/*! 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 6 Vegetation Wild irises are beaut... [FREE SOLUTION] | 91Ó°ÊÓ

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Vegetation Wild irises are beautiful flowers found throughout the United States, Canada, and northern Europe. This problem concerns the length of the sepal (leaf-like part covering the flower) of different species of wild iris. Data are based on information taken from an article by R. A. Fisher in Annals of Eugenics (Vol. 7, part 2, pp. 179-188). Measurements of sepal length in centimeters from andom samples of Iris setosa (I), Iris versicolor (II), and Iris virginica (III) are as follows:$$\begin{array}{ccc}\mathrm{I} & \mathrm{II} & \mathrm{III} \\\5.4 & 5.5 & 6.3 \\\4.9 & 6.5 & 5.8 \\\5.0 & 6.3 & 4.9 \\\5.4 & 4.9 & 7.2 \\\4.4 & 5.2 & 5.7 \\\5.8 & 6.7 & 6.4 \\\5.7 & 5.5 & \\\& 6.1 &\end{array}$$,Shall we reject or not reject the claim that there are no differences among the population means of sepal length for the different species of iris? Use a \(5 \%\) level of significance.

Short Answer

Expert verified
Perform an ANOVA test and compare the F-statistic with the critical F-value; reject \( H_0 \) if the F-statistic is greater than the critical value, indicating significant differences.

Step by step solution

01

State the null and alternative hypotheses

The null hypothesis \( H_0 \) states that there are no differences among the means of sepal length for the different species of iris: \( \mu_1 = \mu_2 = \mu_3 \). The alternative hypothesis \( H_a \) is that at least one population mean is different.
02

Choose the significance level

We are given a significance level \( \alpha = 0.05 \). This is the probability of rejecting the null hypothesis when it is true.
03

Perform an ANOVA test

Analyze the variance among the three groups (Iris setosa, Iris versicolor, and Iris virginica) using ANOVA. Calculate the F-statistic, which will compare the variance among the sample means to within the samples. The computation involves: - Calculate the group means and overall mean.- Compute the Sum of Squares Between (SSB) and the Sum of Squares Within (SSW).- Determine the degrees of freedom and mean squares for both between (MSB) and within (MSW) groups.- The F-statistic is calculated as \( F = \frac{MSB}{MSW} \).- Use statistical software or tables to determine the critical F-value and p-value.
04

Compare the calculated F-statistic and critical F-value

Look up the critical F-value in an F-distribution table for \( (k-1, n-k) \) degrees of freedom, where \( k = 3 \) (number of groups) and \( n = 15 \) (total number of observations). Compare the calculated F-statistic to the critical value at \( \alpha = 0.05 \).
05

Make a decision to reject or not reject the null hypothesis

If the calculated F-statistic is greater than the critical F-value, reject the null hypothesis. Otherwise, do not reject the null hypothesis.
06

Interpret the results

Based on the comparison, if we reject the null hypothesis, it indicates there is a statistically significant difference among the means of sepal length for the different iris species. If we do not reject the null hypothesis, it suggests there is no significant evidence to say that the population means are different.

