I have a question on one of my homework problems:
"You conducted a telephone pool and asked the question "Did you vote in the last presidential election?" Of the 250 respondents, 60% said they did. Your friend is suspicious of your results, and finds that the real percentage was 57% who voted in the last election. Explain, using calculations, how the discrepancy could be a result of random chance."
So 60% of 250 = 150. How would I go about calculating the standard error? Is that the next step?
"You conducted a telephone pool and asked the question "Did you vote in the last presidential election?" Of the 250 respondents, 60% said they did. Your friend is suspicious of your results, and finds that the real percentage was 57% who voted in the last election. Explain, using calculations, how the discrepancy could be a result of random chance."
So 60% of 250 = 150. How would I go about calculating the standard error? Is that the next step?
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The thumb is called the pollix - plural, pollices! That is why they call it a poll - originally people voted thumbs up or thumbs down!
Given the "real" percentage was 57% (which he was easily able to determine because he was part of the 1%, I guess!), what is the likelihood that a sample of 250 random people (assuming the population is considerably larger than 250 people) gave you a result of 60% ?
It can be the binomial distribution, or a normal approximation. The normal approximation is probably better to use here, to avoid having to compute with 250! etc. and add up a whole bunch of those, too.
If you do a normal approximation, the variance is N p (1-p) = 250 (0.57) (0.43) which would be the standard deviation in the binomial distribution for that situation.
Standard deviation sqrt(61.27500) = 7.8278
To figure out where your tail is on the distribution, that difference is 250 (.60 - .57) = 250 (0.03) = 7.50
The actual poll number you got was slightly less than 1 standard deviation away from the exact expected value, so it is not a highly unlikely result.
http://stattrek.com/tables/binomial.aspx
put in
P = 0.57
N = 250
X (successes) = 0.6 x 250 = 150
It easily finds P(X>=150) = 0.18577...
The other way is to use a normal distribution table, you can enter mean 0, std. dev. 1, X = 7.50 / 7.8278
The other way is to divide the standard deviations
Given the "real" percentage was 57% (which he was easily able to determine because he was part of the 1%, I guess!), what is the likelihood that a sample of 250 random people (assuming the population is considerably larger than 250 people) gave you a result of 60% ?
It can be the binomial distribution, or a normal approximation. The normal approximation is probably better to use here, to avoid having to compute with 250! etc. and add up a whole bunch of those, too.
If you do a normal approximation, the variance is N p (1-p) = 250 (0.57) (0.43) which would be the standard deviation in the binomial distribution for that situation.
Standard deviation sqrt(61.27500) = 7.8278
To figure out where your tail is on the distribution, that difference is 250 (.60 - .57) = 250 (0.03) = 7.50
The actual poll number you got was slightly less than 1 standard deviation away from the exact expected value, so it is not a highly unlikely result.
http://stattrek.com/tables/binomial.aspx
put in
P = 0.57
N = 250
X (successes) = 0.6 x 250 = 150
It easily finds P(X>=150) = 0.18577...
The other way is to use a normal distribution table, you can enter mean 0, std. dev. 1, X = 7.50 / 7.8278
The other way is to divide the standard deviations