A third of 20 I don't have a clue how to do it
1 answer:
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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.
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:
The upper limit of this interval is:
The 95% confidence interval for the true proportion of mice that will test positive under similar conditions is (0.5291, 0.6429).
using Pythagorean triplet