91Ó°ÊÓ

Please provide the following information for Problems. (a) What is the level of significance? State the null and alternate hypotheses. (b) Check Requirements What sampling distribution will you use? What assumptions are you making? Compute the sample test statistic and corresponding \(z\) or \(t\) value as appropriate. (c) Find (or estimate) the \(P\) -value. Sketch the sampling distribution and show the area corresponding to the \(P\) -value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \(\alpha ?\) (e) Interpret your conclusion in the context of the application. Note: For degrees of freedom \(d . f .\) not in the Student's \(t\) table, use the closest \(d . f .\) that is smaller. In some situations, this choice of \(d . f .\) may increase the \(P\) -value a small amount and therefore produce a slightly more "conservative" answer. Wildlife: Fox Rabies A study of fox rabies in southern Germany gave the following information about different regions and the occurrence of rabies in each region (Reference: \(\mathbf{B}\). Sayers et al., "A Pattern Analysis Study of a Wildlife Rabies Epizootic," Medical Informatics, Vol. 2, pp. 1 1-34). 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 $$\begin{array}{rlrrrrrrr} x_{1}: \text { Region I data } & 1 & 8 & 8 & 8 & 7 & 8 & 8 & 1 \\ & 3 & 3 & 3 & 2 & 5 & 1 & 4 & 6 \end{array}$$ 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. $$\begin{array}{lllllllll} x_{2} \text { : Region II data } & 1 & 1 & 3 & 1 & 4 & 8 & 5 & 4 \\ & 4 & 4 & 2 & 2 & 5 & 6 & 9 & \end{array}$$ 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 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

Understand the Level of Significance and Hypotheses

The level of significance is given as \( \alpha = 0.05 \). The null hypothesis \( H_0 \) states that there is no difference in the mean number of fox rabies cases between the two regions (\( \mu_1 - \mu_2 = 0 \)). The alternative hypothesis \( H_a \) states that there is a difference (\( \mu_1 - \mu_2 eq 0 \)).

02

Identify Sampling Distribution and Assumptions

We will use a two-sample t-test since the sample sizes are relatively small and the population standard deviations are unknown. Assumptions include normally distributed samples with unknown but equal variances.

03

Calculate the Test Statistic

First, compute the pooled standard deviation \( s_p \) using the formula:\[s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}}\]With \( s_1 = 2.82 \), \( s_2 = 2.43 \), \( n_1 = 16 \), \( n_2 = 15 \),\[s_p = \sqrt{\frac{(15)(2.82)^2+(14)(2.43)^2}{29}} \approx 2.63\]Next, find the test statistic \( t \) using:\[t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\]Substitute \( \bar{x}_1 = 4.75 \) and \( \bar{x}_2 = 3.93 \):\[t = \frac{4.75 - 3.93}{2.63 \sqrt{\frac{1}{16} + \frac{1}{15}}} \approx 0.933\]

04

Find the P-value and Sketch the Distribution

For a two-tailed test with our calculated \( t \)-value of approximately 0.933 and degrees of freedom \( df = n_1 + n_2 - 2 = 29 \), we look up the p-value in a t-distribution table or use a calculator. The t-distribution with 29 degrees of freedom gives a p-value of approximately 0.36. The area under the curve for this p-value represents the likelihood of finding results as extreme as those observed, under the null hypothesis.

05

Decision on the Null Hypothesis

Since the p-value of 0.36 is greater than the significance level of 0.05, we fail to reject the null hypothesis. Thus, the data do not provide sufficient evidence to suggest a statistically significant difference in the mean number of fox rabies cases between the two regions at the given level of significance.

06

Interpretation of Results

In the context of this study, our analysis indicates that there is no statistically significant difference in the mean number of fox rabies cases between Region I and Region II. Therefore, based on this sample, we cannot conclude that the rabies occurrence rates differ meaningfully between these two regions.

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Key Concepts

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

Level of Significance

The level of significance, often denoted as \( \alpha \), is a critical value in hypothesis testing that represents the probability of rejecting the null hypothesis when it is, in fact, true. This is known as making a Type I error. Typically, scientists and researchers choose a level of significance before conducting a test to uphold objectivity and rigor in their findings.
In this exercise, the level of significance is set at 0.05. To put it simply, this means that there is a 5% risk of concluding that a difference exists between the two regions (Region I and Region II) concerning the mean number of fox rabies cases when no actual difference exists.
In hypothesis testing, a null hypothesis \( H_0 \) is usually constructed to indicate no effect or no difference. Here, \( H_0: \mu_1 - \mu_2 = 0 \) implies no difference in the mean rabies cases between the two regions. The alternative hypothesis \( H_a: \mu_1 - \mu_2 eq 0 \) suggests that there is a difference.
By setting your \( \alpha \) at 0.05, you establish a threshold to determine whether observed data lead to the rejection or acceptance of the null hypothesis. Choosing the right level of significance is crucial because it balances potential risks of errors and impacts the decisions in the context of your research.

Two-Sample t-test

When comparing the means from two different groups, and you do not have both population standard deviations, the two-sample t-test is often the best method. This test is perfect for relatively small sample sizes, provided that the data distribution meets certain conditions.
In this problem, the data from Region I and Region II is assumed to be normally distributed. We also consider the variances to be equal, although unknown - a critical assumption for using the pooled variance in the two-sample t-test.
To compute the t-test value, you first need the pooled standard deviation \( s_p \). The formula takes the variances from both samples and accounts for their sample sizes: \[s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}}\] Once \( s_p \) is found, determine the t-test statistic using: \[t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\] Here, you substitute \( \bar{x}_1 = 4.75 \), \( \bar{x}_2 = 3.93 \), and \( s_p = 2.63 \), which results in about 0.933.
This calculated t-value helps determine how far the observed difference in sample means is from zero, assuming \( H_0 \) is true. The closer the t-value to zero, the closer the data fit the null hypothesis, suggesting no substantial difference in the datasets.

P-value Calculation

The P-value plays a crucial role in hypothesis testing. It indicates the probability of observing results as extreme as those seen, under the null hypothesis. In simpler words, it tells us how compatible our data is with the assumption that the null hypothesis is correct.
For our two-sample t-test of the fox rabies data, the calculated t-value was approximately 0.933. Given the degrees of freedom \( df = 29 \), you would refer to a t-distribution table, or employ a statistical calculator to find the corresponding P-value.
In this case, the P-value turns out to be approximately 0.36. This value represents the probability that the observed difference (or one more extreme) between region means is purely by chance.
In hypothesis testing, you compare the P-value against the pre-determined level of significance \( \alpha = 0.05 \). If the P-value is lower than \( \alpha \), it suggests evidence against \( H_0 \), warranting its rejection. If higher, as it is in this scenario (0.36 vs. 0.05), there is insufficient evidence to reject the null hypothesis. Thus, we conclude that any observed difference in fox rabies cases across regions is not statistically significant at the \( 5\% \) level.