A retail store manager is trying to improve and control the rate at which cashiers sign customers up
for store credit cards. Suppose the manager takes 10 samples, each with 100 observations.
The p-bar
value is found to be .05, and the manager does not want a lower limit below .0064. What z-value would this imply, and how confident can she be that the true lower limit is greater than or equal to .0064?
Sigma-p = sqrt (.05*(1 - .05 )/100 ) = .0217945. Using the equation that LCL = p-bar - z*sigma-p gives .0064 = .05 - z*(.0217945 ). Solving gives z=2. Using table S6.2 shows that for z = 2 the manager can be 95.45% confident.
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Indicate whether the statement is true or false
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