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Find the 90%, 95% and 99% confidence intervals for a survey of 2223 U.S. adults when...

Find the 90%, 95% and 99% confidence intervals for a survey of 2223 U.S. adults when 1334 say an occupation of as an athlete is prestigious.

Math   Statistics-And-Probability   
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Answer #1

Sample proportion = 1334 / 2223 = 0.600

a)

90% confidence interval for p is

- Z * sqrt ( ( 1 - ) / n) < p < + Z * sqrt ( ( 1 - ) / n)

0.60 - 1.645 * sqrt( 0.60 * 0.40 / 2223) < p < 0.60 + 1.645 * sqrt( 0.60 * 0.40 / 2223)

0.583 < p < 0.617

90% CI is ( 0.583 , 0.617 )

b)

95% confidence interval for p is

- Z * sqrt ( ( 1 - ) / n) < p < + Z * sqrt ( ( 1 - ) / n)

0.60 - 1.96 * sqrt( 0.60 * 0.40 / 2223) < p < 0.60 + 1.96 * sqrt( 0.60 * 0.40 / 2223)

0.580 < p < 0.620

95% CI is ( 0.580 , 0.620)

c)

99% confidence interval for p is

- Z * sqrt ( ( 1 - ) / n) < p < + Z * sqrt ( ( 1 - ) / n)

0.60 - 2.5758 * sqrt( 0.60 * 0.40 / 2223) < p < 0.60 + 2.5758 * sqrt( 0.60 * 0.40 / 2223)

0.573 < p < 0.627

995 CI is ( 0.573 , 0.627 )

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