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Chapter 8: Problem 1

A random sample of 110 lightning flashes in a region resulted in a sample average radar echo duration of 81 s and a sample standard deviation of \(.34\) s ("Lightning Strikes to an Airplane in a Thunderstorm," J. Aircraft, 1984: 607-611). Calculate a \(99 \%\) (two-sided) confidence interval for the true average echo duration \(\mu\), and interpret the resulting interval.

Short Answer

Expert verified
The 99% confidence interval for the true average echo duration is (80.9165, 81.0835) seconds.

Step by step solution

01

Identify the Known Variables

We are given a sample size \( n = 110 \), a sample mean \( \bar{x} = 81 \) s, and a sample standard deviation \( s = 0.34 \) s. We need to calculate a 99% confidence interval for the population mean \( \mu \).
02

Determine the Critical Value

Since the sample size is large (\( n \geq 30 \)), we can use the z-distribution. For a 99% confidence interval, the critical value \( z^* \) is 2.576, as obtained from a standard normal distribution table.
03

Calculate the Standard Error

The standard error (SE) of the sample mean is calculated using the formula: \[ SE = \frac{s}{\sqrt{n}} = \frac{0.34}{\sqrt{110}} \approx 0.0324.\]
04

Calculate the Margin of Error

The margin of error (ME) can be calculated using the formula: \[ ME = z^* \times SE = 2.576 \times 0.0324 \approx 0.0835.\]
05

Compute the Confidence Interval

The confidence interval is given by \[ \bar{x} \pm ME = 81 \pm 0.0835.\]Thus, the confidence interval is approximately \( (80.9165, 81.0835) \).
06

Interpret the Confidence Interval

This means we are 99% confident that the true average radar echo duration for all lightning flashes in this region lies between approximately 80.9165 seconds and 81.0835 seconds.

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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, often denoted as \( \bar{x} \), represents the average of a set of data points collected from a sample. In our exercise, the sample mean is calculated as 81 seconds, which means that the average radar echo duration from the sample of 110 lightning flashes is 81 seconds. It is a crucial step because it helps us make inferences about the population mean, \( \mu \), based on the sample data.
To find the sample mean, you add up all the observed values (in this case, radar echo durations) and divide by the total number of observations, here 110. It's your starting point in constructing a confidence interval.
The more representative your sample is of the population, the more reliable your sample mean will be.
Standard Error
Standard error (SE) quantifies the amount of variation or dispersion of the sample mean from the true population mean. It gives insight into how accurately your sample mean predicts the population mean.
The formula to calculate the standard error is \( SE = \frac{s}{\sqrt{n}} \), where \( s \) is the sample standard deviation and \( n \) is the sample size. For our lightning flash data, the standard error is approximately 0.0324 seconds.
This small SE suggests that the sample mean is a reliable estimator of the population mean.
  • Larger sample sizes tend to decrease the SE, leading to more precise estimates of the population mean.
  • Smaller SE indicates greater precision in estimating the population mean from the sample mean.
Margin of Error
The margin of error (ME) is an expression of the range within which the true population mean is expected to fall in relation to your sample data, with a certain level of confidence (e.g., 99%). It measures how far off your sample mean might be from the true population mean, allowing us to create a confidence interval.
The formula is \( ME = z^* \times SE \), where \( z^* \) is the critical value corresponding to the desired confidence level. In this exercise, the margin of error is approximately 0.0835 seconds. This means that the true average radar echo duration is expected to differ from the sample mean by no more than this amount.
The margin of error helps communicate the potential error in the sample mean estimate, due to sampling variability.
Critical Value
The critical value, typically found using statistical tables, corresponds to the desired level of confidence (like 99%) and helps determine the range for the confidence interval. In a normally distributed population, this value specifies the number of standard deviations you should go above and below the sample mean to capture the middle portion of the distribution.
Using the z-distribution because our sample size exceeds 30, we got a critical value of 2.576 for a 99% confidence interval. This value signifies the distance from the sample mean in which the true mean is likely to fall.
  • You choose the critical value based on your confidence level; a higher confidence level (e.g., 99%) requires a larger critical value.
  • The critical value ensures that the margin of error is appropriately adjusted to encapsulate the desired range on either side of the sample mean.
This value, represented using the z-score in this exercise, plays a pivotal role in determining the span of your confidence interval.

