Answer:
Step-by-step explanation:
From the given data;
Suppose the level of significance [tex]\alpha = 0.99[/tex]
∴
[tex]\alpha = 1 - C.I[/tex]
[tex]\alpha = 1 - 0.99[/tex]
[tex]= 0.01[/tex]
[tex]\text{Critical Value} \ \ Z_{\alpha/2}= Z_{0.01/2}} \\ \\ = 2.576[/tex]
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:
[tex]n = \hat p (1- \hat p ) (\dfrac{Z}{E})^2[/tex]
[tex]n = 0.5 (1- 0.5 ) (\dfrac{2.576}{0.028})^2[/tex]
[tex]n = 0.5 (0.5 ) (8464)[/tex]
[tex]\mathbf{n = 2116}[/tex]
(b)
Suppose 53% of the adult are involved since it is not given, Then:
[tex]n = \hat p (1- \hat p ) (\dfrac{Z}{E})^2[/tex]
[tex]n = 0.53 (1- 0.53 ) (\dfrac{2.576}{0.028})^2[/tex]
[tex]n = 0.53 (0.47) (8464)[/tex]
[tex]\mathbf{n = 2109}[/tex]
(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.
