A random sample of 30 executives from companies with assets over $1 million was selected and asked for their annual income and level of education.The ANOVA comparing the average income among three levels of education rejected the null hypothesis.The Mean Square Error (MSE)was 243.7.The following table summarized the results: When comparing the mean annual incomes for executives with Undergraduate and Master's Degree or more,the following 95% confidence interval can be constructed:
A)2.0 $\pm$2.052 * 6.51
B)2.0 $\pm$ 3.182 * 6.51
C)2.0 $\pm$ 2.052 * 42.46
D)2.0 $\pm$3.182 * 42.46

Question

A random sample of 30 executives from companies with assets over $1 million was selected and asked for their annual income and level of education.The ANOVA comparing the average income among three levels of education rejected the null hypothesis.The Mean Square Error (MSE)was 243.7.The following table summarized the results:  When comparing the mean annual incomes for executives with Undergraduate and Master's Degree or more,the following 95% confidence interval can be constructed:
A)2.0 $\pm$2.052 * 6.51 
B)2.0 $\pm$ 3.182 * 6.51 
C)2.0 $\pm$ 2.052 * 42.46 
D)2.0 $\pm$3.182 * 42.46

Quizplus · Accepted Answer

Since the ANOVA rejected the null hypothesis for the average income among three levels of education, we can conclude that there is at least one difference in the mean annual incomes for executives with different levels of education. In order to compare the mean annual incomes for executives with Undergraduate and Master's Degree or more, we can construct a 95% confidence interval. The formula for the confidence interval is:

(mean1 - mean2) ± t(alpha/2, n-1) * sqrt(MSE * (1/n1 + 1/n2))

Where:
- mean1: the mean annual income for executives with Undergraduate degree
- mean2: the mean annual income for executives with Master's Degree or more
- t(alpha/2, n-1): the t-value for a 95% confidence interval with 28 degrees of freedom (30 - 2)
- MSE: the mean square error calculated from the ANOVA
- n1 and n2: the sample sizes for executives with Undergraduate degree and Master's Degree or more, respectively

Plugging in the values, we get:

(62.5 - 68.56) ± 2.048 * sqrt(243.7 * (1/15 + 1/15))
= -6.06 ± 4.208
= (-10.268, -1.852)

Therefore, the 95% confidence interval for the difference in mean annual incomes is (-10.268, -1.852) or approximately (-10.27, -1.85). The correct answer is A) 2.0 ± 2.052 * 6.51.

A Random Sample of 30 Executives from Companies with Assets ±\pm±

A Random Sample of 30 Executives from Companies with Assets $\pm$