Res 342 Week 2. Ex.

9.54 The firm examined 35 randomly chosen fax transmissions during the succeeding(a) year, teachable a sample mean of 14.44 with a specimen departure of 4.45 pages. (a) At the .01 level of significance, is the original mean greater than 10? Ho: u <= 10 Ha: u > 10 exemplar mean = 14.44 S = 4.45 CX = .01 test statistic: t = 5.7582.. p-value = P(t > 5.9028.. with df=34) = 0.0000005758 (14.44 10 / 4.45 sqrt 35 = 4.45/.7522 = 5.90) Since the p-value is less than 1%, reject Ho. Hypotheses: H0: ? ? 10 Ha: ? > 10 (claim) Critical Value: ? = 0.01, for a one-tailed test, zcrit = 2.236 mental testing Value: ztest = (xbar - ?) / [ s/?n ] ztest = (14.44 - 10) / [ 4.45/?35 ] ztest = 5.9027 null assumption = rejected analysis: insufficient evidence to support the true mean is greater than 10. 9.62 The Web-based confederation Oh Baby! Gifts has a tendency of processing 95 per centum of its orders on the like day they argon received. I f 485 out of the next 500 orders are processsed on the same day, would this prove that they are exceeding their goal, using a = .025? Hypotheses: H0: p ? 0.95 Ha: p > 0.95 (claim) Critical Value: ? = 0.025, for a one-tailed test, zcrit = 1.96 N = 500 tt = 0.95 p = 485/500 = .97 x = .025 Z = P tt / sqrt tt (1-tt) / N .97 - .95 sqrt .95 (1-.95)/500 = .02/.009 ? = np = (500)(0.95) = 475 ? = ?npq = ?(0.95)(0.05)(500) = 4.8734 ztest = (X - ?) / ? ztest = (485 - 475) / 4.8734 ztest = 2.06 Decision: ztest > Reject the null hypothesis Summary: Statistics foundert (prove) anything, they can only provide a probability.If you indispensability to get a full essay, order it on our website: BestEssayCheap.com

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