A researcher at a major hospital wishes to estimate the proportion of the adult population of the United States that has high blood pressure. How large a sample is needed in order to be 98% confident that the sample proportion will not differ from the true proportion by more than 6%

Respuesta :

Answer:

The sample size 'n' = 347

Step-by-step explanation:

Explanation:-

Step(i):-

Given Margin of error = 6% =0.06

The Margin of error is determined by

[tex]M.E=\frac{Z_{\frac{\alpha }{2} }\sqrt{p(1-p)} }{\sqrt{n} }[/tex]  

98% of Z-value

[tex]Z_{\frac{\alpha }{2} } = Z_{\frac{0.02}{2} } = Z_{0.01} =2.326[/tex]

Step(ii):-

We know that       [tex]\sqrt{p(1-p} \leq \frac{1}{2}[/tex]

[tex]0.06 = \frac{2.236X\frac{1}{2} }{\sqrt{n} }[/tex]

Cross multiplication , we get

[tex]0.06 \sqrt{n} = 2,236 X \frac{1}{2}[/tex]

[tex]0.06 \sqrt{n} = 1.118[/tex]

[tex]\sqrt{n} = \frac{1.118}{0.06} = 18.63[/tex]

Squaring on both sides, we get

n = 347.20≅347

Final answer:-

The sample size 'n' = 347