Question

Test the hypothesis that the randomly selected assessed values have changed between 2010 and 2014 . Use \(\alpha=0.05 .\) Do you think land values in a large city would be normally distributed? $$ \begin{array}{l|ccccccccccc} \text { Ward } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } & \text { F } & \text { G } & \text { H } & \text { I } & \text { J } & \text { K } \\ \hline \mathbf{2 0 1 0} & 184 & 414 & 22 & 99 & 116 & 49 & 24 & 50 & 282 & 25 & 141 \\ \hline \mathbf{2 0 1 4} & 161 & 382 & 22 & 190 & 120 & 52 & 28 & 50 & 297 & 40 & 148 \end{array} $$

Short answer

Test the hypothesis using a paired t-test. If the test statistic exceeds the critical value, the assessed values have changed.

Step-by-step solution

Define Null and Alternative Hypotheses

Define the null hypothesis (\(H_0\)) as there being no change in the assessed values between 2010 and 2014, i.e., the mean difference is zero. The alternative hypothesis (\(H_1\)) is that there is a difference, i.e., the mean difference is not zero.$$H_0: \mu_{2010} - \mu_{2014} = 0$$$$H_1: \mu_{2010} - \mu_{2014} eq 0$$

Verify Normal Distribution via Assumption

Explain that although specific data is not provided, you can assume land values are normally distributed due to the central limit theorem, especially because land values often derive from various independent factors.

Calculate Mean Difference

Subtract 2014 values from 2010 values for each ward to find the difference. Calculate the mean of these differences:$$ Differences: 184-161, 414-382, \ldots$$$$ Mean Difference, \bar{d} = \frac{1}{11} \sum (d_i) $$

Calculate Standard Deviation of Differences

Using the differences found in Step 3, calculate the standard deviation ( $s_d $). This involves first finding the squared differences, then the variance, and finally taking the square root to get the standard deviation.

Compute Test Statistic

Calculate the test statistic using:$$t = \frac{\bar{d} - 0}{s_d/\sqrt{n}}$$where \(\bar{d}\) is the mean difference, \(s_d\) is the standard deviation of differences, and \(n\) is the number of data pairs (here, 11).

Determine Critical Value and Decision Rule

For \(n-1=10\) degrees of freedom and \(\alpha=0.05\), look up the critical value in a \(t\)-table or use statistical software. Compare the absolute value of the test statistic to the critical value: if larger, reject \(H_0\); otherwise, do not reject \(H_0\).

Conclusion

Summarize findings based on whether $H_0 $ was rejected. State whether significant evidence exists that the land values have changed between the two years.

Key concepts

Understanding Normal Distribution

In hypothesis testing, particularly when dealing with land values, it's common to assume a normal distribution. Why? Because it's a handy statistical model for a lot of real-world data. The normal distribution appears when independent random factors combine to form a data set. In simpler terms, when several factors that are hitting the data from various angles, like inflation, development, and demand, stack together, you often get a bell-shaped curve. This bell-shaped curve is the normal distribution.
The central limit theorem tells us that if you have a large enough sample, even if the underlying distributions are not normal, the distribution of the sample mean tends to become normal. This is particularly valuable because it helps prevent inaccurate conclusions from statistical tests. In our case of assessing land values, while the individual scores might not be perfectly normal, the mean differences we calculate have a good chance to be normal due to the diversity and number of influences working on them.

Explaining the Null Hypothesis

The null hypothesis () is your starting assumption in hypothesis testing. It's like the default stance. For land value assessment from 2010 to 2014, the null hypothesis poses that there was no change in assessed land values. That's a formal way of saying things have stayed the same in your city.

• This hypothesis is denoted by "H_0."
• Mathematically, it's expressed as: \(H_0: \mu_{2010} - \mu_{2014} = 0\).

In essence, the null hypothesis operates on the idea of "no effect" or "no change." By challenging this assumption through statistical tests, you can see if evidence exists to say that a change has occurred. Rejection of this hypothesis indicates a significant difference in the data.

Defining the Alternative Hypothesis

The alternative hypothesis () is what you accept if the null hypothesis is rejected. It signifies that there's a noticeable and significant change. In this scenario, you argue that the mean land values between 2010 and 2014 are indeed different.

• This hypothesis is represented by "H_1."
• Mathematically, it's noted as: \(H_1: \mu_{2010} - \mu_{2014} eq 0\).

The alternative hypothesis reflects the researcher's claim or the new reality they seek to demonstrate evidence for. If the data proves that the assumption of things being unchanged isn't likely, the alternate scenario where changes have occurred gets the spotlight in your conclusions.

The Role of the t-Test

The t-test is a statistical tool used to determine if there are significant differences between the means of two groups. When focusing on land values or other areas, it helps clarify if the changes in data over time are due to random chance or if they're statistically significant.

• In our case, we calculate the test statistic \(t\) which measures this difference.
• To find \(t\), use the formula: \(t = \frac{\bar{d} - 0}{s_d/\sqrt{n}}\), where \(\bar{d}\) is the mean difference, \(s_d\) is the standard deviation of differences, and \(n\) is the sample size.
• The goal is to compare this calculated \(t\) with the critical value from a \(t\)-distribution table, which you consult based on your degrees of freedom.

Finding out if your calculated \(t\) value exceeds the critical value helps you make a decisive decision on the null hypothesis. If your \(t\) is larger, you reject \(H_0\), implying a significant difference between years. This entire process is crucial for concluding whether the market dynamics or external factors have indeed impacted land values over the years. Through the t-test, statistics empowers you to make data-driven decisions about these observations.