# 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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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

###### Education

- BSc , Wuhan Univ. China
- MA, Shandong Univ.

###### Recent Feedback

- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
- "Thank you"
- "Thank you very much for your valuable time and assistance!"

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