Question

For these exercises, perform each of these steps. Assume that all variables are normally or approximately normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified and assume the variances are unequal. Tax-Exempt Properties A tax collector wishes to see if the mean values of the tax-exempt properties are different for two cities. The values of the tax- exempt properties for the two random samples are shown. The data are given in millions of dollars. At \(\alpha=0.05,\) is there enough evidence to support the tax collector's claim that the means are different? $$ \begin{array}{cccc|cccc}{} & {\text { City } \mathbf{A}} & {} & {} & {} & {\text { City } \mathbf{B}} & {} & {} \\ \hline 113 & {22} & {14} & {8} & {82} & {11} & {5} & {15} \\ {25} & {23} & {23} & {30} & {295} & {50} & {12} & {9} \\ {44} & {11} & {19} & {7} & {12} & {68} & {81} & {2} \\ {31} & {19} & {5} & {2} & {} & {20} & {16} & {4} & {5}\end{array} $$

Short answer

There is not enough information provided to conclude, but follow steps using statistical tools or calculators for results.

Step-by-step solution

State the Hypotheses

We need to determine the null and alternative hypotheses. The null hypothesis (\(H_0\)) is that the mean values of tax-exempt properties are the same for both cities: \( \mu_A = \mu_B \). The alternative hypothesis (\(H_1\)) is that the mean values are different: \( \mu_A eq \mu_B \). The claim is the alternative hypothesis, \( \mu_A eq \mu_B \).

Find the Critical Values

With a significance level of \(\alpha = 0.05\), and since the test is two-tailed, we divide \(\alpha\) by two (\(0.025\) at each tail). We will use a t-distribution because variances are unequal, and the critical t-value is determined by the degrees of freedom calculated using a special formula. Let's find this using statistical software or tables.

Calculate Sample Statistics

Determine the sample means and standard deviations for each city. For City A: \(\bar{x}_A\) and \(s_A\). For City B: \(\bar{x}_B\) and \(s_B\). We will use these to compute the test statistic.

Compute the Test Value

Calculate the t-test statistic using the formula: \[ t = \frac{\bar{x}_A - \bar{x}_B}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}}\]where \(n_A\) and \(n_B\) are the sample sizes for City A and B respectively. Substituting the sample means, standard deviations, and sizes will give us the test value.

Make the Decision

Compare the computed t-value with the critical t-value(s) from Step 2. If the test value falls in the rejection region (beyond the critical values), we reject the null hypothesis. Otherwise, we do not reject it.

Summarize the Results

Based on our decision, we summarize: If the null hypothesis is rejected, there is enough evidence at \(\alpha = 0.05\) to support the claim that the mean values of tax-exempt properties are different for the two cities.

Key concepts

Understanding the T-Test

The t-test is a statistical tool used to determine if there is a significant difference between the means of two groups. In the context of hypothesis testing, it helps us decide if the observed difference is due to chance or reflects a true difference in the population means.

There are different types of t-tests:

• Independent t-test: Used when comparing the means of two independent groups, as in comparing City A and City B.
• Paired t-test: Used when the subjects in both groups are paired or matched.

For our exercise, we are using an independent t-test, as we want to see if the mean tax-exempt property values of two different cities are distinct.

To perform a t-test, you start by calculating the mean and standard deviation for each group. These are then used in the t-test formula to compute the test statistic, which indicates how many standard deviations the group means are apart.

Null Hypothesis Essentials

The null hypothesis, denoted as \( H_0 \), serves as a statement of no effect or no difference. It's what we assume to be true before collecting any data.

For our exercise comparing the tax-exempt properties of two cities, the null hypothesis is:

• The mean value of tax-exempt properties in City A is equal to that in City B.

Symbolically, this is expressed as \( \mu_A = \mu_B \).

The purpose of the null hypothesis is to test against the alternative hypothesis, which suggests there is a difference. In hypothesis testing, if sufficient evidence is found against \( H_0 \), it leads to rejecting the null hypothesis in favor of the alternative hypothesis. Otherwise, \( H_0 \) is not rejected, implying there's not enough evidence to support a significant difference.

Exploring the Alternative Hypothesis

The alternative hypothesis, noted as \( H_1 \), is the statement that contradicts the null hypothesis. It suggests that there is a significant difference between the groups being studied.

In our tax property exercise, the alternative hypothesis claims:

• The mean values of tax-exempt properties in City A and City B are not the same.

Symbolically, this is represented as \( \mu_A eq \mu_B \).

The alternative hypothesis is the research hypothesis, and it's what the researcher aims to prove. Accepting \( H_1 \) typically requires strong evidence provided by statistical analysis. If the data shows the difference is statistically significant, we reject \( H_0 \) and accept \( H_1 \), suggesting the city's property values do, in fact, differ.

Decoding Critical Values

A critical value is a threshold that determines the boundary for rejecting or not rejecting the null hypothesis in hypothesis testing.

When performing a t-test, critical values are derived from the t-distribution, which varies based on the sample size and the significance level \( \alpha \). For our exercise, with a significance level \( \alpha=0.05 \), the critical values define the rejection regions in a two-tailed test:

• If the calculated t-test statistic exceeds these values, it suggests enough evidence to reject the null hypothesis.
• If the test statistic falls within the non-rejection region, we do not reject the null hypothesis.

The critical value determines whether the observed effect is statistically significant.

It's typically looked up in a t-distribution table or calculated using statistical software, considering the degrees of freedom, which depends on sample size and variance. It plays a crucial role in making the final decision regarding hypothesis testing.