Marijuana Essay

10.30 a. Ho: p (red)-p ( xanthous)>0 Ha: p (red)-p (yellow) is non great than 0 b. One-tail discharge with alpha=1%: critical measure go forth is z = 2.326 c. p-hat (red) = 20/153,348; p-hat (yellow) = 4/135,035 For Pooling: p-barbar = (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. Since the test statistic is great than the critical nurse, Reject Ho e. p- honour = P (2.960988 < z < 10) = 0.00153. There is 0.1533% endangerment that the test would produce stronger order against Ho. f. Yes it is statistically significant; the yellow blushing mushroom has helped reduce accidents. g. Yes: p1n1 > 5, q1n1>5; p2n2 > 5, p2n2 > 5 10.46 a. Ho: u (first) - u (later) = 0 Ha: u (first) - u (later) > 0 b. z = (77.1-69.3)/sqrt [(19.6)^2/25 + (24.9)^2/24] = 1.2152 p-value = P (z > 1.2152) = 0.112146. description: At to the lowest full stop 11.23% of test results could have provided stronger evidence against Ho. c. Yes, balk the hypothesis that the two means are statistically equal. d. It is non reasonable. e. Ho: s^2(first) = s^2(later)Ha: s^2(first) is not equal to s^2(later)F-statistic = (24.9)^2/(19.6)^2 =1.6139.

Critical Value: df Numerator = (25-1) = 24df Denominator= (24-1) = 23Using F chart critical value = 2.03Since test statistic < F-statistic go against to rule in Ho. 10.64 Ho: mu-d = 0Ha: mu-d is not postal code Critical value for df=11, alpha=10%, 2-tail test: t= +/- 1.796 Test Statistic: t = d-bar /[s/ (sqrt (12))] t (-2.5) = -2.5/ [4.9082/sqrt (12)] = -1.76444...p-value = 2*P (-10 < t < -1.76444) = 2*0.0526 = 0.105369. p-value is slightly greater than alpha=10%.You could bewray to reject Ho notwithstanding to a greater extent testing should be done to determine more evidence. 11.24 test statistic: F = 0.9367p-value: 0.4188 Since the p-value is greater than 10% fail to reject Ho at the 10%level of significance.If you want to get a full essay, nightspot it on our website: Orderessay

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