Week 2

10.30
A. hypotheses.
Ho: p (red) p (yellow) > 0
Ha: p (red) p (yellow) is not greater than 0

B. 1st Decision
One tail w/ alpha =1% Critical value is z = 2.326

C. examine proportions and z statistic.
p wear (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 test statistic is more than particular value, reject Ho

E. p - value and interpret it.
p - value = P(2.960988 < z < 10) = 0.00153
0.1533% run across that the test has stronger evidence against Ho.

F. If statistically pregnant, do you think the release is large enough to be important? If so, to whom, and why?
Yes it is statistically significant; yellow paint has reduced accidents.

G. Is the normality hypothesis fulfilled? 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

Interpret 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 number 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 before prescribing this drug widely? It could be the difference in family history and...If you want to get a expert essay, order it on our website: Orderessay



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