Question 1138901: It is said that happy and healthy workers are efficient and productive. A company that manufactures exercising machines wanted to know the percentage of large companies that provide on-site health club facilities. A sample of 200 such companies showed that 150 of them provide such facilities on site.
What is the point estimate of the percentage of all such companies that provide such facilities on site and Construct a 98% confidence interval for the percentage of all such companies that provide such facilities on site as well as find the margin of error for this estimate?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The point estimate is 150/200 or 0.75 as the proportion.
a 98% CI is z* sqrt(0.75*0.25/200) and z 0.99 is 2.33
half-interval is 2.33*sqrt(0.0009375) or 0.071, 0.07 here
98%CI is (0.68, 0.82) with margin of error of 7%
(0.679, 0,821) is to 3 decimal places
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