Need statistics help with result interpretation. Confidence interval and proportions

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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 


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

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