Question

4. A study was conducted to estimate the prevalence of mutated gene among residents of a town neighbouring a nuclear reactor. Two hundred and twenty-eight of out 1000 subjects tested positive for some form of mutation. a) Determine a 99% confidence interval to estimate the prevalence of this mutation. 141 b) What information does the interval in (a) convey? [1] c) What sample size would be required for the researcher to estimate the true prevalence rate, such that he/she can be 99% confident, and have a margin of error no greater than 5%? 141

Answer 1

(a)

n = 1000

$\hat{p}$ = 228/1000 = 0.228

$\alpha$ = 0.01

From Table, critical values of Z_C = $\pm$ 2.576

Confidence Interval:

pEZCV px (1-P) 2=0.228+2.576 0.228 x (1 – 0.228) - 2 = (0.194,0.262) 1000 n

So,

Answer is:

(0.194, 0.262)

(b)

The 99% confidence interval (0.194, 0.262) is a range of values we are 99% will contain the unknown true population proportion.

(c)

Sample Size (n) is given by:

$n=\frac{Z^{2}_{\alpha /2}\times p\times (1-p)}{e^{2}}$

Given:

$\alpha$ = 0.01

From Table, critical values of Z_C = $\pm$ 2.576

p = 0.228

e = 0.05

Substituting, we get:

$n=\frac{2.576^{2}\times 0.228\times (1-0.228)}{0.05^{2}}=468$

So,

Answer is:

468