2. A total of 100 rats whose mothers were exposed to high levels of tobacco smok

Question

2. A total of 100 rats whose mothers were exposed to high levels of tobacco smoke during pregnancy were put through a simple maze. The maze required the rats to make a choice between going left or right at the outset. Eighty of the rats went right when running the maze for the first time. Assume that the 100 rats can be considered an SRS from the population of all rats born to mothers exposed to high levels of tobacco smoke during pregnancy. (Note that this assumption may or may not be reasonable, but researchers often assume lab rats are representative of such larger populations because they are often bred to have very uniform characteristics.) Let p be the proportion of rats in this population that would go right when running the maze for the first time. A 90% confidence interval for p is

a. 0.8 ± 0.066.

b. 0.8 ± 0.078.

c.0.8 ± 0.040.

Explanation / Answer

Sample size= n=100

Population proportion = p = 80

Sample proportion =p= 80/100 = 0.8

Standard error of p = SE(p) = p(1- p) / n = 0.8*0.2 / 100 = 0.04

Margin error = z* * SE(p) = 1.645*0.04 = 0.0658

1 – /2 = 1 – 0.1/2 = 0.95

Where z* is the table value from z table for 0.95 = 1.645

Confidence interval = p ± M.E = 0.8±0.066

A 90% confidence interval for p is (a) 0.8 ± 0.066

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