This Test: 200 pts possible This Question: 20 pts common oolds in a test of the
ID: 3296976 • Letter: T
Question
This Test: 200 pts possible This Question: 20 pts common oolds in a test of the effctiveness of echinaces, 43 of the 48 subjects treated with chinces developed iinwinus inections in a placebo group. 77 of the 32 sujets hinovirus inlections Use a 0.0 signficance level to Seat the claim that echinace as an efect on hinovinus infections Complete parts (a) trough (e) below a Test the dlaire using a hypothesis hypothesis tes Round to twe decimal places as meeded) dentily the Pal Round to three decimal giaces as needed) Round to thee decimal places s needed) Click to select your anwer 3 cops lockExplanation / Answer
Question a)
Hypothesis
Answer: Option A
Test Statistics:
z = (p1^ - p2^) / SE (p1-p2)
SE (p1-p2) = sqrt (P*Q*((1/n1)+(1/n2))
P = (x1+x2)/(n1+n2)
x1 = 43
n1 = 48
x2 = 77
n2 = 92
p1^ = 0.895833
p2^ = 0.836957
P = 0.857143
Q = 0.142857
SE (p1-p2) = 0.0623
z = (0.895833-0.836957)/0.0623
z = 0.94
P-value = 0.174
The p-value is greater than the significance level of a = 0.05 so fail to reject the null hypothesis. There is not sufficient evidence to support the claim that ....
Question b)
(p1^ -p2^) (-/+) SE (p1-p2)
(0.895833 - 0.836957) (-/+) (1.96 * 0.0623)
-0.056 < (p1-p2) < 0.174
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