Hi!
Thank you in advance for assisting with my question.
I am working on a probability question and am able to answer almost all of it except one final part. The answer reads: " The probability is 95% that the sample percentage will be contained within 6.94% symmetrically around the population mean." I can not figure out how the 6.94% was obtained. I realize that .0694 equals the Z value -1.48, but that doesn't help me.
The full question is "The probability is 95% that the sample percentage will be contained within what symmetrical limits of the population percentage."
Here is my work:
n = 200,
π = .5
p = .5
1-.95 =.5
.5/2= .0250
Z value of .0250 = -1.96
Z value of .9750 (.95 + .0250 =.9750) is +1.96
-1.96 <Z <+1.96
A = .5 + (1.96 * .0354) =.5694
B = .5 – (1.96* .0354) =.4306
I searched in my textbook, other statistic books and online, but still can not figure this out.
Any help would be greatly appreciated.
Thanks,
Thank you very much of answering this question. I really appreciate it.
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