Question 458:
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Lets first find the components of the population proportion so we can build the margin of error. We're told that 1143 were positive out of 86,991 for a sample proportion of .01314.

The margin of error for a proportion is found by multiplying the standard error of the proportion times a critical value from the normal distribution for the confidence level. The confidence level we're given is 95%. So we need the zscore which accounts for 95% of the area under the normal curve. We can find this in a ztable in the back of a statistics book or using the percentile to zscore calculator and entering .95 2sided. We get 1.96. So that's the first part.

Now we need to find the standard error of the proportion (SEP) which we can find as the standard deviation divided by the square root of the sample size. For a proportion, the standard deviation is found by taking the square root of p*q where p is .01314 and q is 1.01314 or .9869. So the standard deviation = SQRT(.01314 *.9869) = .1139. The square root of the sample size is SQRT (86991) = 294.94. So we divide these two values .1139/294.94 = .000386, which is our SEP. The margin of error is then .00386*1.96 = .000757.

If we wanted to build the confidence interval we'd just add and subtract the margin of error to the proportion .0131+.000757 and .0131.000757 getting us a 95% CI between .01238 and .013896 positive. We can convert this to a percents, which would make the interval between 1.24% and 1.39%.
We can check our works by using the confidence interval around a proportion calculator to construct a 1sample binomial confidence interval. With a sample size as large as 86k the normal Wald confidence interval and AdjustedWald intervals will be virtually identical (to a few decimal places) and we see the results between 1.24% and 1.39%.
The normality assumption is not a problem since while distribution here is binomial, the Central limit theorem shows that as sample sizes get larger (in this case very large) the binomial distribution begins to resemble the normal distribution. A crude rule of thumb to see if the Normality assumption is ok is if both n*p >5 and n*q >5. In this case thats' 86991*.013 = 1130 and 86991*.98 = 86121 which exceeds 5 easily.