Question 1069518: In a survey, 498 out of 1,397 US adults said they drink at least 4 cups of coffee a day. Find a point estimate (P) for the population proportion of US adults who drink at least 4 cups of coffee a day, then construct a 90% confidence interval for the proportion of adults who drink at least 4 cups of coffee a day.
P=0.3565; (0.4454, 0.3776)
P=0.3565; (0.3354, 0.3336)
P=0.3475; (0.3354, 0.3776)
P=0.3565; (0.3354, 0.3776)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! point estimate is the proportion one finds.
It is 0.3565
Only the last can be right, because the interval should contain the proportion in the middle.
The interval is z(0.95)*sqrt(p*(1-p))/n; the z(0.95) is where I look on the table. Because it is a two sided interval, 0.05 on both sides is a CI of 90%
=1.645*sqrt(.03565)(0.6435)/1397; sqrt term is 0.0128
=0.211 interval
{0.3354, 0.3776}
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