problem-7.11
problem-7.11
We have z* \SE(\phat) £ 0.01. We know z* as a is
given, so we have
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\frac\phat(1-\phat)n
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£ \frac0.01z*. |
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Solving for n we get
(\fracz*0.01)2 \phat(1-\phat) £ n. |
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No matter what \phat is, the expression \phat(1-\phat) is
largest when \phat = 1/2, so we have n is large enough if
(\fracz*0.01)2 \frac12(1 - \frac12) £ n. |
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For our example, we have
## for 90% confidence
> alpha = 0.1; zstar = qnorm(1 - alpha/2)
> (zstar/0.01)^2/4
[1] 6764
## for 80% confidence
> alpha = 0.2; zstar = qnorm(1 - alpha/2)
> (zstar/0.01)^2/4
[1] 4106