Unformatted text preview: a fair coin. The coin really was fair, but none of them knew that. They each counted the number of heads they got and divided by 100 to get the sample average and then computed 95% confidence intervals for the true average (using 1.96 SE’s instead of 2 SE’s). Since the coin was completely fair the true average is 0.5. a) How many of the 100 confidence intervals would you expect to miss the true average of 0.5? ____5___
b) If the 100 experimenters each computed 90% confidence intervals (instead of 95% confidence intervals), how many
would you expect to miss the true average of 0.5? ___10______ (90% CI means about 10% would miss)
c) If each experimenter tossed the coin 400 times, instead of 100 would the width of each confidence interval change, and
if so, how? (Hint, how would the SE ave change if n changes from 100 to 400? )
i)
It wouldn’t change
ii)
It would increase by multiplying by 4
iii)
It would decrease by dividing by 4
iv)
It would increase by multiplying by 2
v)
It would decrease by dividing by 2 (n ↑ by x 4, SEave ↓ by ÷ 2, b/c of sqrt law) d) 5 of the 100 intervals in the figure above missed the true average of 0.5. If there were 1000 experimenters each tossing
the coin 100 times and computing 95% confidence intervals, how many of the 1000 would you expect to miss?
i) 5% of 1000 = 50 ii) 5 € iii) 5 *€ € iv) 5/ 4) In a pre
election poll in a close race, how many people would you have to poll be 95% sure that your sample percent is right to within 3%. In other words, how many people do you need to poll to get a margin of error of 3%? (Assume SD = .5) a) For a 3% margin of error, how many people would you need to poll? 1,111 3% = 2 × 0.5
100
>>> 3 n = 100 >>> n = 33.33 >>> n = 1111 × 100%...
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 Spring '08
 Kim
 Statistics, the00

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