/*! 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 12 A random sample of 110 lightning... [FREE SOLUTION] | 91Ó°ÊÓ

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A random sample of 110 lightning flashes in a certain region resulted in a sample average radar echo duration of \(.81 \mathrm{sec}\) and a sample standard deviation of \(34 \mathrm{sec}\) ("Lightning Strikes to an Airplane in a Thunderstorm," \(J .\) of 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 \([-7.68, 9.30]\) sec.

Step by step solution

01

Identify the Given Information

We are given the sample size \( n = 110 \), the sample mean \( \bar{x} = 0.81 \) sec, and the sample standard deviation \( s = 34 \) sec. We need to calculate a 99% confidence interval for the true average echo duration \( \mu \).
02

Identify the Confidence Level and Degrees of Freedom

The confidence level is 99%, which corresponds to \( \alpha = 0.01 \). Since the sample size is 110, the degrees of freedom for a t-distribution is \( n - 1 = 109 \).
03

Find the Critical Value

Using a t-distribution table, find the critical value \( t^* \) for \( \alpha/2 = 0.005 \) and 109 degrees of freedom. The approximate value is \( t^* = 2.62 \).
04

Calculate the Standard Error

Compute the standard error of the mean, given by \( SE = \frac{s}{\sqrt{n}} \). Substituting in the given values: \[ SE = \frac{34}{\sqrt{110}} \approx 3.24 \] sec.
05

Calculate the Margin of Error

The margin of error (ME) is found by multiplying the critical value by the standard error: \[ ME = t^* \times SE = 2.62 \times 3.24 \approx 8.49 \] sec.
06

Determine the Confidence Interval

The confidence interval is given by \( \bar{x} \pm ME \). Substituting the values: \[ 0.81 \pm 8.49 \] results in the interval: \([-7.68, 9.30]\) sec.
07

Interpret the Confidence Interval

The 99% confidence interval means that we are 99% confident that the true average echo duration \( \mu \) is between -7.68 sec and 9.30 sec. However, negative echo duration does not make sense physically, indicating the model may need revision.

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Key Concepts

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

t-distribution
The t-distribution is a probability distribution used to estimate population parameters when the sample size is small and the standard deviation is unknown. It is especially useful in situations where data follows a normal distribution but the sample size is limited. In the context of the problem, we must rely on the t-distribution because although we have a decent sample size of 110, we're estimating the population mean with an unknown standard deviation. The t-distribution resembles the normal distribution but has thicker tails, meaning it better accounts for the variability inherent when sampling. These thicker tails result in wider confidence intervals compared to those calculated using the normal distribution, which is why it is particularly helpful in hypothesis testing and constructing confidence intervals, as shown in the solution. When comparing the normal and t-distributions:
  • The smaller the sample size, the more the t-distribution differs from the normal distribution.
  • As the sample size increases, the t-distribution approaches the normal distribution.
standard error
The standard error (SE) is a measure of the amount by which the sample mean is expected to vary from the true population mean. It's calculated by dividing the sample standard deviation by the square root of the sample size. In this exercise, SE helps us understand how much the sample mean of 0.81 sec could differ from the actual average radar echo duration. The formula to calculate SE is straightforward: \[ SE = \frac{s}{\sqrt{n}} \] where \( s \) is the sample standard deviation (34 sec) and \( n \) is the sample size (110). Calculating this gives us an SE of approximately 3.24 sec. This represents the inference variability—how far off the sample mean might be from the population mean. It helps us in creating the confidence interval, as the smaller the SE, the less variable our sample is, improving the precision of the confidence interval.
margin of error
The margin of error (ME) indicates the range within which we expect the population parameter to lie, with a given level of confidence. It accounts for potential errors due to sample variability and uses the standard error and the critical t-value for its calculation. The formula used is: \[ ME = t^* \times SE \] where \( t^* \) is the critical value based on the t-distribution, and SE is the standard error. For our example, with a critical value of 2.62 and an SE of 3.24 sec, the ME computes to about 8.49 sec. This means there is a 99% confidence that the true mean radar echo duration is within approximately 8.49 seconds plus or minus the measured mean of 0.81 seconds.The calculation of the margin of error is crucial because:
  • It gives an upper and lower bound for estimates, informing how much error might be associated with a survey or study result.
  • A larger margin indicates lower confidence in the estimate precision.
degrees of freedom
Degrees of freedom (df) are a statistical concept that describes the number of independent observations in a data set that are available to estimate another parameter. It is essentially the amount of independent information we have for estimating statistical parameters. In t-distributions, df influences the shape of the distribution. More degrees of freedom result in a distribution that more closely approximates a normal curve. For a sample size \( n \), the degrees of freedom for estimating the mean is \( n-1 \). In our problem with a sample size of 110, the degrees of freedom is 109. This affects our t-distribution and the critical value we use for calculating the confidence interval. Understanding degrees of freedom is vital because:
  • It helps determine the appropriate distribution to use for hypothesis tests.
  • Influences the critical values, and subsequently, the width of confidence intervals.
With higher degrees of freedom, confidence intervals generally become narrower, reflecting increased precision of the estimate.

