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3) 3) A supplier of digital memory cards claims that no more than 1% of the cards are defective. In a random sample of 600 me

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Answer #1

H0: p \leq 0.01

Ha: p > 0.01

Test statistics

z = ( \hat{p} - p ) / sqrt [ p ( 1 - p) / n ]

= ( 0.03 - 0.01) / sqrt ( 0.01 (1 - 0.01) / 600)

= 4.92

p-value = P(Z > z)

= P(Z > 4.92)

= 1 - P(Z < 4.92)

= 1 - 1

= 0

Since p-value < 0.01 level, reject the null hypothesis.

We have sufficient evidence to support the suppliers claim.

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