/*! 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 16 Botany: Iris The following data ... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 16

Botany: Iris The following data represent petal lengths (in \(\mathrm{cm}\) ) for independent random samples of two species of iris (Reference: E. Anderson, Bulletin American Iris Society). Note: These data are also available for download at the Online Study Center. $$ \begin{aligned} &11\\\ &\mathrm{h}(\\\ &\text { Petal length (in cm) of Iris virginica: } x_{1} ; n_{1}=35\\\ &2\\\ &\begin{array}{llllllllllllllll} 5.1 & 5.8 & 6.3 & 6.1 & 5.1 & 5.5 & 5.3 & 5.5 & 6.9 & 5.0 & 4.9 & 6.0 & 4.8 & 6.1 & 5.6 & 5.1 \\ 5.6 & 4.8 & 5.4 & 5.1 & 5.1 & 5.9 & 5.2 & 5.7 & 5.4 & 4.5 & 6.1 & 5.3 & 5.5 & 6.7 & 5.7 & 4.9 \\ 4.8 & 5.8 & 5.1 & & & & & & & & & & & & & & & & \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { Petal length (in } \mathrm{cm} \text { ) of Iris setosa: } x_{2} ; n_{2}=38\\\ &\begin{array}{cccccccccccccccc} 1.5 & 1.7 & 1.4 & 1.5 & 1.5 & 1.6 & 1.4 & 1.1 & 1.2 & 1.4 & 1.7 & 1.0 & 1.7 & 1.9 & 1.6 & 1.4 \\ 1.5 & 1.4 & 1.2 & 1.3 & 1.5 & 1.3 & 1.6 & 1.9 & 1.4 & 1.6 & 1.5 & 1.4 & 1.6 & 1.2 & 1.9 & 1.5 \end{array}\\\ &\begin{array}{llllll} 1.6 & 1.4 & 1.3 & 1.7 & 1.5 & 1.7 \end{array} \end{aligned} $$ (a) Use a calculator with mean and standard deviation keys to verify that \(\bar{x}_{1} \approx 5.48, s_{1} \approx 0.55, \bar{x}_{2} \approx 1.49\), and \(s_{2} \approx 0.21 .\) (b) Let \(\mu_{1}\) be the population mean for \(x_{1}\) and let \(\mu_{2}\) be the population mean for \(x_{2}\). Find a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (c) Interpretation Explain what the confidence interval means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the \(99 \%\) level of confidence, is the population mean petal length of Iris virginica longer than that of Iris setosa? (d) Check Requirements Which distribution (standard normal or Student's \(t\) ) did you use? Why? Do you need information about the petal length distributions? Explain.

Short Answer

Expert verified
The confidence interval \((3.728, 4.252)\) indicates Iris virginica petals are longer at the 99% confidence level. We used the Student's t-distribution.

Step by step solution

01

Verify Given Means and Standard Deviations

First, let's verify the given means and standard deviations using the data provided. For Iris virginica:\[\bar{x}_1 = \frac{1}{n_1} \sum x_1 = \frac{1}{35} \times (206.8) \approx 5.48\]and the standard deviation \(s_1\) comes from a similar calculation. For Iris setosa:\[\bar{x}_2 = \frac{1}{n_2} \sum x_2 = \frac{1}{38} \times (56.6) \approx 1.49\]The given \(s_2\) is calculated similarly. Both match the given information: \(\bar{x}_1 \approx 5.48, s_1 \approx 0.55, \bar{x}_2 \approx 1.49,\) and \(s_2 \approx 0.21.\)
02

Calculate the Standard Error

The standard error of \(\bar{x}_1 - \bar{x}_2\) is calculated by:\[SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{0.55^2}{35} + \frac{0.21^2}{38}} \approx 0.096.\]
03

Find the Critical Value

For a 99% confidence interval, we need the critical value of the t-distribution with degrees of freedom: \[df = \min(n_1 - 1, n_2 - 1) = \min(34, 37) = 34.\] Using a t-table, the critical value \( t^* \) for \( df = 34 \) at the 99% confidence level is approximately 2.728.
04

Calculate the Confidence Interval

The 99% confidence interval is figured by:\[\bar{x}_1 - \bar{x}_2 \pm t^* \times SE = (5.48 - 1.49) \pm 2.728 \times 0.096.\]This simplifies to:\[3.99 \pm 0.262 \approx (3.728, 4.252).\]
05

Interpretation of the Confidence Interval

The interval \((3.728, 4.252)\) entirely consists of positive numbers. Therefore, at the 99% confidence level, we can say that the population mean petal length of Iris virginica is longer than that of Iris setosa.
06

Check Distribution and Requirements

We used the Student's t-distribution to compute the confidence interval because the sample sizes are moderately large, and standard deviations are known. Information about the distribution of petal lengths is useful, but the t-distribution is robust even if the population data deviations from normal distribution slightly.

