In a random sample of 1000 students, p= 0.80 (or 80%) were in favor of longer hours at the school library. The standard error of p(the sample proportion) is (approximately)
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Given : In a random sample of 1000 students,
p= 0.80 (or 80%) were in favor of longer hours at the school library.
To Find : The standard error of p(the sample proportion) is (approximately)
Solution:
The standard error of p = √ [p (1-p) / n)]
p = 0.8
1 - p = 1 - 0.8 = 0.2
n = 1000
=> The standard error of p = √ [0.8 (0.2) / 1000)]
=> The standard error of p ≈ 0.01265
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