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Deer Where are the deer? Random samples of square-kilometer plots were taken in different ecological locations of Mesa Verde National Park. The deer counts per square kilometer were recorded and are shown in the following table (Source: The Mule Deer of Mesa Verde National Park, edited by G. W. Mierau and J. L. Schmidt, Mesa Verde Museum Association).$$\begin{array}{ccc}\text { Mountain Brush } & \text { Sagebrush Grassland } & \text { Pinon Juniper } \\\30 & 20 & 5 \\\29 & 58 & 7 \\\20 & 18 & 4 \\\29 & 22 & 9\end{array}$$.Shall we reject or accept the claim that there is no difference in the mean number of deer per square kilometer in these different ecological locations? Use a \(5 \%\) level of significance..

Short Answer

Expert verified
Do not reject the null hypothesis; no significant difference in mean deer counts among locations.

Step by step solution

01

Identify the Hypotheses

Our null hypothesis (H_0) is that there is no difference in the mean number of deer among the three ecological locations. This means \( \mu_1 = \mu_2 = \mu_3 \), where \( \mu_1 \), \( \mu_2 \), and \( \mu_3 \) represent the mean deer count per square kilometer for Mountain Brush, Sagebrush Grassland, and Pinon Juniper, respectively. The alternative hypothesis (H_a) is that at least one mean is different.
02

Perform ANOVA Test for Means

For comparing means across multiple groups, we use the ANOVA test. Begin by calculating group means, overall mean, and the sum of squares within (SSW) and between (SSB) groups. These statistics will help calculate the F-statistic. Given the data, compute the group means: Mountain Brush: 27, Sagebrush Grassland: 29.5, Pinon Juniper: 6.25.Overall mean is calculated as: \((30 + 29 + 20 + 29 + 20 + 58 + 18 + 22 + 5 + 7 + 4 + 9) / 12 = 20\).
03

Calculate Sum of Squares

Calculate SSW and SSB:- SSB = \( n_k \sum (\bar{x}_k - \bar{x})^2 \) per group, where \( n_k \) is the sample size, \( \bar{x}_k \) is the group mean, and \( \bar{x} \) is the overall mean.- SSW = \( \sum_{i=1}^3 \sum (x_{ij} - \bar{x}_k)^2 \), where \( x_{ij} \) are the data points in each group.Plug in the values to get:- SSB = 46.5 for all groups.- SSW = 1106.5.
04

Determine F-statistic

Compute the F-statistic:\[ F = \frac{SSB / (k - 1)}{SSW / (N - k)} \]Where \( k = 3 \) (number of groups) and \( N = 12 \) (total data points). This yields:- Mean SSB = 23.25- Mean SSW = 110.65Calculate the F-statistic:\[ F = \frac{23.25}{110.65} = 0.21 \].
05

Compare F-statistic to Critical Value

The critical F-value can be obtained from F-distribution tables based on \( \alpha = 0.05 \), \( df1 = 2 \) (numerator), and \( df2 = 9 \) (denominator). For these degrees of freedom and significance level, the critical F-value is approximately 4.26. Since the calculated F-statistic (0.21) is less than the critical value, we fail to reject the null hypothesis.
06

Conclusion

Since the F-statistic calculated is less than the critical F-value, there is insufficient evidence to suggest a significant difference in the mean number of deer per square kilometer among the different ecological locations. Therefore, we do not reject the null hypothesis at the 5% significance level.

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Key Concepts

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

ANOVA Test
The ANOVA test, short for Analysis of Variance, is a statistical method used to compare the means of three or more samples to understand if at least one of the group means is significantly different from the others. It's particularly useful when dealing with multiple groups and when trying to determine if group differences are due to an actual variance in means or just by chance.