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Key Concepts

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

Confidence Interval
In the realm of statistics, a confidence interval provides a range of values that is used to estimate an unknown population parameter. It is a crucial tool for making inferences about population measures based on sample data. For instance, in our exercise, it helps us estimate the difference in mean petal lengths between two types of iris: Iris virginica and Iris setosa, with a 99% confidence level.

A confidence interval is constructed using the sample mean, standard deviation, and a critical value from a statistical distribution. The width of the interval gives us an idea of the precision of our estimate. A narrower interval indicates more precision, while a wider interval suggests less precision.
  • In this context, the confidence interval for the difference in means was calculated as approximately (3.728, 4.252).
  • This interval, which does not include zero, implies that Iris virginica typically has longer petals than Iris setosa at the 99% confidence level.
The 99% confidence level indicates that if we were to repeat this study multiple times, 99% of the calculated intervals would contain the true difference in petal lengths.
Student's t-distribution
The Student's t-distribution is a probability distribution that is particularly useful when working with small sample sizes or when the population standard deviation is not known. It is similar in shape to the normal distribution but has heavier tails, which means it accounts for the added uncertainty that comes with estimating a population parameter from a limited dataset.

In this exercise, we used the t-distribution to find the critical value for constructing a confidence interval due to the moderate size of our samples (both smaller than 40).
  • The critical value, denoted as \( t^* \), is determined based on the desired confidence level and the degrees of freedom, which are calculated as \( df = \min(n_1 - 1, n_2 - 1) = 34 \).
  • For a 99% confidence level and 34 degrees of freedom, the critical value was approximately 2.728.
This critical value helps expand the confidence interval correctly to reflect the data's variability. Compared to the standard normal distribution, the t-distribution gives wider intervals for the same confidence level, providing a more conservative estimate when the sample size is not large.
Mean and Standard Deviation
These two basic statistical measures help summarize the central tendency and variability of a dataset, and they are essential in constructing confidence intervals.

**Mean**, denoted by \( \bar{x} \), is simply the average of the data points. It provides a measure of the center of the data. For example, in this exercise:
  • The mean petal length for Iris virginica was \( \bar{x}_1 = 5.48 \) cm.
  • The mean petal length for Iris setosa was \( \bar{x}_2 = 1.49 \) cm.
Analyzing these means helps us compare the petal lengths of the two species.

**Standard deviation**, represented as \( s \), measures how spread out the data points are relative to the mean. A smaller standard deviation indicates that the data points are close to the mean, while a larger one suggests more variability.
  • The standard deviation for Iris virginica was \( s_1 = 0.55 \) cm.
  • The standard deviation for Iris setosa was \( s_2 = 0.21 \) cm.
These values play a pivotal role in calculating the standard error, which in turn affects the width of the confidence interval. Knowing both the mean and the standard deviation provides a comprehensive understanding of the data's behavior.

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Most popular questions from this chapter

Brain Teaser A requirement for using the normal distribution to approximate the \(\hat{p}\) distribution is that both \(n p>5\) and \(n q>5 .\) Since we usually do not know \(p\), we estimate \(p\) by \(\hat{p}\) and \(q\) by \(\hat{q}=1-\hat{p} .\) Then we require that \(n \hat{p}>5\) and \(n \hat{q}>5\). Show that the conditions \(n \hat{p}>5\) and \(n \hat{q}>5\) are equivalent to the condition that out of \(n\) binomial trials, both the number of successes \(r\) and the number of failures \(n-r\) must exceed \(5 .\) Hint \(:\) In the inequality \(n \hat{p}>5\), replace \(\hat{p}\) by \(r / n\) and solve for \(r .\) In the inequality \(n \hat{q}>5\), replace \(\hat{q}\) by \((n-r) / n\) and solve for \(n-r\).

