Question

You want to estimate the percentage of adults who believe that passwords should be replaced with biometric security​ (such as​ fingerprints). How many randomly selected adults must you​ survey? Assume that you want to be ​% confident that the sample percentage is within percentage of the true population percentage. Complete parts​ (a) through​ (c) below.
a. Assume that nothing is known about the percentage of adults who believe that passwords should be replaced with biometric security.
b. Assume that a prior survey suggests that about 53 % of adults believe that biometric security should replace passwords.
c. Does the additional survey information from part (b) have much of an effect on the sample size that is required?
The additional survey information from part (b) causes the required sample size to change by ______10%. Based on this, the additional survey information causes ________in the sample size that is required.

Answer 1

Answer:

Step-by-step explanation:

From the given data;

Suppose the level of significance $\alpha = 0.99$

∴

$\alpha = 1 - C.I$

$\alpha = 1 - 0.99$

$= 0.01$

$\text{Critical Value} \ \ Z_{\alpha/2}= Z_{0.01/2}} \\ \\ = 2.576$

This shows that 99% of the area under the standard normal curve falls within 2.576 S.D of the mean.

(a)

Using the knowledge of M.O.E (margin of error); we can make "n" the subject and have a derived formula:

$n = \hat p (1- \hat p ) (\dfrac{Z}{E})^2$

$n = 0.5 (1- 0.5 ) (\dfrac{2.576}{0.028})^2$

$n = 0.5 (0.5 ) (8464)$

$\mathbf{n = 2116}$

(b)

Suppose 53% of the adult are involved since it is not given, Then:

$n = \hat p (1- \hat p ) (\dfrac{Z}{E})^2$

$n = 0.53 (1- 0.53 ) (\dfrac{2.576}{0.028})^2$

$n = 0.53 (0.47) (8464)$

$\mathbf{n = 2109}$

(c)

Here; we realize that the additional information brings about a change by less than 10%. Hence, the additional information had NO significant change in the required sample.