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Sample Sizes to limit the margin of error (assume O = 0
Sample Sizes to limit the margin of error (assume O = 0.2676) 67 (a) What was the margin of error for the 95% Cl calculated in this lab? 68 margin of error, m = 69 70 (b) How large a sample is needed to cut this margin down to .05? 71 Use the formula: n = (Z x o / m) Round to next highest number 72 for m = 0.05, then n = 73 74 (c) How large a sample is needed to cut this margin of error down to .025? 75 for m = 0.025, then n = 76 77 (d) How large a sample is needed to cut the margin of error down to .0125? 78 for m = 0.0125, then n = 79 80
Expert Solution
a. (no sample size given)
b. n = 110
c. n = 440
d. n = 1761
Step-by-step explanation
a. Please provide the sample size(n) for part a
b.
Given:
95 % confidence interval
Zα/2 at 95% = 1.960
σ = 0.2676
m = 0.05
Formula:
n = (Zα/2 x σ)/m)2
Substituting to the formula we have
n = (Zα/2 x σ)/m)2
n = (1.96 x 0.2676)/0.05)2
n = 110
c.
b.
Given:
95 % confidence interval
Zα/2 at 95% = 1.960
σ = 0.2676
m = 0.0125
Formula:
n = (Zα/2 x σ)/m)2
Substituting to the formula we have
n = (Zα/2 x σ)/m)2
n = (1.96 x 0.2676)/0.025)2
n = 440
d.
Given:
95 % confidence interval
Zα/2 at 95% = 1.960
σ = 0.2676
m = 0.05
Formula:
n = (Zα/2 x σ)/m)2
Substituting to the formula we have
n = (Zα/2 x σ)/m)2
n = (1.96 x 0.2676)/0.0125)2
n = 1761
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