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

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 $\pi$, and a confidence interval $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

Z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

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 $\pi = \frac{44}{75} = 0.586$.

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 $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

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

The upper limit of this interval is:

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

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