/*! 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 38 Forest rangers wanted to better ... [FREE SOLUTION] | 91Ó°ÊÓ

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Forest rangers wanted to better understand the rate of growth for younger trees in the park. They took measurements of a random sample of 50 young trees in 2009 and again measured those same trees in 2019. The data below summarize their measurements, where the heights are in feet: $$\begin{array}{cccc} \hline & 2009 & 2019 & \text { Differences } \\ \hline \bar{x} & 12.0 & 24.5 & 12.5 \\ s & 3.5 & 9.5 & 7.2 \\ n & 50 & 50 & 50 \\ \hline \end{array}$$ Construct a \(99 \%\) confidence interval for the average growth of (what had been) younger trees in the park over \(2009-2019\).

Short Answer

Expert verified
The 99% confidence interval for average growth is [9.77, 15.23] feet.

Step by step solution

01

Identify the Information Given

We have a sample of 50 young trees where the mean height in 2009 is \(\bar{x}_{2009} = 12.0\) feet, in 2019 is \(\bar{x}_{2019} = 24.5\) feet, with mean difference \(\bar{x}_{\text{diff}} = 12.5\) feet. The standard deviation of the differences is \(s_{\text{diff}} = 7.2\) feet. The sample size is \(n = 50\).
02

Determine the Confidence Level and degrees of freedom

We want a 99% confidence interval, which means our significance level \(\alpha = 1 - 0.99 = 0.01\). The degrees of freedom (df) for this problem is \(n - 1 = 49\).
03

Find the Critical Value

Using a t-distribution table or calculator for df = 49 and a two-tailed test at \(\alpha = 0.01\), the critical value \(t^*\) is approximately 2.68.
04

Calculate the Margin of Error

The margin of error (ME) is calculated using the formula: \[ ME = t^* \times \frac{s_{\text{diff}}}{\sqrt{n}}. \] Substituting the given values, we have \( ME = 2.68 \times \frac{7.2}{\sqrt{50}} \approx 2.73 \).
05

Construct the Confidence Interval

The 99% confidence interval for the mean difference in tree heights is given by: \[ \bar{x}_{\text{diff}} \pm ME = 12.5 \pm 2.73. \] This results in the interval \([9.77, 15.23]\).

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Key Concepts

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

t-distribution
When working with small sample sizes or when the population standard deviation is unknown, the t-distribution is used instead of the normal distribution. This distribution accounts for the smaller sample size by having thicker tails, which means there's a higher probability of values far from the mean.
For this exercise, the sample size is 50, which is fairly large but often the t-distribution is still used in practice for sample sizes under 30. However, since the exact population variance remains unknown, it is still appropriate for larger samples.
The t-distribution is applicable when dealing with sample means and becomes more similar to the normal distribution as the sample size increases. It is defined by its degrees of freedom, which are directly related to the sample size.
margin of error
The margin of error (ME) quantifies the amount of random sampling error in a survey's results. It tells us how precise our estimate of the population parameter is.
For a confidence interval, the margin of error is added to and subtracted from the sample mean to find the upper and lower bounds of the interval.
In this problem, the formula to calculate the margin of error is:
  • ME = \( t^* \times \frac{s_{\text{diff}}}{\sqrt{n}} \)
Here, \( t^* \) is the critical value from the t-distribution for the desired confidence level, \( s_{\text{diff}} \) is the standard deviation of the sample differences, and \( n \) is the sample size. This calculation provides the ME value, which is then used to construct the confidence interval around the sample mean.
degrees of freedom
Degrees of freedom (df) in statistics are the number of values in a calculation that are free to vary. They are crucial for deriving the critical value of a statistic, such as \( t^* \) in the t-distribution.
For a sample, df is typically the sample size minus one:
  • df = n - 1
In our tree example, calculating degrees of freedom involves the sample size, which is 50. Thus, the df = 49.
The degrees of freedom impact the shape of the t-distribution, especially how spread out it is. Fewer degrees of freedom result in a curve that is more spread out, which reflects greater uncertainty in estimating the population parameter.
sample mean
A sample mean (\( \bar{x} \)) is the average of a set of observations from a sample, representing the center of the dataset. It serves as an estimate for the population mean when the population mean is unknown.
In this exercise, the sample mean difference, denoted as \( \bar{x}_{\text{diff}} \), is 12.5 feet. This value represents the average growth between 2009 and 2019 for the selected trees.
The sample mean is fundamental when calculating confidence intervals, such as the one estimated here, as it forms the basis around which the interval is constructed. Its accuracy is dependent on having a representative sample.
standard deviation
Standard deviation (\( s \)) is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range.
In our context, the standard deviation of the differences, \( s_{\text{diff}} \) = 7.2 feet, expresses the variability in height differences of the trees from 2009 to 2019.
Knowing the standard deviation is vital for calculating the margin of error and the confidence interval as it reflects how much the sample mean could vary from the true population mean. This measure helps gauge how uncertain or certain the estimate is.

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Most popular questions from this chapter

Determine if the following statements are true or false, and explain your reasoning for statements you identify as false. (a) When comparing means of two samples where \(n_{1}=20\) and \(n_{2}=40,\) we can use the normal model for the difference in means since \(n_{2} \geq 30\). (b) As the degrees of freedom increases, the \(t\) -distribution approaches normality. (c) We use a pooled standard error for calculating the standard error of the difference between means when sample sizes of groups are equal to each other.

The standard deviation of SAT scores for students at a particular Ivy League college is 250 points. Two statistics students, Raina and Luke, want to estimate the average SAT score of students at this college as part of a class project. They want their margin of error to be no more than 25 points. (a) Raina wants to use a \(90 \%\) confidence interval. How large a sample should she collect? (b) Luke wants to use a \(99 \%\) confidence interval. Without calculating the actual sample size, determine whether his sample should be larger or smaller than Raina's, and explain your reasoning. (c) Calculate the minimum required sample size for Luke.

A \(90 \%\) confidence interval for a population mean is \((65,77) .\) The population distribution is approximately normal and the population standard deviation is unknown. This confidence interval is based on a simple random sample of 25 observations. Calculate the sample mean, the margin of error, and the sample standard deviation.

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For a given confidence level, \(t_{d f}^{\star}\) is larger than \(z^{\star}\). Explain how \(t_{d f}^{*}\) being slightly larger than \(z^{*}\) affects the width of the confidence interval.

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