A laboratory technician wants to compare the breaking strength of three kinds of

Question

A laboratory technician wants to compare the breaking strength of three kinds of thread, and originally he planned to repeat each determination six times. Not having enough time, however, he has to base his analysis on the following results: #2 A laboratory technician wants to compare the breaking strength of three kinds of thread, and originally he planned to repeat each determination six times. Not having enough time, however, he has to base his analysis on the following results (in ounces): Thread 1: 18.0, 16.4, 15.7, 19.6, 16.5 18.2 Thread 2: 21.1, 17.8, 18.6, 20.8, 17.9, 19.0 Thread 3: 16.5, 17.8, 16.1 Use the following table to perform an analysis of variance to test at the 0.05 level of significance whether the differences among the sample means are significant Source of Degrees of Sum of Squares Mean Square Variation freedom Treatment Values of the Test Statistic F F: Variance) 15.12 Error Total 37.46 a) Find the missing values and complete the ANOVA table. b) Specify the null and alternative hypotheses to investigate whether a difference exists among the treatment means c) Give the rejection region for this test. Use .01. d) what is the critical value of F for = .01?

Explanation / Answer

Solution:

a. There are three treatments, Thread 1, 2 and 3. k = 3

Total sample size = 6 + 6 + 3 = 15, n = 15

b. Null Hypothesis (Ho): There is no difference among the treatment means.

Alternative Hypothesis (Ha): There is difference among the treatment means.

c. If F-statistics > F-critical value, we reject Ho.

d. Using F-tables, the critical value at a = 0.01 with (2, 12) degrees of freedom

F (0.01, 2, 12) = 6.927

Source of variation Degrees of freedom Sum of squares Mean square (variance) Test Statistic F Treatment k - 1 = 3 - 1 = 2 15.12 15.12/2 = 7.56 7.56/1.862 = 4.061 Error n - k = 15 - 3 = 12 37.46-15.12 = 22.34 22.34/12 = 1.862 Total n - 1 = 15 - 1 = 14 37.46

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