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A random sample of \(n_{1}=10\) regions in New England gave the following violent crime rates (per million population). \(x_{1}\) : New England Crime Rate \(\begin{array}{llll}3.5 & 3.7 & 4.0 & 3.9\end{array}\) \(\begin{array}{llllll}3.3 & 4.1 & 1.8 & 4.8 & 2.9 & 3.1\end{array}\) Another random sample of \(n_{2}=12\) regions in the Rocky Mountain states gave the following violent crime rates (per million population). \(x_{2}\) : Rocky Mountain States \(\begin{array}{llllllllllll}3.7 & 4.3 & 4.5 & 5.3 & 3.3 & 4.8 & 3.5 & 2.4 & 3.1 & 3.5 & 5.2 & 2.8\end{array}\) (Reference: Crime in the United States, Federal Bureau of Investigation.) Assume that the crime rate distribution is approximately normal in both regions. i. Use a calculator to verify that \(\bar{x}_{1} \approx 3.51, s_{1} \approx 0.81, \bar{x}_{2} \approx 3.87\), and \(s_{2} \approx 0.94\). ii. Do the data indicate that the violent crime rate in the Rocky Mountain region is higher than in New England? Use \(\alpha=0.01\).

Step by step solution

01

State the Hypotheses

We begin by setting up our null and alternative hypotheses. To determine if the crime rate in the Rocky Mountain region is higher than that in New England, the hypotheses are:- Null hypothesis, \(H_0: \mu_1 = \mu_2\) (There is no difference in crime rates between New England and Rocky Mountain regions.)- Alternative hypothesis, \(H_a: \mu_2 > \mu_1\) (The crime rate in the Rocky Mountain region is higher than in New England.)

02

Verify Given Data

We verify the given sample means and standard deviations:- For New England, \(\bar{x}_1 = \frac{3.5 + 3.7 + 4.0 + 3.9 + 3.3 + 4.1 + 1.8 + 4.8 + 2.9 + 3.1}{10} = 3.51\) and \(s_1 = 0.81\).- For the Rocky Mountain states, \(\bar{x}_2 = \frac{3.7 + 4.3 + 4.5 + 5.3 + 3.3 + 4.8 + 3.5 + 2.4 + 3.1 + 3.5 + 5.2 + 2.8}{12} = 3.87\) and \(s_2 = 0.94\).The given data are consistent with these calculations.

03

Calculate the Test Statistic

We use a two-sample t-test for means with unequal variances to determine the test statistic.The formula for the test statistic is:\[ t = \frac{\bar{x}_2 - \bar{x}_1}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]where \(\bar{x}_1 = 3.51\), \(\bar{x}_2 = 3.87\), \(s_1 = 0.81\), \(s_2 = 0.94\), \(n_1 = 10\), and \(n_2 = 12\).Substituting the values:\[ t = \frac{3.87 - 3.51}{\sqrt{\frac{0.81^2}{10} + \frac{0.94^2}{12}}} \approx \frac{0.36}{0.3659} \approx 0.984 \]

04

Determine the Critical Value

Using a t-distribution table and a significance level of \(\alpha = 0.01\), we find the critical t-value for a one-tailed test with degrees of freedom calculated approximately using the formula:\[ df \approx \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}} \approx 19.59 \]Checking a t-table, \(t_{0.01, 19} = 2.539\) is the critical value for \(df \approx 20\).

05

Compare and Conclude

Since the calculated t-value \(0.984\) is less than the critical value \(2.539\), we fail to reject the null hypothesis at \(\alpha = 0.01\).This means there is not enough statistical evidence to support the claim that the violent crime rate is higher in the Rocky Mountain region compared to New England.

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Key Concepts

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

Two-sample t-test

The two-sample t-test is a statistical method used when we want to compare the means of two independent groups to see if there is a significant difference between them. In the context of our example, we're comparing the crime rates of New England to those of the Rocky Mountain states. When we conduct a two-sample t-test, we assume that both samples are drawn from normally distributed populations and that they have equal or similar variances. In our specific exercise, the test is focused on determining whether there is a significant difference in the average violent crime rates between the two groups. The formula used in a two-sample t-test when variances are unequal is:

• \[ t = \frac{\bar{x}_2 - \bar{x}_1}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]
• Here, \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means for the two groups, while \(s_1\) and \(s_2\) are the standard deviations.

This test helps us determine if any observed difference in means is statistically significant or just due to random chance.

Significance Level

The significance level, often denoted by \(\alpha\), is a critical threshold in hypothesis testing that helps determine whether to reject the null hypothesis. It is the probability of rejecting a true null hypothesis, essentially the risk of observing a result by chance. In our example, a significance level of \(\alpha = 0.01\) implies a high standard, meaning we want to be 99% confident in our test results. A lower significance level means we have stricter criteria for finding an effect.

• In context: if our test statistic exceeds the critical value determined by our significance level, we would reject the null hypothesis.
• Why choose a significance level? It guides how conservative or aggressive your test should be. A lower \(\alpha\) means you're more cautious and less likely to make a type I error (wrongly rejecting a true null).

The significance level is an essential part of understanding what conclusions can confidently be drawn from statistical tests.

Null Hypothesis

The null hypothesis, denoted as \(H_0\), is a statement of no effect or no difference. It serves as the baseline assumption in hypothesis testing. For our exercise, \(H_0: \mu_1 = \mu_2\), which means that the mean crime rate in New England is equal to the mean crime rate in the Rocky Mountain states. This is where our statistical testing begins, providing a point of reference against which we measure observed data.

• Why do we use it? It helps control for random variation and serves as a benchmark in hypothesis testing.
• The objective is to gather sufficient evidence to determine whether the null hypothesis can be rejected.

Rejection of the null hypothesis suggests that there is a statistically significant difference between the groups, while failure to reject suggests any observed differences are likely due to random sampling variability.

Alternative Hypothesis

The alternative hypothesis, denoted as \(H_a\), offers an opposing view to the null hypothesis. It proposes that there is a difference or effect. In the exercise, \(H_a: \mu_2 > \mu_1\) suggests that the violent crime rate in the Rocky Mountain states is higher than in New England. It is what researchers aim to support with their data.

• Purpose: It represents the effect we are testing for. We are looking to find evidence that supports this hypothesis over the null.
• If our test statistic meets the criteria to reject the null hypothesis at our chosen significance level, we accept the alternative hypothesis as more likely.

We must always remember that failing to reject the null doesn’t mean the alternative hypothesis is false—it simply means there's not enough evidence to support it given the data.