A drug company has introduced a new test for the flu. In a random sample of 150 US adults, the new test is accurate for 125 adults.
a. Find the 99% confidence interval for the proportion of accurate tests for all US adults.
b. Interpret the confidence interval in the words of the problem.
c. Find the error bound.
d. Does this data and analysis provide evidence so that the company can claim its new test is accurate in more than 80% of all US adults? Justify!
Answer:
Given,
sample n = 150
x = 125
Sample proportion p^ = x/n
= 125/150
= 0.833
a)
Here for 99% CI, z value is 2.58
Consider,
Interval = p^ +/- z*sqrt(p^(1-p^)/n)
substitute the known values
= 0.833 +/- 2.58*sqrt(0.833(1-0.833)/150)
= 0.833 +/- 0.0786
= (0.7544 , 0.9116)
b)
Here we are 99% confident that the true population is i.e., lies b/w (0.7544 , 0.9116)
c)
Margin of error = E = sqrt(p^(1-p^)/n)
substitute values
= sqrt(0.833(1-0.833)/150)
= 0.0305
d)
Yes, the company can claim its new test which is more than 80% of all US adults.
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