Question 1121206: In a random sample of 200 observations, we found the proportion of successes to be 48 %.
1.2.1 Use a 90 % degree of confidence to estimate the population proportion of
successes. (5)
1.2.2 Repeat with n = 500. (2)
1.2.3 Repeat with n = 1000. (2)
1.2.4 Comment on the difference in answers of 1.2.1 ; 1.2.2 and 1.2.3.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The half interval for 90% CI is +/- z*SE, where z is 1.645 and SE is sqrt(.48*.52/200)
1.645*0.0353 or 0.0581
So the interval is 48%+/-5.81%
or (0.4219, 0.5381)
For n=500 the SE is 0.0223 and the half interval 0.0368
(0.4432, 0.5168)
Fon n=1000, the SE is 0.0158 and the half interval 0.026
(0.454, 0.506)
As the sample size increases, the SE decreases meaning that the estimate of the true number has less uncertainty the larger the sample. As the sample approaches the population size, the uncertainty approaches 0. That would occur only at extremely large sample sizes (it would be +/-0.001 for a sample size of a million).
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