What is the best estimate for the width of your index finger
1 answer:
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Answer:
Standard deviation of the students = 0.408
The students would you have to poll to be 95% confident of the outcome within /- 2% of the vote
= 0.408 X 1600
= 652.8
Step-by-step explanation:
<u><em>Step(i):</em></u>-
Given sample size 'n' = 1600
95% confidence interval of the margin error is determined by
Level of significance = 0.05
Z₀.₀₅ = 1.96
Given Margin of error = 2% = 0.02
0.02 X √1600 = 1.96 X S.D
Standard deviation of the students = 0.408
The students would you have to poll to be 95% confident of the outcome within /- 2% of the vote
= 0.408 X 1600
= 652.8