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Most popular questions from this chapter

The alternating current (AC) breakdown voltage of an insulating liquid indicates its dielectric strength. The article "Testing Practices for the AC Breakdown Voltage Testing of Insulation Liquids" (IEEE Electrical Insulation Magazine, 1995: 21-26) gave the accompanying sample observations on breakdown voltage (kV) of a particular circuit under certain conditions. \(\begin{array}{llllllllllllllll}62 & 50 & 53 & 57 & 41 & 53 & 55 & 61 & 59 & 64 & 50 & 53 & 64 & 62 & 50 & 68 \\ 54 & 55 & 57 & 50 & 55 & 50 & 56 & 55 & 46 & 55 & 53 & 54 & 52 & 47 & 47 & 55 \\ 57 & 48 & 63 & 57 & 57 & 55 & 53 & 59 & 53 & 52 & 50 & 55 & 60 & 50 & 56 & 58\end{array}\) a. Construct a boxplot of the data and comment on interesting features. b. Calculate and interpret a \(95 \%\) CI for true average breakdown voltage \(\mu\). Does it appear that \(\mu\) has been precisely estimated? Explain. c. Suppose the investigator believes that virtually all values of breakdown voltage are between 40 and 70 . What sample size would be appropriate for the \(95 \% \mathrm{Cl}\) to have a width of \(2 \mathrm{kV}\) (so that \(\mu\) is estimated to within \(1 \mathrm{kV}\) with \(95 \%\) confidence)?

Determine the \(t\) critical value for a two-sided confidence interval in each of the following situations: a. Confidence level \(=95 \%\), df \(=10\) b. Confidence level \(=95 \%\), df \(=15\) c. Confidence level \(=99 \%\), df \(=15\) d. Confidence level \(=99 \%, n=5\) e. Confidence level \(=98 \%\), df \(=24\) f. Confidence level \(=99 \%, n=38\)

The Pew Forum on Religion and Public Life reported on Dec. 9,2009 , that in a survey of 2003 American adults, \(25 \%\) said they believed in astrology. a. Calculate and interpret a confidence interval at the \(99 \%\) confidence level for the proportion of all adult Americans who believe in astrology. b. What sample size would be required for the width of a \(99 \%\) CI to be at most \(.05\) irrespective of the value of \(\hat{p}\) ?

A sample of 14 joint specimens of a particular type gave a sample mean proportional limit stress of \(8.48 \mathrm{MPa}\) and a sample standard deviation of .79 MPa (" Characterization of Bearing Strength Factors in Pegged Timber Connections," J. of Structural Engr, 1997:326-332). a. Calculate and interpret a \(95 \%\) lower confidence bound for the true average proportional limit stress of all such joints. What, if any, assumptions did you make about the distribution of proportional limit stress? b. Calculate and interpret a \(95 \%\) lower prediction bound for the proportional limit stress of a single joint of this type.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a continuous probability distribution having median \(\tilde{\mu}\) (so that \(\left.P\left(X_{l} \leq \tilde{\mu}\right)=P\left(X_{l} \geq \tilde{\mu}\right)=.5\right)\). a. Show that $$ P\left(\min \left(X_{i}\right)<\tilde{\mu}<\max \left(X_{j}\right)\right)=1-\left(\frac{1}{2}\right)^{n-1} $$ so that \(\left(\min \left(x_{i}\right), \max \left(x_{i}\right)\right)\) is a \(100(1-\alpha) \%\) confidence interval for \(\tilde{\mu}\) with \(\alpha=\left(\frac{1}{2}\right)^{n-1}\). [Hint: The complement of the event \(\left\\{\min \left(X_{j}\right)<\widetilde{\mu}<\max \left(X_{j}\right)\right\\}\) is \(\left\\{\max \left(X_{j}\right) \leq\right.\) \(\tilde{\mu}\\} \cup\left\\{\min \left(X_{i}\right) \geq \tilde{\mu}\right\\}\). But \(\max \left(X_{i}\right) \leq \tilde{\mu}\) iff \(X_{i} \leq \tilde{\mu}\) for all \(i .]\) b. For each of six normal male infants, the amount of the amino acid alanine \((\mathrm{mg} / 100 \mathrm{~mL}\) ) was determined while the infants were on an isoleucine-free diet, resulting in the following data: \(\begin{array}{llllll}2.84 & 3.54 & 2.80 & 1.44 & 2.94 & 2.70\end{array}\) Compute a \(97 \%\) CI for the true median amount of alanine for infants on such a diet ("The Essential Amino Acid Requirements of Infants," Amer. \(J\). of Nutrition, 1964: 322-330). c. Let \(x_{(2)}\) denote the second smallest of the \(x_{i}\) 's and \(x_{(1 n-1)}\) denote the second largest of the \(x_{i}\) 's. What is the confidence level of the interval \(\left(x_{(2)}, x_{(n-1)}\right)\) for \(\tilde{\mu}\) ?

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