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O e. helpful variance QUESTION 18 Discussion file #18 Our existing process produces golf balls that have an average carry distance of, 250.4164 yards. We have developed a new ball that we feel goes farther than the existing ball. We test new ball and the test results are shown below. Summary statistics: Column n Mean Variance Std. Dev. Std. Err. Brand B 250 251.11317 18.375103 4.286619 0.27110958 Based on the sample data and the proper hypothesis test at the alpha = .005 level, what is your conclusion and why? Conclusion is to reject the null. The mean for the new ball is not the same as the existing because the test `statistic is not as extreme as the critical value Conclusion is to reject the null. The mean for the new ball is greater than that of the existing because the test statistic is more extreme than the critical value Conclusion is to reject the null. The mean for the new ball is not the same as the existing because the critical value is more extreme than the test statistic d Conclusion is to accept the null. The mean for the new ball is the same as the existing, because the test statistic is not as extreme as the critical value Oe. Conclusion is to fail to reject the null. The mean for the new ball is the same as the existing, because the "test statistic is not as extreme as the critical value QUESTION 19 Click Save and Submit to save and submit. Click Save All Answers to save all answers. Save All Answe