1)
Sample size estimation for proportion
Preliminary estimate for proportion (Known) = 0.51
Margin of error =E=0.03
Level of significance = 0.04
Z critical value is (by using Z table)=2.054
Sample size
=1171.45
Therefore, sample size approximately was 1171
2)
given a confidence interval (15,27)
point estimate of the population mean = (15+27)/2 = 21
The margin of error = uppper bound - point estimate
= 27 - 21
= 6
3)
given xbar = 120.4
s = 12.9
n = 40
t critical value for 99% confidence with 39 df is = 2.707
120.4 +/- 2.707*12.9/sqrt(40)
120.4 +/- 5.52
lower bound = 114.88
upper bound = 125.92
In a poll, 51% of the people polled answered yes to the question "Are you in...
In a poll, 51% of the people polled answered yes to the question "Are you in favor of the death penalty for a person convicted of murder?" The margin of error in the poll was 2%, and the estimate was made with 94% confidence. At least how many people were surveyed? The minimum number of surveyed people was? (Round up to the nearest integer.)
a) b) Determine the point estimate of the population mean and margin of error for the confidence interval Lower bound is 20, upper bound is 24. The point estimate of the population mean is . The margin of error for the confidence interval is In a poll, 51% of the people polled answered yes to the question "Are you in favor of the death penalty for a person convicted of murder?" The margin of error in the poll was 2%,...
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In a poll, 69% of the people polled answered yes to the question "Are you in favor of the death penalty for a person convicted of murder?" The margin of error in the poll was 2%, and the estimate was made with 95% confidence. At least how many people were surveyed?
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In a poll, 37% of the people polled answered yes to the question "Are you in favor of the death penalty for a person convicted of murder?" The margin of error in the poll was 33%, and the estimate was made with 94% confidence. At least how many people were surveyed? Please show work, and how you'd use a TI84 calc. to solve it.
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