1083 Americans over 23 drawn 867 don't smoke p=0.199. Construct the 90% confidence interval for the population proportion of Americans over 23 who smoke lower bound and higher bound round to 3 decimal points
To find the 90% confidence interval we first find Z_crit, which is given by:
N.inv ( p = 0.95 , mean = 0, sd = 1 ) = 1.645
We then find the standard error, which so long as your population is large can be estimated by:
SE = sqrt[ p*( 1 - p ) / n ] where p = 0.199 and n = 1083:
SE = sqrt(0.199*(0.801)/1083) = 0.0121
Multiply SE by Z_crit:
0.0121 * 1.645 = 0.020
So the interval is given by 0.199-0.020 , 0.199+0.020 = (0.179, 0.219) (to 3 decimal places)
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