SOLUTION: A RANDOM SAMPLE OF 400 MEMBERS OF LABOR FORCE IN A FIVE STATE REGION SHOWED THAT 32 WERE UNEMPLOYED. CONSTRUCT THE 95% CONFIDENCE INTERVAL FOR THE PROPORTION UNEMPLOYED IN THE REGI
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Question 1085207: A RANDOM SAMPLE OF 400 MEMBERS OF LABOR FORCE IN A FIVE STATE REGION SHOWED THAT 32 WERE UNEMPLOYED. CONSTRUCT THE 95% CONFIDENCE INTERVAL FOR THE PROPORTION UNEMPLOYED IN THE REGION. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! The interval width is z(0.975)*sqrt (0.08*0.92)/400; sqrt term=0.01356.
The 0.975 is the z-value with 0.025 on each end or 95%. The sqrt is of p*(1-p)/n, and p is 32/400=0.08, the point estimate.
0.0266 is the interval width.
That is above and below the point estimate
(0.0534, 0.1066) is 95% CI
