Question

When testing a random sample of 180 patients with a disease, the procedure yielded 16 “false negatives”.

a. Estimate the true proportion of all cases the diagnostic will yield a false negative, using 95% confidence. (18)

b. Verify that the sample size used in part A is large enough for the procedure to be considered valid. (3)

c. How many cases would need to be sampled to estimate the desired proportion to within a margin of error of 2%, using 95% confidence? (8)

Answer 1

a)
sample proportion, = 0.0889
sample size, n = 180
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.0889 * (1 - 0.0889)/180) = 0.0212

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0212
ME = 0.042

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.0889 - 1.96 * 0.0212 , 0.0889 + 1.96 * 0.0212)
CI = (0.047 , 0.130)

b)

np >= 5
180 * 0.0889 = 16

nq > =5

180 * (1-0.0889) = 164

c)

The following information is provided,
Significance Level, α = 0.05, Margin of Error, E = 0.02

The provided estimate of proportion p is, p = 0.0889
The critical value for significance level, α = 0.05 is 1.96.

The following formula is used to compute the minimum sample size required to estimate the population proportion p within the required margin of error:
n >= p*(1-p)*(zc/E)^2
n = 0.0889*(1 - 0.0889)*(1.96/0.02)^2
n = 777.89

Therefore, the sample size needed to satisfy the condition n >= 777.89 and it must be an integer number, we conclude that the minimum required sample size is n = 778
Ans : Sample size, n = 778

if ou take p value upto 3 or 2 decimal answer would be change