Respuesta :
SOLUTION
Step 1 :
We need to know how large a sample is necessary.
Let p denotes the United States Homes have a direct satellite TV Receiver is 25 % or 0. 25.
From the information, we know that E is 3 % or 0.03 and the confidence interval of 95%.
When p is known, then the sample size formula is :
[tex]\begin{gathered} n\text{ = }^{}\frac{\lbrack Z_{\frac{\alpha}{2}}\rbrack^2}{E^2}\text{ X p x q} \\ \text{where p + q = 1} \end{gathered}[/tex]When p is unknown, then the sample size formula is:
[tex]n\text{ = }\frac{\lbrack Z_{\frac{\propto}{2}}\rbrack^2}{E}\text{ X 0.25}[/tex]Now, we have to determine:
[tex]Z_{\frac{\alpha}{2}}[/tex]The confidence level is :
[tex]1\text{ - }\alpha[/tex]We know that:
[tex]\begin{gathered} \alpha\text{ = 95\% = 0.95} \\ 1\text{ - }\alpha\text{ = 1 - 0.95} \\ \text{1 - }\alpha\text{ = 0.05} \\ Z_{\frac{\alpha}{2}\text{ }}=\text{ 0.05/ 2} \end{gathered}[/tex][tex]Z_{\frac{0.05}{2}\text{ }}=\text{ 1.96 ( using table value )}[/tex]If we use the known p, then the sample size is :
[tex]\begin{gathered} n\text{ = }\frac{(1.96)^2}{(0.03)^2}\text{ X ( 0.25 ) ( 0.75 )} \\ n\text{ = }\frac{3.8416}{0.0009}\text{ x 0.1875} \\ \text{n = 4268.44444 x 0.1875} \\ n\text{ = 800.333} \\ n\text{ }\approx\text{ 801} \end{gathered}[/tex]If we use the known p, then the sample size is 801.
Therefore, a sample n is 801.
Step 2 :
If we use the un-known p , then the sample size is :
[tex]\begin{gathered} n\text{ = }\frac{(1.96)^2}{(0.03)^2}\text{ x 0. 25} \\ \text{n = }\frac{3.8416}{0.0009}\text{ x 0.25} \\ n\text{ = 4268.44444 x 0.25} \\ \text{n = 1067. 11111} \\ n\approx\text{ 1068} \end{gathered}[/tex]If we use the unknown p, then the sample size is approximately 1068.
Therefore, a sample n is 1068.
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