A random sample of 20 female executives from companies with assets over $1 milli

Question

A random sample of 20 female 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 Error (MSE) was 250. The following table summarized an Square the results: High School Undergraduate D Masters Degree or More or Les Number Sam 93 100 Mean Salary (S1,000s) To compare the mean annual incomes of female executives with a high school education or less and female a master's degree or more, compute the executives with 90% confidence interval. Given the following ANovA table for three treatments each with six observations: Source Sum of Squares dif Mean Square 1116 Treatment 1068 Error 2184 Total What is the decision regarding the null hypothesis?

Explanation / Answer

I am able to give you answer for second problem.

Using Minitab,

Test and CI for Two Variances

Method

Null hypothesis (First) / (Second) = 1
Alternative hypothesis (First) / (Second) 1
Significance level = 0.01

F method was used. This method is accurate for normal data only.

Statistics

99% CI for
Sample N StDev Variance StDevs
First 10 4.733 22.400 (2.923, 10.780)
Second 10 3.464 12.000 (2.140, 7.890)

Ratio of standard deviations = 1.366
Ratio of variances = 1.867

99% Confidence Intervals

CI for
CI for StDev Variance
Method Ratio Ratio
F (0.534, 3.494) (0.285, 12.210)

Tests

Test
Method DF1 DF2 Statistic P-Value
F 9 9 1.87 0.366

Decision: For the given data, the high p-values for tests 0.366 (P-value <0.01) indicate that the standard deviations are not different.

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