Question 844860: In a recent poll, 45% of survey respondents said that, if they only had one child, they would prefer the child to be a boy. Suppose you conducted a survey of 190 randomly selected students on your campus and find that 86 of them would prefer a boy.
a) Use the normal approximation to the binomial to approximate the probability that, in a random sample of 190 students, at least 86 would prefer a boy, assuming the true percentage is 45%.
b) Does this result contradict the poll? Explain
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! normal approximation to binomial is:
mean = n * p
standard deviation = square root of (n * p * (1-p))
n is equal to 190
p is equal to .45
n * p is equal to .45 * 190 = 85.5
this is the probability that you would expect based on the knowledge that 45% of survey respondents said they would prefer a boy than a girl.
in your survey you got 86 out of 190 and you want to know if this is statistically significant or if you can expect this deviation by chance.
standard deviation is calculated as square root of (n * p * (n-p)) which becomes square root of (190 * .45 * .55) which is equal to 6.8575 rounded to 4 decimal places which is more than enough accuracy for what is required.
without a continuity correction, your formula would be:
z (86 > 85.5) = (population mean - sample mean) / population standard deviation which becomes (86 - 85.5) / 6.8575 which is equal to .0729.
this is so close to a z factor of 0 that there's no question that the results are well within limits and can very easily be due to chance.
with a continuity correction, your formula would be:
z (85.5 > 85.5) = 0
a z factor of zero means that your sample mean is exactly the population mean.
the probability that your score will be above 0 is 50%.
since 5% is the cutoff, your well within the confidence interval limits.
the following video might help you with the process as it demonstrates solving a similar type problem.
http://www.youtube.com/watch?v=mdIUHzWydr8
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