Res 342 Week 2. Ex.

9.54 The firm examined 35 randomly chosen fax transmissions during the succeeding(a) year, obedient a sample mean of 14.44 with a specimen discrepancy of 4.45 pages. (a) At the .01 level of significance, is the professedly 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 strain Value: ztest = (xbar - ?) / [ s/?n ] ztest = (14.44 - 10) / [ 4.45/?35 ] ztest = 5.9027 null venture = rejected analysis: insufficient evidence to support the true mean is greater than 10. 9.62 The Web-based confederation Oh Baby! Gifts has a terminus of processing 95 per centum of its orders on the similar day they atomic number 18 rec eived. If 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 wear downt (prove) anything, they can only provide a probabilit y.If you urgency to get a full essay, order! it on our website: BestEssayCheap.com

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