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gaby4567
28.01.2020 •
Mathematics
I'll give you the brainliest! a random sample of 3000 people was chosen for a survey. 47% of them had children under 18 living at home. find the 95% confidence interval - see picture
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Ответ:
0.4521 - 0.4879
Step-by-step explanation:
1) Find the standard deviation with the given information:
n=3000
p=47% ⇒ 0.47
1-p= 1 - 0.47 = 0.53
2) Find the margin of error (ME) with the given information:
C=95% ⇒ Z = 1.960
σ=0.009112
ME = Z × σ
= 1.96 (0.009112)
= 0.01786
3) Find the confidence interval with the given information:
p = 0.47
ME = 0.01786
CI = p ± ME
= 0.47 ± 0.01786
= (0.4521, 0.4879)
Ответ: