SOLUTION: I would like to know if I did the problem correctly: Percentage of adults who drink alcohol. Of the 1516 adults interviewed, 985 said that they drank. a. Determine a 95% confidnc

Question 352085: I would like to know if I did the problem correctly:
Percentage of adults who drink alcohol. Of the 1516 adults interviewed, 985 said that they drank.
a. Determine a 95% confidnce interval for the proportion, p, of all Americans who drink alcohol.
b. Interpret results.
This is what I came up with:
N=1516, x=985 and n-x=531 Both x and n-x are at least 5
985/1516=0.6497; za/2=z0.025=1.96
(a) 0.6497-1.96sqrt0.6497(1-0.6497)/1516 to 0.6497+1.96sqrt0.6497
)/1516=6.35,0079
(b) we can be 95% confident that the proprotion of adults who drink alcohol is somewhere between those numbers.
Thank you for any help you can give me.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
I would like to know if I did the problem correctly:
Percentage of adults who drink alcohol. Of the 1516 adults interviewed, 985 said that they drank.
a. Determine a 95% confidnce interval for the proportion, p, of all Americans who drink alcohol.
b. Interpret results.
This is what I came up with:
N=1516, x=985 and n-x=531 Both x and n-x are at least 5
985/1516=0.6497; za/2=z0.025=1.96
-----------------------------------
a)
p-hat = 0.6497
E = 1.96sqrt(0.6497*(1-0.5497)/1516) = 0.0240
95%CI: 0.6497-0.0240 < p < 0.6497+0.0240
95%CI: 0.6257 < p < 0.6737
-----------------------------------
(a) 0.6497-1.96sqrt0.6497(1-0.6497)/1516 to 0.6497+1.96sqrt0.6497
)/1516=6.35,0079
(b) we can be 95% confident that the proportion of adults who drink alcohol is somewhere between 0.6257 and 0.6737.
----
Cheers,
Stan H.
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