Question

Medical scientists study the effect of acute infection on tissue-specific immunity. In a collection of experiments under the same conditions, 44 of 75 mice test positive for lymphadenopathy. Compute a 95% confidence interval for the true proportion of mice that will test positive under similar conditions.

Answer 1

The 95% confidence interval for the true proportion of mice that will test positive under similar conditions is (0.5291, 0.6429).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence interval
`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:

In a collection of experiments under the same conditions, 44 of 75 mice test positive for lymphadenopathy. This means that
`n = 75` and
`\pi = (44)/(75) = 0.586`.

Compute a 95% confidence interval for the true proportion of mice that will test positive under similar conditions.

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.586 - 1.96\sqrt{(0.586*0.414)/(75)} = 0.5291`

The upper limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.586 + 1.96\sqrt{(0.586*0.414)/(75)} = 0.6429`

The 95% confidence interval for the true proportion of mice that will test positive under similar conditions is (0.5291, 0.6429).