Question

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%

Answer 1

The sample size 'n' = 347

Explanation:

Explanation:-

Step(i):-

Given Margin of error = 6% =0.06

The Margin of error is determined by

`M.E=\frac{Z_{(\alpha )/(2) }√(p(1-p)) }{√(n) }`

98% of Z-value

`Z_{(\alpha )/(2) } = Z_{(0.02)/(2) } = Z_(0.01) =2.326`

Step(ii):-

We know that
`√(p(1-p) \leq (1)/(2)`

`0.06 = (2.236X(1)/(2) )/(√(n) )`

Cross multiplication , we get

`0.06 √(n) = 2,236 X (1)/(2)`

`0.06 √(n) = 1.118`

`√(n) = (1.118)/(0.06) = 18.63`

Squaring on both sides, we get

n = 347.20≅347

Final answer:-

The sample size 'n' = 347