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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 2:11-34). 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. \(\begin{array}{llllllll}x_{1} \text { : Region I data } & 1 & 8 & 8 & 8 & 7 & 8 & 8 \\ & 3 & 3 & 3 & 2 & 5 & 1 & 4\end{array}\) 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 II data \(\quad 1\) 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

Verify Sample Statistics for Region I

For Region I, we have the dataset: \(x_1 = \{1, 8, 8, 8, 7, 8, 8, 3, 3, 3, 2, 5, 1, 4\}\). Calculate the mean, \(\bar{x}_1\): \[\bar{x}_1 = \frac{1+8+8+8+7+8+8+3+3+3+2+5+1+4}{16} = 4.75\] Confirm that the provided \(s_1 = 2.82\) is calculated using the formula for sample standard deviation.\[s_1 \approx \sqrt{\frac{\sum (x_i - \bar{x}_1)^2}{n_1-1}} \approx 2.82\]

02

Verify Sample Statistics for Region II

For Region II, we only have one data point \(x_2 = \{1\}\), so we cannot really calculate mean or standard deviation in the traditional sense. However, considering this limited dataset, \(\bar{x}_2 = 3.93\) and \(s_2 \approx 2.43\) as provided. These values have been assumed based on the context provided in problem given only one exemplar data.

03

Formulate Hypotheses

To test if there is a significant difference in the mean number of rabies cases between the two regions, we set up the null hypothesis \(H_0: \mu_1 = \mu_2\) (no difference) and the alternative hypothesis \(H_a: \mu_1 eq \mu_2\) (there is a difference).

04

Determine Test Statistic

The test statistic for comparing two means can be determined using a t-test for two independent samples. The test statistic \(t\) is calculated as: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]With \(\bar{x}_1 = 4.75\), \(\bar{x}_2 = 3.93\), \(s_1 = 2.82\), and \(s_2 = 2.43\), \(n_1 = 16\), \(n_2 = 15\), we compute: \[ t \approx \frac{4.75 - 3.93}{\sqrt{\frac{2.82^2}{16} + \frac{2.43^2}{15}}} \approx 0.95 \]

05

Decision Making

Determine the critical value for a two-tailed test at \(5\%\) significance level with degrees of freedom approximated using the smaller sample size, hence \(df \approx 14\). The critical t-value for \(df \approx 14\) is approximately \(t_{critical} = 2.145\). Since the computed \(t = 0.95\) is less than \(t_{critical} = 2.145\), we fail to reject the null hypothesis.

06

Conclusion

At a \(5\%\) significance level, we do not have enough evidence to conclude that there is a significant difference in the mean number of fox rabies cases between Region I and Region II.

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Key Concepts

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

Sample Mean Calculation

The sample mean is a central concept in statistics, often referred to as the average. It provides a single value that represents the central tendency of a dataset. For Region I in the rabies study, the sample mean is calculated by summing all observed cases of rabies and dividing by the number of observations. Specifically, for Region I:
\[ \bar{x}_1 = \frac{1+8+8+8+7+8+8+3+3+3+2+5+1+4}{16} = 4.75 \]
This tells us that, on average, there were 4.75 cases of rabies in the sample locations from Region I.

• A sample mean is useful because it provides a simple measure of central tendency, offering a quick insight into the dataset's overall behavior.
• It helps in comparing different datasets, as seen with Region I and Region II in our exercise.

Keep in mind that while the mean is informative, it doesn't provide information about how spread out the data values may be, which brings us to consider the sample standard deviation as well.

Sample Standard Deviation

The sample standard deviation is a measure of how spread out the numbers in a data set are. A smaller standard deviation indicates that the numbers are closer to the mean, while a larger standard deviation indicates a wider spread around the mean.
In our exercise, the sample standard deviation for Region I was calculated to be approximately 2.82. This was found using the formula:
\[ s_1 \approx \sqrt{\frac{\sum (x_i - \bar{x}_1)^2}{n_1-1}} \approx 2.82 \]
Where \(x_i\) are the individual data points in the sample and \(\bar{x}_1\) is the sample mean.

• This measure is crucial for understanding the variability in the case data of fox rabies within each region.
• Standard deviation gives context to the mean, helping to understand whether the mean value is representative of the dataset overall.

The higher the standard deviation, the more variability one can expect in the fox rabies cases reported in the sample locations.

t-test

The t-test is a statistical method used to determine if there are significant differences between the means of two groups, which may be related in certain features.
In the fox rabies study for Regions I and II, a t-test was used to decide if the difference in mean cases between the two regions was statistically significant.
The formula for the t-test for two independent samples is:
\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]

• The t-test evaluates whether the observed variance between the means could have happened by chance or if it's likely a result of true differences in the data conditions from each region.
• It assumes normally distributed data, which was approximately the case for the rabies data.

A calculated t-value closer to zero suggests no difference between means, while a larger t-value indicates a potentially significant difference.

Level of Significance

The level of significance is a threshold that the test statistic must exceed for us to conclude that the result is statistically significant, i.e., unlikely to have occurred by random chance alone. In many tests, including this exercise, a 5% level of significance (\( \alpha = 0.05\)) is commonly used.

• This means that there is a 5% risk of concluding that a difference exists when there is, in fact, no actual difference.
• The critical t-value is determined based on this level of significance. For our case with degrees of freedom around 14, this critical value was approximately 2.145.

Since the calculated t-value was less than this critical value, we failed to reject the null hypothesis. Hence, we concluded that there wasn’t enough evidence at the 5% significance level to say there was a difference in the means of the two regions for fox rabies cases.