Question

One hundred rats with mothers that were exposed to high levels of tobacco smoke during pregnancy were put through a simple maze. At the outset, the maze required the rats to make a choice between going left and going right. Eighty of the tats 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 who were exposed to high levels of tobacco smoke during pregnancy. (Note that this assumption may or may not be reasonable, but researchers often assume that lab rats are representative of large populations because they are often bred to have uniform characteristics.) Let p be the proportion of rats in this population that would go right when running the maze for the first time. Determine a 90% confidence interval for p. Show as much work as is possible.

Answer 1

The 90% confidence interval for pis (0.7342, 0.8658).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

For this problem, we have that:

100 rats mothers, 80 went right. So
`n = 100, \pi = (80)/(100) = 0.8`

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a pvalue of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.8 - 1.645\sqrt{(0.8*0.2)/(100)} = 0.7342`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.8 + 1.645\sqrt{(0.8*0.2)/(100)} = 0.8658`

The 90% confidence interval for pis (0.7342, 0.8658).