Question

Pinworm: In a prior sample of U.S. adults, the Center for Disease Control (CDC), found that 10% of the people in this sample had pinworm but the margin of error for the population estimate was too large. They want an estimate that is in error by no more than 2.5 percentage points at the 90% confidence level. Enter your answers as whole numbers.

(a) What is the minimum sample size required to obtain this type of accuracy? Use the prior sample proportion in your calculation.

The minimum sample size is U.S. adults.

(b) What is the minimum sample size required to obtain this type of accuracy when you assume no prior knowledge of the sample proportion?

The minimum sample size is U.S. adults.

Answer 1

Solution :

Given that,

margin of error = E = 2.5% = 0.025

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

(a)

= 10% = 0.10

1 - = 1 - 0.10 = 0.90

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.645 / 0.025)² * 0.10 * 0.90

= 389.66 = 390

The minimum sample size is U.S. 390 adults .

(b)

= 10% = 0.50

1 - = 1 - 0.50 = 0.50

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.645 / 0.025)² * 0.50 * 0.50

= 1082.41 = 1083

The minimum sample size is U.S. 1083 adults .