Question

In an experiment to compare the tensile strengths of I = 4 different types of copper wire, ) = 3 samples of each type were used. The average between group and average within group estimates were computed as MST = 2676.3 and MSE = 1190.2, respectively. Often hypothesis tests are conducted using what is referred to as the rejection region method. This compares the calculated F test statistic to an F value that cuts off a "rejection region". If the calculated test statistic falls within the rejection region, the null is rejected. The F value that represents the cut off for the rejection region, is the F value that cuts off the top a% of area in an F distribution with 1-1 and 1(3-1) degrees of freedom. In other words, if a = 0.05, the F rejection region cut off will be the 95th percentile from the appropriate F distribution. Find the F value that represents the cut off for the rejection region when a = 0.05. Use the percentile command in R qf(percentile, dfi, df2). Round your answers to two decimal places. rejection region f2 Calculate the F test statistic F to test Ho: u1 = u2 = ... = 44 versus Ha: at least two uj's are unequal. Use a significance level 0.05. Round your answers to two decimal places. test statistic f= Based on the rejection region method, what can we conclude from the Single Factor ANOVA F test? Fail to reject Ho. The data indicates a difference in the mean tensile strengths. . Fail to reject Ho. The data indicates there is no difference in the mean tensile strengths. . Reject Ho. The data indicates a difference in the mean tensile strengths. Reject Ho. The data indicates there is no difference in the mean tensile strengths.

Answer 1

for (4-1=3,4*3-4=8) degree of freedom and 0.05 level, rejection region f>=4.07

test statistic f =2676.3/1190.2 =2.25

fail to reject Ho, the data indicates there is no difference,,,,,,,