The annual dividend rates for a random sample of 16 companies in three different industries, utilities, banking, and insurance were recorded. The ANOVA comparing the mean annual dividend rate among three industries rejected the null hypothesis that the dividend rates were equal. The Mean Square Error (MSE) was 3.36. The following table summarized the results:
Compare the mean annual dividend rate for companies in the utilities and insurance industries and construct the 95% confidence interval.
A) 5.78 $\pm$ 2.160 * 2.40
B) 5.78 $\pm$ 2.120 * 2.40
C) 5.78 $\pm$ 2.160 * 1.11
D) 5.78 $\pm$ 2.120 * 1.11

Question

The annual dividend rates for a random sample of 16 companies in three different industries, utilities, banking, and insurance were recorded. The ANOVA comparing the mean annual dividend rate among three industries rejected the null hypothesis that the dividend rates were equal. The Mean Square Error (MSE) was 3.36. The following table summarized the results:  
Compare the mean annual dividend rate for companies in the utilities and insurance industries and construct the 95% confidence interval.
A) 5.78 $\pm$ 2.160 * 2.40 
B) 5.78 $\pm$ 2.120 * 2.40 
C) 5.78 $\pm$ 2.160 * 1.11 
D) 5.78 $\pm$ 2.120 * 1.11

Quizplus · Accepted Answer

To compare the mean annual dividend rate for companies in the utilities and insurance industries, we need to calculate the difference in means and construct the confidence interval. 
The difference in means is: 
5.78 - 3.50 = 2.28 
The standard error of the difference in means is: 
sqrt[(MSE/n1) + (MSE/n2)] = sqrt[(3.36/8) + (3.36/8)] = 1.11 
Using a t-distribution with 14 degrees of freedom (16-2), the critical value for a 95% confidence level is 2.145. 
The 95% confidence interval is: 
2.28 ± (2.145 * 1.11) = (0.96, 3.60) 
Therefore, the best choice is C, 5.78 ± 1.11 * 2.160, which provides the correct difference in means and confidence interval.

The Annual Dividend Rates for a Random Sample of 16 ±\pm±

The Annual Dividend Rates for a Random Sample of 16 $\pm$