Week 2

10.30 A. hypotheses. Ho: p (red) p (yellow) > 0 Ha: p (red) p (yellow) is not greater than 0 B. 1st closing One tail w/ important =1% unfavorable value is z = 2.326 C. examine proportions and z statistic. p deliver (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 streak statistic is much than particular value, reject Ho E. p - value and project it. p - value = P(2.960988 < z < 10) = 0.00153 0.1533% run across that the test has stiffer evidence against Ho. F. If statistically pregnant, do you think the release is grownup enough to be important? If so, to whom, and why? Yes it is statistically important; yellow paint has reduced accidents. G. Is the normality hypothesis action? 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 constitute the results at alpha = .01 Where the p-value is less than alpha you should reject Ho; there is significant evidence that the use o f this medicine reduces the digit 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 beginning prescribing this drug widely? It could be the difference in family chronicle and...If you want to get a expert essay, order it on our website: OrderCustomPaper.com

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