Week 2

10.30 A. hypotheses. Ho: p (red) p (yellow) > 0 Ha: p (red) p (yellow) is not greater than 0 B. 1st finis One tail w/ alpha =1% tiny value is z = 2.326 C. examine proportions and z statistic. p offend (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 demonstrate statistic is to a greater extent than particular value, reject Ho E. p - value and generate it. p - value = P(2.960988 < z < 10) = 0.00153 0.1533% run across that the test has gruellinger evidence against Ho. F. If statistically pregnant, do you think the release is walloping 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 carry through? Explain. Yes: p 1n1 > 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 take the results at alpha = .01 Where the p-value is less than alpha you should reject Ho; there is significant evidence th at the use of this medicine reduces the far! e of deaths. 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 in the lead prescribing this drug widely? It could be the difference in family record and...If you want to get a expert essay, order it on our website: OrderCustomPaper.com

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