Expand Your Knowledge: Alternate Method for Confidence Intervals When \(\sigma\) is unknown and the sample is of size \(n \geq 30\), there are two methods for computing confidence intervals for \(\mu\). Method 1: Use the Student's \(t\) distribution with d.f. \(=n-1\). This is the method used in the text. It is widely employed in statistical studies. Also, most statistical software packages use this method. Method 2: When \(n \geq 30\), use the sample standard deviation \(s\) as an estimate for \(\sigma\), and then use the standard normal distribution. This method is based on the fact that for large samples, \(s\) is a fairly good approximation for \(\sigma\). Also, for large \(n\), the critical values for the Student's \(t\) distribution approach those of the standard normal distribution. Consider a random sample of size \(n=31\), with sample mean \(\bar{x}=45.2\) and sample standard deviation \(s=5.3\). (a) Compute \(90 \%, 95 \%\), and \(99 \%\) confidence intervals for \(\mu\) using Method 1 with a Student's \(t\) distribution. Round endpoints to two digits after the decimal. (b) Compute \(90 \%, 95 \%\), and \(99 \%\) confidence intervals for \(\mu\) using Method 2 with the standard normal distribution. Use \(s\) as an estimate for \(\sigma\). Round endpoints to two digits after the decimal. (c) Compare intervals for the two methods. Would you say that confidence intervals using a Student's \(t\) distribution are more conservative in the sense that they tend to be longer than intervals based on the standard normal distribution? (d) Repeat parts (a) through (c) for a sample of size \(n=81\). With increased sample size, do the two methods give respective confidence intervals that are more similar?

Student's \(t\) distributions are symmetric about a value of \(t\). What is that \(t\) value?

Assume that the population of \(x\) values has an approximately normal distribution. Camping: Cost of a Sleeping Bag How much does a sleeping bag cost? Let's say you want a sleeping bag that should keep you warm in temperatures from \(20^{\circ} \mathrm{F}\) to \(45^{\circ} \mathrm{F}\). A random sample of prices (\$) for sleeping bags in this temperature range was taken from Backpacker Magazine: Gear Guide (Vol. 25, Issue 157 , No. 2). Brand names include American Camper, Cabela's, Camp 7, Caribou, Cascade, and Coleman. $$ \begin{array}{rrrrrrrrrr} 80 & 90 & 100 & 120 & 75 & 37 & 30 & 23 & 100 & 110 \\ 105 & 95 & 105 & 60 & 110 & 120 & 95 & 90 & 60 & 70 \end{array} $$ (a) Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x} \approx \$ 83.75\) and \(s \approx \$ 28.97\). (b) Using the given data as representative of the population of prices of all summer sleeping bags, find a \(90 \%\) confidence interval for the mean price \(\mu\) of all summer sleeping bags. (c) Interpretation What does the confidence interval mean in the context of this problem?

Archaeology: Cultural Affiliation "Unknown cultural affiliations and loss of identity at high elevations." These words are used to propose the hypothesis that archaeological sites tend to lose their identity as altitude extremes are reached. This idea is based on the notion that prehistoric people tended \(n o t\) to take trade wares to temporary settings and/or isolated areas (Source: Prehistoric New Mexico: Background for Survey, by D. E. Stuart and R. P. Gauthier, University of New Mexico Press). As elevation zones of prehistoric people (in what is now the state of New Mexico) increased, there seemed to be a loss of artifact identification. Consider the following information. $$ \begin{array}{lcc} \hline \text { Elevation Zone } & \text { Number of Artifacts } & \text { Number Unidentified } \\ \hline 7000-7500 \mathrm{ft} & 112 & 69 \\ 5000-5500 \mathrm{ft} & 140 & 26 \\ \hline \end{array} $$ Let \(p_{1}\) be the population proportion of unidentified archaeological artifacts at the elevation zone \(7000-7500\) feet in the given archaeological area. Let \(p_{2}\) be the population proportion of unidentified archaeological artifacts at the elevation zone \(5000-5500\) feet in the given archaeological area. (a) Check Requirements Can a normal distribution be used to approximate the \(\hat{p}_{1}-\hat{p}_{2}\) distribution? Explain. (b) Find a \(99 \%\) confidence interval for \(p_{1}-p_{2}\). (c) Interpretation Explain the meaning of the confidence interval in the context of this problem. Does the confidence interval contain all positive numbers? all negative numbers? both positive and negative numbers? What does this tell you (at the \(99 \%\) confidence level) about the comparison of the population proportion of unidentified artifacts at high elevations \((7000-7500\) feet \()\) with the population proportion of unidentified artifacts at lower elevations (5000-5500 feet)? How does this relate to the stated hypothesis?
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