When conducting an ANOVA test, our focus is on:
  • Calculating the group means and the overall mean.
  • Determining the variation within and between the groups, referred to as Sum of Squares Within (SSW) and Sum of Squares Between (SSB).
This method helps us to test hypotheses regarding group means and provides a statistical basis for comparison. If the ANOVA test yields a significant result, it implies that not all of the group means are equal; some are indeed different.
Null and Alternative Hypotheses
In hypothesis testing, specifying the null and alternative hypotheses is the first and perhaps the most critical step.

The null hypothesis, denoted as \( H_0 \), is a statement of no effect or no difference. It asserts that the situation measured is just by chance. In the context of the deer study, the null hypothesis states that there is no difference in mean deer counts across different ecological locations, symbolically represented as \( \mu_1 = \mu_2 = \mu_3 \).

The alternative hypothesis, represented as \( H_a \), states the opposite of the null hypothesis. It suggests that there is a difference, meaning at least one of the group means is different. This helps researchers understand whether their data provides enough evidence to reject the null hypothesis in favor of the alternative hypothesis, hence supporting the claim they are testing.
F-statistic
The F-statistic is a crucial component in the ANOVA test. It acts as a ratio that compares the model's variance explained by the groups (SSB) to the variance due to random error or individual differences within the groups (SSW). Essentially, it tells us whether the between-group differences are large relative to the within-group differences.

The formula for computing the F-statistic in an ANOVA context is:
\[ F = \frac{SSB / (k - 1)}{SSW / (N - k)} \]
where \( k \) represents the number of groups and \( N \) is the total number of observations across all groups. A larger F-statistic implies a more significant difference between group means.

Once calculated, the F-statistic is compared to a critical value derived from the F-distribution table, considering factors like degrees of freedom and significance level. In our deer example, if the calculated F-statistic is less than the critical value, the null hypothesis is not rejected, signaling no substantial difference in the deer count means across the locations.

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Most popular questions from this chapter

The following table shows the Myers-Briggs personality preferences for a random sample of 406 people in the listed professions (Atlas of Type Tables by Macdaid, McCaulley, and Kainz). E refers to extroverted and I refers to introverted. $$\begin{array}{l|c|c|c} \hline \multirow{2}{*} {\text { Occupation }} & \multicolumn{3}{|c} {\text { Personality Preference Type }} \\ \\)\cline { 2 - 5 } & \\( \mathrm{E} & \mathrm{I} & \text { Row Total } \\ \hline \text { Clergy (all denominations) } & 62 & 45 & 107 \\ \hline \text { M.D. } & 68 & 94 & 162 \\ \hline \text { Lawyer } & 56 & 81 & 137 \\ \hline \text { Column Total } & 186 & 220 & 406 \\ \hline \end{array}$$Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.05 level of significance.

When using the \(F\) distribution to test variances from two populations, should the random variables from each population be independent or dependent?

A random sample of leading companies in South Korea gave the following percentage yields based on assets (see reference in Problem 7): $$\begin{array}{cc} 2.5 & 2.0 & 4.5 & 1.8 & 0.5 & 3.6 & 2.4 \\ 0.2 & 1.7 & 1.8 & 1.4 & 5.4 &1.1 \end{array}$$ Use a calculator to verify that \(s^{2}=2.247\) for these South Korean companies. Another random sample of leading companies in Sweden gave the following percentage yields based on assets: $$\begin{array}{ccccccccc} 2.3 & 3.2 & 3.6 & 1.2 & 3.6 & 2.8 & 2.3 & 3.5 & 2.8 \end{array}$$ Use a calculator to verify that \(s^{2}=0.624\) for these Swedish companies. Test the claim that the population variance of percentage yields on assets for South Korean companies is higher than that for companies in Sweden. Use a \(5 \%\) level of significance. How could your test conclusion relate to an economist's question regarding volatility of corporate productivity of large companies in South Korea compared with that in Sweden?

In general, how do the hypotheses for chi-square tests of independence differ from those for chi-square tests of homogeneity? Explain.

In general, is the \(F\) distribution symmetric? Can values of \(F\) be negative?

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