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Most popular questions from this chapter

Aphid infestation of fruit trees can be controlled either by spraying with pesticide or by inundation with ladybugs. In a particular area, four different groves of fruit trees are selected for experimentation. The first three groves are sprayed with pesticides 1,2 , and 3 , respectively, and the fourth is treated with ladybugs, with the following results on yield: \begin{tabular}{llll} \hline Treatment & \multicolumn{1}{l}{\(n_{i}\) (number of } & \({\bar{x}_{i}}\) (bushels/ \\ & trees) & tree) & \\ \hline 1 & 100 & \(10.5\) & \(1.5\) \\ 2 & 90 & \(10.0\) & \(1.3\) \\ 3 & 100 & \(10.1\) & \(1.8\) \\ 4 & 120 & \(10.7\) & \(1.6\) \\ \hline \end{tabular} Let \(\mu_{i}=\) the true average yield (bushels/tree) after receiving the \(i\) th treatment. Then $$ \theta=\frac{1}{3}\left(\mu_{1}+\mu_{2}+\mu_{3}\right)-\mu_{4} $$ measures the difference in true average yields between treatment with pesticides and treatment with ladybugs. When \(n_{1}, n_{2}, n_{3}\), and \(n_{4}\) are all large, the estimator \(\hat{\theta}\) obtained by replacing each \(\mu_{i}\) by \(\bar{X}_{i}\) is approximately normal. Use this to derive a large-sample \(100(1-\alpha) \%\) CI for \(\theta\), and compute the \(95 \%\) interval for the given data.

Nine Australian soldiers were subjected to extreme conditions, which involved a 100 -min walk with a 25 -lb pack when the temperature was \(40^{\circ} \mathrm{C}\left(104^{\circ} \mathrm{F}\right)\). One of them overheated (above \(39^{\circ} \mathrm{C}\) ) and was removed from the study. Here are the rectal Celsius temperatures of the other eight at the end of the walk ("Neural Network Training on Human Body Core Temperature Data," Combatant Protection and Nutrition Branch, Aeronautical and Maritime Research Laboratory of Australia, DSTO TN-0241, 1999): \(\begin{array}{llllllll}38.4 & 38.7 & 39.0 & 38.5 & 38.5 & 39.0 & 38.5 & 38.6\end{array}\) We would like to get a \(95 \%\) confidence interval for the population mean. a. Compute the \(t\)-based confidence interval of Section 8.3. b. Check for the validity of part (a). c. Generate a bootstrap sample of 999 means. d. Use the standard deviation for part (c) to get a \(95 \%\) confidence interval for the population mean. e. Investigate the distribution of the bootstrap means to see if part (d) is valid. f. Use part (c) to form the \(95 \%\) confidence interval using the percentile method. g. Compare the intervals and explain your preference. h. Based on your knowledge of normal body temperature, would you say that body temperature can be influenced by environment?

Here are the lengths (in minutes) of the 63 nineinning games from the first week of the 2001 major league baseball season: \(\begin{array}{llllllllll}194 & 160 & 176 & 203 & 187 & 163 & 162 & 183 & 152 & 177 \\ 177 & 151 & 173 & 188 & 179 & 194 & 149 & 165 & 186 & 187 \\ 187 & 177 & 187 & 186 & 187 & 173 & 136 & 150 & 173 & 173 \\ 136 & 153 & 152 & 149 & 152 & 180 & 186 & 166 & 174 & 176 \\ 198 & 193 & 218 & 173 & 144 & 148 & 174 & 163 & 184 & 155 \\ 151 & 172 & 216 & 149 & 207 & 212 & 216 & 166 & 190 & 165 \\ 176 & 158 & 198 & & & & & & & \end{array}\) Assume that this is a random sample of nineinning games (the mean differs by \(12 \mathrm{~s}\) from the mean for the whole season). a. Give a \(95 \%\) confidence interval for the population mean. b. Give a \(95 \%\) prediction interval for the length of the next nine-inning game. On the first day of the next week, Boston beat Tampa Bay 3-0 in a nine- inning game of \(152 \mathrm{~min}\). Is this within the prediction interval? c. Compare the two intervals and explain why one is much wider than the other. d. Explore the issue of nomality for the data and explain how this is relevant to parts (a) and (b).

If you go to a major league baseball game, how long do you expect the game to be? From the 2,429 games played in 2001 , here is a random sample of 25 times in minutes: \(\begin{array}{llllllllll}352 & 150 & 164 & 167 & 225 & 159 & 142 & 182 & 229 & 163 \\ 188 & 197 & 189 & 235 & 161 & 195 & 177 & 166 & 195 & 160 \\ 154 & 130 & 189 & 188 & 225 & & & & & \end{array}\) This is one of those rare instances in which we can do a confidence interval and compare with the true population mean. The mean of all 2,429 lengths is \(178.29\) (almost \(3 \mathrm{~h}\) ). a. Compute the \(t\)-based confidence interval of Section \(8.3 .\) b. Use a normal plot to see if part (a) is valid. c. Generate a bootstrap sample of 999 means. d. Use the standard deviation for part (c) to get a \(95 \%\) confidence interval for the population mean. e. Investigate the distribution of the bootstrap means to see if the CI of part (d) is valid. f. Use part (c) to form the \(95 \%\) confidence interval using the percentile method. g. Say which interval should be used and explain why. Does your interval include the true value, \(178.29\) ?

A manufacturer of college textbooks is interested in estimating the strength of the bindings produced by a particular binding machine. Strength can be measured by recording the force required to pull the pages from the binding. If this force is measured in pounds, how many books should be tested to estimate the average force required to break the binding to within . \(1 \mathrm{lb}\) with \(95 \%\) confidence? Assume that \(\sigma\) is known to be .8.
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