Question 855169: A recent poll of 1,235 randomly sampled likely votes show your favourite candidate ahead with 52.1% in favour. There are two candidates . Use hypotheses testing to infer to the larger groupg of all likely voters
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! this looks like a proportion type problem.
the candidate needs to get greater than 50% of the population to win so that's the assumed population percentage that he needs to be higher than.
so:
population proportion is 50% / 100 = .5
sample size is 1235.
population standard deviation is assumed to be sqrt(p*q/n).
this is a standard formula used in proportion hypothesis testing.
p is .5
q is .5
sqrt(p*q/n) equals sqrt((.5*.5)/1235) = .0142278 rounded to 6 decimal places.
sample proportion is 52.1% / 100 = .521
z-score = (x-m) / se
se = standard error = standard deviation of the mean distribution = .0142278
x = sample proportion = .521
m = population mean = .5
z-score = (.521 - .5) / .0142278 = 1.48 rounded to 2 decimal places.
since the z-score tables accuracy is to 2 decimal places, rounding to 2 decimal places is about as good as you can get unless you want to do interpolations which are not really necessary since anything that close is a tossup anyway.
z-score is equal to 1.48
probability of getting a z-score is greater than 1.48 is equal to .0694 rounded to 4 decimal places.
this is greater than .05 alpha for a one sided hypothesis test so the results are not considered to be statistically significant, but the candidate is doing pretty good.
if the results of the sample were based on random chance alone, it would happen only 6.94% of the time that the candidate would get a score greater than .521.
this indicates that, while it is possible that the candidate would do as well based on random sampling error alone, that wouldn't happen that often, so there is a likelihood that the voters really like him over the other opponent.
|
|
|
