Answer:
The 95% confidence interval for the true proportion of mice that will test positive under similar conditions is (0.5291, 0.6429).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence interval
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
Z is the zscore that has a pvalue of
.
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
and
.
Compute a 95% confidence interval for the true proportion of mice that will test positive under similar conditions.
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
![\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.586 - 1.96\sqrt{\frac{0.586*0.414}{75}} = 0.5291](https://tex.z-dn.net/?f=%5Cpi%20-%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.586%20-%201.96%5Csqrt%7B%5Cfrac%7B0.586%2A0.414%7D%7B75%7D%7D%20%3D%200.5291)
The upper limit of this interval is:
![\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.586 + 1.96\sqrt{\frac{0.586*0.414}{75}} = 0.6429](https://tex.z-dn.net/?f=%5Cpi%20-%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.586%20%2B%201.96%5Csqrt%7B%5Cfrac%7B0.586%2A0.414%7D%7B75%7D%7D%20%3D%200.6429)
The 95% confidence interval for the true proportion of mice that will test positive under similar conditions is (0.5291, 0.6429).