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STATISTICS QUESTION: If a sample consisting of 64 working people indicates that 20 percent of them...

STATISTICS QUESTION: If a sample consisting of 64 working people indicates that 20 percent of them skip lunch, what is the 95 percent confidence interval for that proportion?

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Answer #1

95% confidence interval for p is

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

0.20 - 1.96 * sqrt( 0.20 * 0.80 / 64) < p < 0.20 + 1.96 * sqrt( 0.20 * 0.80 / 64)

0.102 < p < 0.298

95% CI is ( 0.102 , 0.298 )

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