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    Bounce Back Glass Backboards

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    My problem is that I have been doing a group project with no one else participating. So I have formulated my own results, but think that I might be a little off track. So I am trying to seek some help with the problem and see if I am going in the right direction. I have attached a copy of the problem.

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    Solution Summary

    The makers of Bounce Back glass backboards for basketball gymnasiums have claimed that their board is at least as durable, on the average, as the leading backboard made by Swoosh Company. Products Testing Services of Des Moines, Iowa, was hired to verify this claim. It selected a random sample of 50 backboards of each type and subjected the boards to a pressure test to determine how much weight hung from a basketball rim it would take to break the fiberglass backboard. The following results were determined from the testing process.
    SWOOSH BOUNCEBACK
    = 691 lb  = 653 lb
     = 112 lb  = 105 lb
    In your groups, complete the following tasks.
    a. Assuming that the more pounds needed to break the backboard, the better it is, state the appropriate null and alternative hypotheses.
    b. At a significance level of 0.01, what conclusion should be reached with respect to the claim made by the Bounce Back Company? Discuss.
    c. Suppose the hypothesis test was conducted at a significance level of 0.10 instead of 0.01. Would this change the conclusion reached based on the sample data? If so, discuss why; if not, discuss why not.
    Now suppose the Bounce Company also claims that its boards are more consistent with the Swoosh boards in terms of strength. Specifically, the Bounce Back Company claims the standard deviation in weight required to break its boards is less than that for the Swoosh boards.
    d. State the appropriate null and alternative hypotheses.
    e. Based on the sample data and an alpha of 0.05, what should Products Testing Services conclude about the variability of the two boards? Discuss your results in a letter to the two companies.
    You may use the following Web resources to find the critical F values:
    F distribution calculator for alpha=0.01, 0.05

    $2.49

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