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posted by  Anonymous on 10/5/2008 9:31:14 PM  |  status: Live  

homework for stats class

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 In a bottling process, a manufacturer will lose money if the bottles contain either more or, less than is claimed on the label. The quality manager of a plant is interested in testing whether the mean ounces of catsup per family size bottle differ from the label by amount of 20 ounces. A sample of 9 bottles yields a mean fill of 19.7 ounces with a standard deviation of 0.3 ounces. Does the sample evidence indicate that the catsup-dispensing machines need adjustment at a 98% confidence?

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posted by "DONALD" on 10/6/2008 1:36:09 PM  |  status: Live
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A sample of 9(n) bottles yields a mean fill of 19.7(x-bar) ounces with a standard deviation of 0.3(s) ounces.
The sample evidence indicate that the catsup-dispensing machines need adjustment at a 98% confidence.
Null Hypothesis: μ = 20
Alternative Hypothesis: μ ≠ 20
The test statistic is given by
The critical value of 't' at 0.02, the level of significance is 2.896
Since the calculated value of '|t|' is greater than the critical value of 't' so we reject the null hypothesis and conclude that the mean ounces of catsup per family size bottle differ from the label by amount of 20 ounces.
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