A set of middle school student heights are normally distributed with a mean of 150150150 centimeters and a standard deviation of
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Let p be the proportion. Let c be the given confidence level , n be the sample size.
Given: p=0.3, n=1180, c=0.99
The formula to find the Margin of error is
ME =
Where z (α/2) is critical value of z.
P(Z < z) = α/2
where α/2 = (1- 0.99) /2 = 0.005
P(Z < z) = 0.005
So in z score table look for probability exactly or close to 0.005 . There is no exact 0.005 probability value in z score table. However there two close values 0.0051 and 0.0049 . It means our required 0.005 value lies between these two probability values.
The z score corresponding to 0.0051 is -2.57 and 0.0049 is -2.58. So the required z score will be average of -2.57 and -2.58
(-2.57) + (-2.58) = -5.15
-5.15/2 = -2.575
For computing margin of error consider positive z score value which is 2.575
The margin of error will be
ME =
=
= 2.575 * 0.0133
ME = 0.0342
The margin of error is 0.0342
