Question

4. For a 99% confidence level, how large of a sample size is needed for a margin of error of 0.03 for the es-timate of the population proportion? Past studies are not available

Answer #1

Solution :

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E = 0.03

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.58 ( Using z table ( see the 0.005 value in standard normal (z) table corresponding z value is 2.58 )

Sample size = n = (Z/2 / E)2 * * (1 - )

= (2.58 / 0.03)2 * 0.5 * 0.5

=1849

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a planning value for p*. Round up to the next whole number.

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confidence interval for the proportion
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available.
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n =
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n =

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the 99.74% confidence level for a pharmaceutical drug study in
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7. What is the minimum sample size needed to ensure a survey to
create a confidence interval estimate of a population proportion
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μ : Mean of variable
sample size 94
99% confidence interval results:
Variable
Sample Mean
Std. Err.
DF
L. Limit
U. Limit
original
3.0989362
0.017739741
93
3.0522854
3.1455869
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population proportion?
Write down the confidence interval that you obtained. Interpret
the result.
What is the margin of error?
Using the same data, construct a 98% confidence interval for
the population proportion. Then, answer the following three
questions:
(i) What happens...

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proportion? Assume that past data are not available for developing
a planning value for p*. Round up to the next whole number.

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a margin of error of .015 for the estimation of a population
proportion? Assume that past data are not available for developing
a planning value for P* . Round up to the next whole number.

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Also what is your alpha value / z score / standard deviation
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