Mathematics, 09.11.2020 16:40 therealpr1metime45
In late June 2012, Survey USA published results of a survey stating that 55% of the 579 randomly sampled Kansas residents planned to set off fireworks on July 4th. Determine the margin of error for the 55% point estimate using a 95% confidence level. The margin of error is: % (please round to the nearest percent)
Answers: 2
4.1 %
= 0.55
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In late June 2012, Survey USA published results of a survey stating that 55% of the 579 randomly sam...
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