question archive Using 69% confidence level, find the confidence interval using your sample proportion
Using 69% confidence level, find the confidence interval using your sample proportion
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Using 69% confidence level, find the confidence interval using your sample proportion. (HINT: Use the sample proportion, and sample size in the example - though change the z-critical value for your new confidence level)
95% confidence level
Sample Proportion less than Sample Mean: 4/10 or .40
P: .40
Q: .60
Sample Size: 10
Z-Critical Value: 1.959963985
Margin Error: 0.28402577
Lower Margin: 0.41597423= 42%
Higher Margin: 0.9840257= 98%
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CONFIDENCE INTERVAL FOR SAMPLE PORTION
Example 1:
Formula for confidence interval of sample portion:
p ± z sqrt( ((p)(1-p))/n )
where:
p = sample portion
z = z score of confidence interval
n = sample size
Given:
p = 0.4
n = 10
significance level = 1 - confidence interval
= 1 - 0.69
= 0.31
α/2 = significance level/2
= 0.31/2
= 0.155
*find the z-value given p = 0.155
zα/2 =1.015
upper interval: 0.40 + (1.015)*[sqrt( ((0.4)(0.6))/10 )] = 0.4 + 0.1572 = 0.5572
lower interval: 0.40 - (1.015)*[sqrt( ((0.4)(0.6))/10 )] = 0.4 - 0.1572 = 0.2428
