Question

In a clinical study, 3900 healthy subjects aged 18-49 were vaccinated with a vaccine against a seasonal illness. Over a period of roughly 28 weeks, 25 of these subjects developed the illness Complete parts a through e below. Click here to view the table of standard normal cumulative. Rrobabilities (page 1). Click here to view the table of standard normal cumulative probabilities Ypage 2). a. Find the point estimate of the population proportion that were vaccinated with the vaccine but still developed the illness. The point estimate is (Round to five decimal places as needed.) b. Find the standard error of this estimate. The standard error of this estimate is (Round to five decimal places as needed.) c. Find the margin of error for a 95% confidence interval. The margin of error is (Round to five decimal places as needed.) d. Construct the 95% confidence interval for the population proportion. Interpret the interval. The 95% confidence interval for the population proportion is 0. (Round to five decimal places as needed.)

Answer 1

Given n=3900 , developed the illness, X=25

a.) Estimate of the population proportion that were vaccinated with the vaccine but still developed the illness is given by sample proportion i.e.

$p=\frac{X}{n}=\frac{25}{3900}$

р 0.00641

b.)Now standard error of this estimate is given by

$S.E(p)=\sqrt\frac{PQ}{n}$

Since P is not known taking p as an estimate of P then

$S.E(p)=\sqrt\frac{pq}{n}$

$S.E(p)=\sqrt\frac{0.00641*(1-0.00641)}{3900}$

$S.E(p)=\sqrt\frac{0.006359}{3900}$

$S.E(p)=0.00128$

c.) Margin of error is given by

M.E 20/2 * S.E(P)

For 95% level of significance

a = 1-0.95 = 0.05

Critical value at alpha 0.05

$z_{\alpha/2}=\pm 1.96$

Therefore,

$M.E=\pm 1.96*0.00128$

$M.E=\pm0.00251$

d.) 95% confidence interval for population proportion is given by

$C.I=p\pm M.E$

$C.I=0.00641\pm0.00251$

C.I = 0.00390, 0.00892

Hence we estimate with 95% confidence that true population proportion will lie between 0.00390 and 0.00892.