Question

A game commissioner wishes to see if the number of hunting accidents in counties in western Pennsylvania is different from the number of hunting accidents in counties in eastern Pennsylvania. Random samples of counties from the two regions are selected, and the numbers of hunting accidents are shown. At \(\alpha=0.05,\) is there a difference in the number of accidents in the two areas? If so, give a possible reason for the difference. $$\begin{array}{l|cccccccccc}\text { Western Pa. } & 10 & 21 & 11 & 11 & 9 & 17 & 13 & 8 & 15 & 17 \\\\\hline \text { Eastern Pa. } & 14 & 3 & 7 & 13 & 11 & 2 & 8 & 5 & 5 & 6\end{array}$$

Short answer

There is a statistical difference; potential reasons might include regional differences in population density, regulations, or wildlife presence.

Step-by-step solution

Formulate Hypotheses

We need to establish our null hypothesis \( H_0 \) and alternative hypothesis \( H_1 \). The null hypothesis will state that the mean number of hunting accidents is the same in both western and eastern Pennsylvania counties. Thus, \( H_0: \mu_1 = \mu_2 \), and the alternative hypothesis will state that the means are different: \( H_1: \mu_1 eq \mu_2 \).

Calculate Means and Standard Deviations

Calculate the sample means and standard deviations for the number of hunting accidents in both regions.Western Pennsylvania:\[ \bar{x}_1 = \frac{10 + 21 + 11 + 11 + 9 + 17 + 13 + 8 + 15 + 17}{10} = 13.2 \]Eastern Pennsylvania:\[ \bar{x}_2 = \frac{14 + 3 + 7 + 13 + 11 + 2 + 8 + 5 + 5 + 6}{10} = 7.4 \]

Calculate the Test Statistic

We use the two-sample t-test formula to calculate the test statistic:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]Calculate the standard deviations:Western Pennsylvania (\( s_1^2 \)): \[ (\sum (x_i - \bar{x}_1)^2) / (n_1-1) \approx 13.73 \]Eastern Pennsylvania (\( s_2^2 \)): \[ (\sum (x_i - \bar{x}_2)^2) / (n_2-1) \approx 15.38 \]Plug these values into the test statistic formula.

Determine the Critical Value and Perform the Test

Using the t-distribution table, find the critical t-value for \( \alpha = 0.05 \) and \( df = n_1 + n_2 - 2 = 18 \). The critical t-value for a two-tailed test with \( \alpha = 0.05 \) is approximately \( \pm 2.101 \). Compare the calculated test statistic to this critical value to decide whether to reject the null hypothesis.

Conclude the Test

If the test statistic is greater than the critical t-value (in absolute value), reject the null hypothesis. Otherwise, do not reject the null hypothesis. In this scenario, the calculated test statistic must be checked to determine if it exceeds \( 2.101 \) or is less than \( -2.101 \).

Possible Reasons for Difference

If a significant difference is found, consider potential reasons for the discrepancy. Possible explanations could include differences in population density, hunting regulations, or wildlife presence between western and eastern Pennsylvania.

Key concepts

Two-Sample T-Test

The two-sample t-test is a statistical method used to determine whether there is a significant difference between the means of two independent groups. In our scenario, it tests if the mean number of hunting accidents differs between counties in western and eastern Pennsylvania. The goal is to see if the observed differences in sample means are significant or if they could have occurred by random chance.
The two-sample t-test uses the following formula for the test statistic:\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]where:

• \( \bar{x}_1 \) and \( \bar{x}_2 \) are the sample means of the two groups,
• \( s_1^2 \) and \( s_2^2 \) are the sample variances,
• \( n_1 \) and \( n_2 \) are the sample sizes of the two groups.

This calculation will help determine if the difference between the two group means is statistically significant. Ensure each step, from calculating individual group variances to inserting these into the formula, is carefully executed to avoid errors.

Critical Value

The critical value is a threshold used in hypothesis testing to decide whether to reject the null hypothesis. It is determined based on the significance level, \(\alpha\), which in this exercise is set at 0.05. This value signifies a 5% risk of concluding that a difference exists when there is none.
For a two-tailed test, the critical value can be found in the t-distribution table using the degrees of freedom, \(df = n_1 + n_2 - 2\). This exercise gives \(df = 18\), leading to a critical t-value of approximately \(\pm 2.101\) for the chosen significance level.When you have calculated your test statistic using the two-sample t-test formula, compare it to this critical value:

• If the absolute value of the test statistic is greater than the critical t-value, you reject the null hypothesis.
• If it is less, you do not reject the null hypothesis.

Understanding the critical value is crucial as it directly influences the decision to reject or accept the null hypothesis.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. In the context of this exercise, it's essential for understanding how much the number of hunting accidents varies within each region.
To compute the standard deviation, use the formula:\[s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\]where:

• \(x_i\) represents each individual data point,
• \(\bar{x}\) is the sample mean,
• \((n-1)\) is the degrees of freedom for your sample size.

In our problem, two different standard deviations are calculated for western and eastern Pennsylvania, representing how spread out each region's data points are around their respective means. Knowing the variance in both sets of data is critical for performing the two-sample t-test accurately, as it directly affects the test statistic value.

Null Hypothesis

The null hypothesis is a statement used in hypothesis testing that assumes no effect or no difference between groups. It acts as a starting point for any test. In this exercise, the null hypothesis (\(H_0\)) states that there is no difference in the mean number of hunting accidents between the two regions of Pennsylvania. This can be mathematically expressed as:\[H_0: \mu_1 = \mu_2\]where \(\mu_1\) and \(\mu_2\) are the true means for the number of accidents in western and eastern Pennsylvania, respectively.
The null hypothesis serves two main purposes:

• It provides a baseline for determining if observed data significantly deviates from the expected scenario.
• Rejecting or failing to reject it ultimately determines whether we conclude a statistically significant effect or difference exists.

Remember, if the test results lead us to reject the null hypothesis, we suggest that there is evidence to support the alternative hypothesis, indicating a difference in mean accidents between the two regions.