Need statistics help with proportions and confidence intervals

Statistics
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In a random sample, 136 of 400 persons given a flu vaccine experienced some discomfort.

I was asked to set up a 95% confidence interval for the true proportion of persons who will experience some discomfort. The problem is solved by I need help with interpretation.


We have p = 0.34, n = 400, and z = 1.96 for 95% confidence. 

Interval = 0.34 ± 1.96√[(0.34)(0.66)/400] 

= 0.34 ± 1.96√0.000561 

= 0.34 ± 1.96(0.023685439) 

= (0.29357654, 0.38642346)


Nov 23rd, 2015

Thank you for the opportunity to help you with your question!

plus or minus a margin of error that our proportion is occurring . The result is called a confidence interval for the population proportion, p.

The formula for calculating confidence interval for a population proportion is given as the foollowing

       p+_z*Squareroot (p(1-p)/n)

 is the sample proportion, n is the sample size, and z* is the appropriate value from the standard normal distribution for your desired confidence level.  certain confidence levels.

p =136/400=0.34 which is sample proportion from our sample n which is 400 z is the value from tables that shows the margin of error occurring which is from 95% confidence level 

The z will offer a range in which the error may occur thus you get  two answers 

Please let me know if you need any clarification. I'm always happy to answer your questions.
Nov 23rd, 2015

This is not what I asked . I've already resolved the problem and I know  which formula to use. I need to find out what my answer means (interpretation )

Nov 23rd, 2015

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Nov 23rd, 2015
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Nov 23rd, 2015
Dec 2nd, 2016
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