Solve the Problem $95 \%$ Confident That $p$ Is Greater Than $\frac { 1 } { 7 }$

question 20

Multiple Choice

Solve the problem.
-Suppose that n trials of a binomial experiment result in no successes. According to the "Rule of Three", we have 95% confidence that the true population proportion has an upper bound of 3/n. If a manufacturer randomly selects 21 computers for quality control and finds no defective computers, what statement can you make by using the rule of three, about the proportion p, of all its computers which are defective?

• A) We are \(95 \%\) confident that \(p\) is greater than \(\frac { 1 } { 7 }\).
• B) We are \(95 \%\) confident that \(p\) does not exceed \(\frac { 1 } { 7 }\).
• C) We are \(95 \%\) confident that \(p\) lies between \(\frac { 1 } { 21 }\) and \(\frac { 1 } { 7 }\).
• D) The value of \(p\) cannot be greater than \(\frac { 1 } { 7 }\).

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