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A study of fox rabies in southern Germany gave the following information about different regions and the occurrence of rabies in each region (Reference: B. Sayers et al., "A Pattern Analysis Study of a Wildlife Rabies Epizootic," Medical Informatics, Vol. 2, pp. 11-34). Based on information from this article, a random sample of \(n_{1}=16\) locations in region I gave the following information about the number of cases of fox rabies near that location. $$ \begin{array}{lllllllll} x_{1}: \text { Region I data } & 1 & 8 & 8 & 8 & 7 & 8 & 8 & 1 \\ & 3 & 3 & 3 & 2 & 5 & 1 & 4 & 6 \end{array} $$ A second random sample of \(n_{2}=15\) locations in region II gave the following information about the number of cases of fox rabies near that location. \(x_{2}:\) Region I $$ \begin{array}{lllllllll} \text { II data } & 1 & 1 & 3 & 1 & 4 & 8 & 5 & 4 \\ & 4 & 4 & 2 & 2 & 5 & 6 & 9 & \end{array} $$ i. Use a calculator with sample mean and sample standard deviation keys to verify that \(\bar{x}_{1}=4.75\) with \(s_{1} \approx 2.82\) in region \(\mathrm{I}\) and \(\bar{x}_{2} \approx 3.93\) with \(s_{2} \approx 2.43\) in region II. ii. Does this information indicate that there is a difference (either way) in the mean number of cases of fox rabies between the two regions? Use a \(5 \%\) level of significance. (Assume the distribution of rabies cases in both regions is mound-shaped and approximately normal.)

Step by step solution

01

Calculate Sample Mean for Region I

To find the sample mean for Region I, sum up all the values of \(x_1\) and divide by the number of observations, \(n_1 = 16\).\[\bar{x}_1 = \frac{1 + 8 + 8 + 8 + 7 + 8 + 8 + 1 + 3 + 3 + 3 + 2 + 5 + 1 + 4 + 6}{16} = 4.75\]

02

Calculate Sample Standard Deviation for Region I

To calculate the sample standard deviation, first find the sum of squared deviations from the mean for each observation in Region I, divide by \(n_1 - 1\), and take the square root.\[s_1 = \sqrt{\frac{(1 - 4.75)^2 + (8 - 4.75)^2 + \ldots + (6 - 4.75)^2}{15}} \approx 2.82\]

03

Calculate Sample Mean for Region II

Add all values for \(x_2\) and divide by the number of observations \(n_2 = 15\).\[\bar{x}_2 = \frac{1 + 1 + 3 + 1 + 4 + 8 + 5 + 4 + 4 + 4 + 2 + 2 + 5 + 6 + 9}{15} \approx 3.93\]

04

Calculate Sample Standard Deviation for Region II

Compute the standard deviation for Region II using a similar method as for Region I.\[s_2 = \sqrt{\frac{(1 - 3.93)^2 + (1 - 3.93)^2 + \ldots + (9 - 3.93)^2}{14}} \approx 2.43\]

05

Formulate Hypotheses for Difference in Means

Set up the null hypothesis \(H_0: \mu_1 = \mu_2\) and the alternative hypothesis \(H_a: \mu_1 eq \mu_2\) to test whether there is a significant difference between the means of the two regions.

06

Calculate Test Statistic

Use the formula for the test statistic \(t\) for two independent samples: \[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\] Substitute the values: \[t = \frac{4.75 - 3.93}{\sqrt{\frac{2.82^2}{16} + \frac{2.43^2}{15}}}\approx 0.97\]

07

Determine Critical t-value and Compare

Find the critical \(t\)-value for a two-tailed test at a 5% significance level with \(n_1 + n_2 - 2 = 29\) degrees of freedom. The critical \(t\)-value \( \approx 2.045 \). Since \(|0.97| < 2.045\), we fail to reject the null hypothesis.

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Key Concepts

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

Sample Mean

The sample mean is a fundamental concept in statistics. It's the average value from a set of data. In the context of our fox rabies study, the sample mean for each region helps us understand the central tendency of rabies cases reported.

Calculating the sample mean involves adding up all individual data points and then dividing by the number of observations. For Region I, this means summing all rabies cases and dividing by 16, the number of observations, yielding a sample mean of 4.75. Similarly, for Region II, the mean is about 3.93 from 15 observations.

This value provides a basic idea of what's typical in each region regarding rabies cases. It's crucial because it sets the stage for further analysis, serving as a baseline for comparison.

Sample Standard Deviation

While the sample mean tells us about the average, the sample standard deviation offers insight into how much variation or spread there is in the dataset. It is a measure of the amount of variability or dispersion of a set of values.

In practical terms, a high standard deviation implies that the data points are spread out over a wider range of values, while a low standard deviation indicates that they cluster closer to the mean.

For example, in Region I, the standard deviation is approximately 2.82. This suggests a relatively moderate variation around the mean of 4.75 cases. In Region II, the standard deviation is about 2.43, indicating slightly less variability around its mean of 3.93 cases. Understanding this spread is vital in the context of hypothesis testing because it affects calculations related to the sample distribution.

T-test

The T-test is a statistical tool used to determine if there is a significant difference between the means of two groups. It helps to infer whether observed differences in sample data suggest genuine differences in population means.

In our exercise, we conduct a two-sample T-test to see if the mean number of rabies cases differs between Region I and Region II. With values of means and standard deviations known, the test statistic is calculated, which in this case is around 0.97.

The test result is compared against a critical value from the T-distribution table. If the absolute value of the test statistic is greater than the critical value, we would reject the null hypothesis. However, here we find it at 0.97, which is less than the critical value of approximately 2.045, we fail to reject the null hypothesis. This indicates that the observed difference in means is not statistically significant at a 5% significance level.

Null Hypothesis

The null hypothesis is a statement used in statistics that proposes no statistical significance exists in a set of given observations. Its primary function is to challenge the alternative hypothesis, which suggests a statistically significant effect.

In context, our null hypothesis posits that the mean number of rabies cases is the same in both regions, symbolized as:

• \(H_0:
  = \)

The alternative hypothesis is that these means are not equal. By using statistical tests, we aim to determine whether there is enough evidence to reject the null hypothesis.

Failing to reject the null hypothesis, as in our exercise, implies that there is not enough statistical evidence to claim a significant difference in mean rabies cases between the two regions. It encourages a careful interpretation, meaning we should not assume there is a difference unless the test strongly suggests otherwise.