Problem 20 - Basic Statistics for Business and Economics with Student

Question

The null and alternate hypotheses are: $$ \begin{array}{l} H_{0}: \mu_{d}=0 \\ H_{1}: \mu_{d} \neq 0 \end{array} $$ The following paired observations show the number of traffic citations given for speeding by Officer Dhondt and Officer Meredith of the South Carolina Highway Patrol for the last five months. At the . 05 significance level, is there a difference in the mean number of citations given by the two officers?

Step-by-Step Solution

Answer

If the test statistic exceeds the critical value, there is a significant difference in citations given by the officers; otherwise, there is not.

Step 1: Calculate Differences for Each Pair

First, calculate the differences in the number of citations given by Officer Dhondt and Officer Meredith for each of the five months. For example, if Officer Dhondt gave 10 citations and Officer Meredith gave 8 citations in a month, the difference is 2. Do this for all pairs.

Step 2: Calculate the Mean Difference

Find the mean of the differences calculated in Step 1. Add all the differences together and divide by the number of pairs (5 in this case). This will give you the mean difference, denoted as \( \bar{d} \).

Step 3: Calculate the Standard Deviation of Differences

Calculate the standard deviation of the differences. Use the formula for standard deviation: \( s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}} \), where \( d_i \) represents each difference, and \( n \) is the number of pairs.

Step 4: Compute the Test Statistic

Use the t-test formula for paired samples: \( t = \frac{\bar{d}}{s_d/\sqrt{n}} \), where \( \bar{d} \) is the mean difference, \( s_d \) is the standard deviation of the differences, and \( n \) is the number of pairs. This will give you the test statistic \( t \).

Step 5: Determine the Critical Value

For a two-tailed test at the 0.05 significance level with \( n-1 \) degrees of freedom (\( n = 5 \)), find the critical t-value from the t-distribution table. This is the value that the test statistic must exceed in absolute terms to reject the null hypothesis.

Step 6: Make a Decision

Compare the calculated t-test statistic with the critical value from Step 5. If the absolute value of the t-test statistic is greater than the critical value, reject the null hypothesis \( H_0 \). Otherwise, do not reject it.

Key Concepts

Null Hypothesis

The null hypothesis is a starting point in hypothesis testing. It is a statement that there is no effect or no difference. In our example, the null hypothesis is that there is no difference in the mean number of traffic citations given by the two officers. We write this as \( H_0: \mu_d = 0 \).
This means that we assume any observed difference is due to random chance. The null hypothesis serves as the default or "no change" statement that we attempt to test.
If the data provides enough evidence against the null hypothesis, it may be rejected, indicating that there is a significant effect or difference. However, if the evidence is not strong enough, we do not reject the null hypothesis.

Alternate Hypothesis

The alternate hypothesis is a statement that contradicts the null hypothesis. It suggests that there is an effect or a difference. In this case, the alternate hypothesis is that the mean number of citations given is not equal between the two officers. We express this as \( H_1: \mu_d eq 0 \).
The alternate hypothesis is usually what the researcher is trying to prove. Finding evidence in favor of the alternate hypothesis suggests that the effect or difference is real and not due to random chance alone.
In a paired sample t-test, proving the alternate hypothesis means showing that the differences in paired observations are consistent with an actual effect.

Significance Level

The significance level, denoted by \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true. It is a threshold for how confident we want to be about rejecting the null hypothesis. Commonly used significance levels are 0.05 and 0.01.
In our scenario, the significance level is 0.05. This means that there is a 5% risk of concluding that a difference exists when there is none. A significance level of 0.05 balances being cautious and allowing for sufficient sensitivity to detect actual differences.

• A higher significance level means a higher chance of making a Type I error (wrongly rejecting a true null hypothesis).
• A lower significance level reduces this risk but can also make the test less sensitive to detecting a real effect.

Choosing a significance level is important as it reflects how much risk of error is tolerable in the specific context.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. In the context of paired sample t-tests, the standard deviation helps understand how varied the pair differences are.
Computing the standard deviation involves calculating how much each observation differs from the mean, squaring those differences, averaging them, and taking the square root. In our step-by-step guide, the formula \( s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}} \) is used.
The standard deviation provides critical insight when interpreting the test statistic. A smaller standard deviation indicates the data points are close to the mean, suggesting consistency in differences. A larger standard deviation implies the differences vary more widely, which might affect the t-test outcome.
Understanding standard deviation helps in assessing the reliability and variability of the presented data.