between Television Viewing and Irregular Sleep Schedules Among Children Less Than 3 Years of Age" (Pediatrics 120051 : \(851-856\) ) reported the accompanying \(95 \%\) confidence intervals for average TV viewing time (in hours per day) for three different age groups. Confidence Interval Age Group \(=\) \(95 \%\) Less than (0.8,1.0) 12 12 to 23 2 months \begin{tabular}{l} (0.8,1.0) \\ (1.4,1.8) \\ (2.1,2.5) \\ \hline \end{tabular} \begin{tabular}{r} 12 to 2 \\ 24 to 3 \\ \hline \end{tabular} n 12 montns 5 months months 35 a. Suppose that the sample sizes for each of the three age group samples were equal. Based on the given confidence intervals, which of the age group samples had the greatest variability in TV viewing time? Explain your choice. b. Now suppose that the sample standard deviations for the three age group samples were equal, but that the three sample sizes might have been different. Which of the three age-group samples had the largest sample size? Explain your choice. c. The interval (.768,1.032) is either a \(90 \%\) confidence interval or a \(99 \%\) confidence interval for the mean TV viewing time computed using the sample data for children less than 12 months old. Is the confidence level for this interval \(90 \%\) or \(99 \%\) ? Explain your choice.

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In June 2009 , Harris Interactive conducted its Great Schools Survey. In this survey, the sample consisted of 1086 adults who were parents of school-aged children. The sample was selected in a way that makes it reasonable to regard it as representative of the population of parents of school-aged children. One question on the survey asked respondents how much time (in hours) per month they spent volunteering at their children's school during the previous school year. The following summary statistics for time volunteered per month were given: \(n=1086 \quad \bar{x}=5.6 \quad\) median \(=1\) a. What does the fact that the mean is so much larger than the median tell you about the distribution of time spent volunteering per month? b. Based on your answer to Part (a), explain why it is not reasonable to assume that the population distribution of time spent volunteering is approximately normal. c. Explain why it is appropriate to use the \(t\) confidence interval to estimate the mean time spent volunteering for the population of parents of school-aged children even though the population distribution is not approximately normal. d. Suppose that the sample standard deviation was \(s=\) 5.2. Compute and interpret a \(98 \%\) confidence interval for \(\mu,\) the mean time spent volunteering for the population of parents of school-aged children.

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