Week 2

10.30 A. hypotheses. Ho: p (red) p (yellow) > 0 Ha: p (red) p (yellow) is not greater than 0 B. 1st conclusion One tail w/ important =1% exact value is z = 2.326 C. examine proportions and z statistic. p fag out (red) = 20/153,348 ; p hat (yellow) = 4/135,035 For Pooling: p relegateber = (20+4) / (153348+135035) = 0.00002.08 z = [(20/153,348) - (4/135,035)] / sqrt [(0.00002.08)(0.999979) / 153348 + (0.00002.08)*0.999979/135035)] = 2.960988... D. 2nd Decision sightedness as testing statistic is more than than particular value, reject Ho E. p - value and stage it. p - value = P(2.960988 < z < 10) = 0.00153 0.1533% run across that the test has loyaler evidence against Ho. F. If statistically pregnant, do you think the release is liberal enough to be important? If so, to whom, and why? Yes it is statistically noteworthy; yellow paint has reduced accidents. G. Is the normality hypothesis finish? Explain. Yes: p1n1 > 5, q1n1 > 5 ; p2n2 > 5, p2n2 > 5 10.44 A. Hypotheses. Ho: p(inactive) - P(active) = 0 Ha: p(inactive) - p(active) > 0 p-hat(inactive) = 97 / 2081 = 0.0466 p-hat(active) = 57 / 2325 = 0.0245 Pool the data you get p - bar = (97 + 57) / (2081 + 2325) = 0.0350 and q - bar = 1 p - bar = 0.9650 alpha = 1%, critical value z = 2.326 B.

Test statistic and p-value. z(0.0466 - 0.0245) = 0.0221 / sqrt [(0.035 * 0.965 / 2081) + (0.035 * 0.965 / 2325)] 3.9849, p-value - 0.0000338 stand for the results at alpha = .01 Where the p-value is less than alpha you should reject Ho; there is significant evidence that the use of this medicine reduces the itemise of dea! ths. C. Is normality aware? 2081 * 0.035 = 72.84 > 5 and 2081 * 0.965 > 5 2325 * 0.035 = 70.86 > 5 and 2325 * 0.965 > 5 D. Is the residue large enough to be important? p-value has strong evidence that Ha is true. E. What else would medical researchers lack to know onwards prescribing this drug widely? It could be the difference in family biography and...If you want to get a expert essay, order it on our website: OrderCustomPaper.com

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