Problem 37 - Basic Statistics for Business and Economics with Student

Question

Fairfield Homes is developing two parcels near Pigeon Fork, Tennessee. In order to test different advertising approaches, it uses different media to reach potential buyers. The mean annual family income for 15 people making inquiries at the first development is \(\$ 150,000\) with a standard deviation of \(\$ 40,000\). A corresponding sample of 25 people at the second development had a mean of \(\$ 180,000,\) with a standard deviation of \(\$ 30,000\). Assume the population standard deviations are the same. At the .05 significance level, can Fairfield conclude that the population means are different?

Step-by-Step Solution

Answer

The population means are significantly different.

Step 1: Define the Hypotheses

Identify the null hypothesis \(H_0\) and the alternative hypothesis \(H_1\). We want to test if the two population means are different. Thus, \(H_0: \mu_1 = \mu_2\) and \(H_1: \mu_1 eq \mu_2\).

Step 2: Determine the Test Statistic

Use a two-sample t-test for the means. The test statistic \(t\) is given by:\[t = \frac{\bar{X}_1 - \bar{X}_2}{s_p \cdot \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\]where \(\bar{X}_1 = 150,000\), \(\bar{X}_2 = 180,000\), \(n_1 = 15\), \(n_2 = 25\). Since the populations' standard deviations are assumed equal, we calculate the pooled standard deviation \(s_p\) as:\[s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}\]

Step 3: Compute the Pooled Standard Deviation

Insert the values into the pooled standard deviation formula:\[s_p = \sqrt{\frac{(15 - 1)\cdot (40,000)^2 + (25 - 1)\cdot (30,000)^2}{15 + 25 - 2}}\]

Step 4: Calculate the Test Statistic

Calculate the test statistic \(t\) using the pooled standard deviation from Step 3:\[t = \frac{150,000 - 180,000}{s_p \cdot \sqrt{\frac{1}{15} + \frac{1}{25}}}\]

Step 5: Determine the Critical Value

At \(\alpha = 0.05\) and degrees of freedom \(df = n_1 + n_2 - 2 = 38\), find the critical t-value from the t-distribution table for a two-tailed test.

Step 6: Compare Test Statistic with Critical Value

Compare the computed t-value with the critical values. If the absolute value of t is greater than the critical value, reject the null hypothesis \(H_0\).

Step 7: Draw Conclusion

Based on the comparison, if we reject \(H_0\), this suggests that there is sufficient evidence to conclude that the population means are different. Otherwise, we do not have enough evidence to claim they are different.

Key Concepts

Pooled Standard Deviation

The pooled standard deviation is a crucial component when comparing two populations. It helps to combine the variances of each group, especially when we assume the variance is equal across them. This assumption is known as 'homogeneity of variance.' By computing the pooled standard deviation, we simplify comparisons between the two samples.To calculate the pooled standard deviation \(s_p\), use the formula:\[ s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}} \]Where:

• \(n_1\) and \(n_2\) are the sample sizes
• \(s_1\) and \(s_2\) are the standard deviations of the samples

Once you have \(s_p\), it becomes a part of calculating the t-statistic to assess the means. This pooled measure is reliable when the population standard deviations are assumed identical.

Null Hypothesis

The null hypothesis, denoted as \(H_0\), serves as a baseline statement that there is no effect or difference between groups. In our context, it posits that the mean income of the two groups is the same. Formally written, it is:\[ H_0: \mu_1 = \mu_2 \]This expression suggests that any observed difference in sample means may be due to random sampling variability. The alternative hypothesis \(H_1\):\[ \mu_1 eq \mu_2 \]indicates a belief that the group means are truly different. When performing a hypothesis test, our objective is to determine whether there is enough statistical evidence to reject \(H_0\) in favor of \(H_1\). It's essential to rigorously test the null hypothesis since making a decision entails the risk of errors.

Critical Value

The critical value is a threshold that helps determine if the test statistic is unusual enough to reject the null hypothesis. We decide on these values based on the desired confidence level. For instance, with \(\alpha = 0.05\), we accept a 5% chance of concluding a difference when there is none (Type I error).At this level of significance, critical values for a t-test are chosen from the t-distribution table for a two-tailed test. This accounts for the possibility of extreme values on both ends of the distribution. The degrees of freedom \(df = n_1 + n_2 - 2\) also guide which row of the table we consult.The total area under the t-distribution corresponds to 1. Therefore, when \(\alpha = 0.05\), each tail holds 2.5% (0.025) probability, meaning:- If the computed t-statistic falls beyond this critical value, we reject \(H_0\).- Otherwise, there isn’t enough evidence to reject the null.

Test Statistic

The test statistic helps summarize the data to compare the two population means. Using a two-sample t-test, the formula for the test statistic \(t\) can be expressed as:\[ t = \frac{\bar{X}_1 - \bar{X}_2}{s_p \cdot \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \]In this expression:

• \(\bar{X}_1\) and \(\bar{X}_2\) are sample means
• \(s_p\) is the pooled standard deviation
• \(n_1\) and \(n_2\) are sample sizes

The test statistic tells you how many standard deviations the sample mean difference is from zero, assuming the null hypothesis is true. A large absolute value for \(t\) suggests that the populations' means differ more than expected by chance, prompting further investigation.

Population Means Comparison

Comparing population means is a common goal in statistics to see if groups differ significantly. In our scenario, we investigated whether the average annual income inquiry between two developments varies meaningfully. To do this effectively:

• First, we determine if assumptions like equal variance are met.
• Next, apply the correct testing method, such as a two-sample t-test.
• Evaluate the test statistic against critical values to verify significance.

This process aids in understanding if the observed differences are due to systematic causes or just randomness. In essence, comparing means enables businesses like Fairfield Homes to decide if different advertising had significant impacts on attracting varying income-level clients.