Question

With 80% confidence, for sample proportion 0.40 and sample size 26, what is the upper confidence limit with 2 decimal places?

Answer #1

Solution :

Given that,

n = 26

= 0.40

1 - = 1 - 0.40 = 0.60

At 80% confidence level the z is ,

= 1 - 80% = 1 - 0.80 = 0.20

= 0.20

Z_{}
= Z_{0.20} = 0.842

Margin of error = E = Z_{
/ 2} * ((
* (1 -
)) / n)

= 0.842 * (((0.40 * 0.60) / 26) = 0.08

A 80 % confidence interval for population proportion p is ,

+ E

0.4+ 0.08

0.48

The 80% confidence interval for the population proportion p is : ( upper limit = 0.48)

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with 80% confidence, for sample proportion 0.56 and a sample
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A population proportion is 0.40. A sample of size 200 will be
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p
will be used to estimate the population proportion. (Round your
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(a)
What is the probability that the sample proportion will be
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(b)
What is the probability that the sample proportion will be
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QUESTION PART A: For a confidence level of 80% with a sample
size of 26, find the critical t value.
QUESTION PART B: Assume that a sample is used to estimate a
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corresponds to a sample of size 13 with a mean of 46.5 and a
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Report ME accurate to one decimal place because the sample
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