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Key Concepts

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

Null Hypothesis
The null hypothesis (\(H_0\)) is a fundamental concept in statistical analysis that acts as a starting point for hypothesis testing. In the context of ANOVA (Analysis of Variance), the null hypothesis asserts that there is no significant difference between the groups being compared. Here, with respect to the exercise on wild iris species, the null hypothesis suggests that the sepal lengths across the different species of iris (Iris setosa, Iris versicolor, and Iris virginica) are equal. Mathematically, it is expressed as follows:\[ \mu_1 = \mu_2 = \mu_3 \]This means there are no differences among the population means of sepal length for these species. The null hypothesis is crucial because unless evidence strongly suggests otherwise, we assume it to be true. A significant result means this assumed status quo is rejected, indicating that at least one group differs from the others.
Alternative Hypothesis
The alternative hypothesis (\(H_a\)) contradicts what is proposed by the null hypothesis. It suggests there is a significant difference between the population groups being studied. For the sepal lengths of the wild iris species, the alternative hypothesis implies that at least one of the species has a different average sepal length compared to the others.In mathematical terms, the alternative hypothesis is expressed as:\[ \mu_1 eq \mu_2 eq \mu_3 \]This means at least one of the population means is different, though it does not specify which one. The aim of hypothesis testing in ANOVA is to determine whether the data provides sufficient evidence to reject the null hypothesis and accept the alternative. If we find the null hypothesis unlikely, the alternative hypothesis becomes a more plausible explanation of our data results.
Significance Level
The significance level (\(\alpha\)) represents the probability of rejecting the null hypothesis when it is true. It is a threshold set by the researcher before conducting an experiment or statistical test. In the wild iris exercise, a significance level of 0.05 or 5% is used. This means there is a 5% chance of incorrectly concluding that there is a difference in sepal lengths among the iris species when in fact, there is none.Choosing the right significance level depends on the consequences of making errors in hypothesis testing. A lower significance level, such as 0.01, reduces the chance of a false positive (Type I error), but may increase the likelihood of missing a significant result (Type II error). The 0.05 level is widely accepted as a balance between sensitivity and reliability in scientific experiments.
F-statistic
The F-statistic is a value calculated in ANOVA tests to determine whether there are any significant differences between the means of multiple groups. It compares the amount of systematic variance (variance between groups) to non-systematic variance (variance within groups).* This ratio helps in determining how many times the variance between groups is greater than the variance within groups.The formula for the F-statistic is:\[ F = \frac{MSB}{MSW} \]Where MSB is the Mean Square Between groups and MSW is the Mean Square Within groups. A higher F-value indicates a larger disparity between group means, thus suggesting significant differences.Once the F-statistic is calculated, it is compared to a critical F-value to decide whether to accept or reject the null hypothesis. The interpretation of the F-statistic is central to understanding whether the group differences observed in the sample data justify a conclusion of significant differences among the population means.
Critical F-value
The critical F-value is a threshold in ANOVA which defines whether the observed differences among means are statistically significant. Calculated through statistical tables or software, it depends on the chosen significance level and the degrees of freedom associated with the data, specifically the number of groups and the total number of observations.For the wild iris exercise, degrees of freedom for the between groups (\(k-1\)) is determined by the number of species types, while the degrees of freedom for within groups (\(n-k\)) considers the total observations minus the number of groups. With these, and a significance level of 0.05, we find the critical F-value from an F-distribution table.Comparing the calculated F-statistic with this critical F-value is a key step in hypothesis testing:
  • If the F-statistic is greater than the critical F-value, it suggests rejecting the null hypothesis, thus there is a significant difference among the means.

  • If it is less, the null hypothesis cannot be rejected.
This approach provides a systematic method to evaluate the significance of observed data variations.

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Most popular questions from this chapter

A new thermostat has been engineered for the frozen food cases in large supermarkets. Both the old and the new thermostats hold temperatures at an average of \(25^{\circ} \mathrm{F}\). However, it is hoped that the new thermostat might be more dependable in the sense that it will hold temperatures closer to \(25^{\circ} \mathrm{F}\). One frozen food case was equipped with the new thermostat, and a random sample of 21 temperature readings gave a sample variance of \(5.1 .\) Another, similar frozen food case was equipped with the old thermostat, and a random sample of 16 temperature readings gave a sample variance of \(12.8 .\) Test the claim that the population variance of the old thermostat temperature readings is larger than that for the new thermostat. Use a \(5 \%\) level of significance. How could your test conclusion relate to the question regarding the dependability of the temperature readings?

For chi-square distributions, as the number of degrees of freedom increases, does any skewness increase or decrease? Do chi-square distributions become more symmetric (and normal) as the number of degrees of freedom becomes larger and larger?

For a chi-square goodness-of-fit test, how are the degrees of freedom computed?

Economics: Profits per Employee How productive are U.S. workers? One way to answer this question is to study annual profits per employee. A random sample of companies in computers (I), aerospace (II), heavy equipment (III), and broadcasting (IV) gave the following data regarding annual profits per employee (units in thousands of dollars) (Source: Forbes Top Companies, edited by J. T. Davis, John Wiley and Sons).$$\begin{array}{cccc}\mathbf{I} & \mathbf{I I} & \mathbf{I I I} & \mathbf{I V} \\\27.8 & 13.3 & 22.3 & 17.1 \\\23.8 & 9.9 & 20.9 & 16.9 \\\14.1 & 11.7 & 7.2 & 14.3 \\\8.8 & 8.6 & 12.8 & 15.2 \\\11.9 & 6.6 & 7.0 & 10.1 \end{array}$$. Shall we reject or not reject the claim that there is no difference in population mean annual profits per employee in each of the four types of companies? Use a \(5 \%\) level of significance.

The following problem is based on information from Archaeological Surveys of Chaco Canyon, New Mexico, by A. Hayes, D. Brugge, and W. Judge, University of New Mexico Press. A transect is an archaeological study area that is \(1 / 5\) mile wide and 1 mile long. A site in a transect is the location of a significant archaeological find. Let \(x\) represent the number of sites per transect. In a section of Chaco Canyon, a large number of transects showed that \(x\) has a population variance \(\sigma^{2}=42.3 .\) In a different section of Chaco Canyon, a random sample of 23 transects gave a sample variance \(s^{2}=46.1\) for the number of sites per transect. Use a \(5 \%\) level of significance to test the claim that the variance in the new section is greater than 42.3 Find a \(95 \%\) confidence interval for the population variance.

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