Question 1192817: A researcher wishes to estimate, with 90% confidence, the proportion of adults who have high speed internet access. Her estimate must be accurate within 8% of the true proportion. Find the minimum sample size needed, using a prior study that found that 67% of the respondents said they have high speed internet access.
Answer by math_tutor2020(3817)   (Show Source):
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At 90% confidence, the z critical value is roughly z = 1.645
Use a Z table to determine this.
Notice that P(-1.645 < Z < 1.645) = 0.90 approximately.
The prior study says phat = 0.67 is the proportion of people who have high speed internet access (i.e. the 67%)
E = 0.08 due to the desired 8% error. The researcher wants to achieve this error or smaller.
The min sample size needed is...
n = phat*(1-phat)*(z/E)^2
n = 0.67*(1-0.67)*(1.645/0.08)^2
n = 93.4847 approximately
n = 94 always round UP to the nearest whole number
Let's say we didn't round up and instead went with n = 93 (it's tempting because 93.4847 is closer to 93 than it is to 94)
The margin of error would be roughly
E = z*sqrt(phat*(1-phat)/n)
E = 1.645*sqrt(0.67*(1-0.67)/93)
E = 0.0802
We get an error larger than 8% which isn't what your teacher wants.
However, if we go with n = 94, then,
E = z*sqrt(phat*(1-phat)/n)
E = 1.645*sqrt(0.67*(1-0.67)/94)
E = 0.0798
Now the error is under 8%
Likely we won't land *exactly* on 8%, so the next best thing is to get under that value.
The larger n gets, the smaller E will get, and vice versa.
This is why we always round up for minimum sample size problems.
Answer: 94
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