Question

Please find either the sample size or maximum error for the following:

a) Sample size is 25, s is 12 and alpha is 5%.

b) Maximum error = 2.3, sigma is 15.4 and CL = 95%.

c) Given 90 nurses, 54 of them have full benefits; confidence level is 98%.

Answer 1

Ansa:

n=25

s=12

=0.05

Maximum error=t/2,(n-1)*s/sqrt(n)

t/2,(n-1) =t0.05/2,24=2.0639

Maximum error=2.0639*12/sqrt(25)

Maximum error=4.95

Ansb:

Maximum error = 2.3, sigma is 15.4

Z value for 95 % CI is 1.96

n=(Z*sigma/ Maximum error)

n=(1.96*15.4/2.3)

n=21.176

sample size=21

Ansc:

confidence level is 98%.

n=90 x=54

p^=x/n=54/90=0.6

Z value for 98% CI is 2.3264

(p^z*sqrt(p^*(1-p^)/n)

(0.62.3264sqrt(0.6*0.4/90)

(0.60.1201)

confidence level is 98%. is (0.48,0.72)