A random sample of 40 companies with assets over $10 million was surveyed and asked to indicate their industry and annual computer technology expense. The ANOVA comparing the average computer technology expense among three industries rejected the null hypothesis. The mean square error (MSE)was 195. The following table summarized the results: When comparing the mean annual computer technology expense for companies in the education and tax services industries,which of the following 95% confidence interval can be constructed?
A)13.5 ± 2.026 * 5.78
B)13.5 ± 2.021 * 5.78
C)13.5 ± 2.026 * 13.96
D)13.5 ± 2.021 * 13.9

Question

A random sample of 40 companies with assets over $10 million was surveyed and asked to indicate their industry and annual computer technology expense. The ANOVA comparing the average computer technology expense among three industries rejected the null hypothesis. The mean square error (MSE)was 195. The following table summarized the results:  When comparing the mean annual computer technology expense for companies in the education and tax services industries,which of the following 95% confidence interval can be constructed?
A)13.5 ± 2.026 * 5.78 
B)13.5 ± 2.021 * 5.78 
C)13.5 ± 2.026 * 13.96 
D)13.5 ± 2.021 * 13.9

Quizplus · Accepted Answer

Since the ANOVA rejected the null hypothesis, we can conclude that there is a significant difference in the mean computer technology expense among the three industries. Thus, we can construct a confidence interval for the difference between the mean annual computer technology expense for the education and tax services industries.

Using the formula for the confidence interval for the difference between two means:

Difference in means ± t* (pooled standard error)

We can calculate the pooled standard error as follows:

sqrt(MSE/n1 + MSE/n2) = sqrt(195/18 + 195/22) = 5.78

where n1 and n2 are the sample sizes for the education and tax services industries, respectively.

Since we are constructing a 95% confidence interval, we can find the corresponding t-value with degrees of freedom equal to the total sample size minus two (38):

t = 2.026

Finally, we can plug in the values to get the interval:

13.5 ± 2.026 * 5.78

which simplifies to:

(2.22, 24.78)

Therefore, the best choice is A.

A Random Sample of 40 Companies with Assets Over $10