SOLUTION: Kara surveyed 200 people, chosen at random on thompson drive, to try to determine what % of thompsonites are skiers. 86 people in her sample said they were skiers.
A.) What is
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-> SOLUTION: Kara surveyed 200 people, chosen at random on thompson drive, to try to determine what % of thompsonites are skiers. 86 people in her sample said they were skiers.
A.) What is
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Question 1207432: Kara surveyed 200 people, chosen at random on thompson drive, to try to determine what % of thompsonites are skiers. 86 people in her sample said they were skiers.
A.) What is the 95% confidence interval for the percent of Thompsonites who are skiers ? (Formula: 95% confidence interval = mean ± 1.96(standard deviation)
You can put this solution on YOUR website! sample size is 200
p = 86/200 = .43.
q = (1 - 86/200) = 114/200 = .57.
s = sqrt(.43 * .57 / 200) = .035007 rounded to 6 decimal places.
critical z-score at 95% confidence interval is z = plus or minus 1.96.
z-score formula is z = (x-m)/s
z is the z-score
x is the critical raw score.
m is the raw mean which is equal to p.
s is the standard error which is equal to s.
on the high side of the confidence interval, the formula becomes 1.96 = (x - .43) / .035007.
solve for x to get x = 1.96 * .035007 + .43 = .49861372.
on the low side of the confidence interval, the formula becomes -1.906 = (x - .43) / .035007.
solve for x to get x = -1.96 * .035007 + .43 = .36138628.
the 95% confidence interval is .36138628 to .49861372.
the margin of error is (.49861372 minus .36138628) / 2 = .06861375.
