<PRE><B>Question 1009834</B>
 In a random sample of 400 clerical employees in a large multinational corporation, 284 said that they would like the company to adopt a flexible scheduling policy, which would permit employees to decide when and how much they should work. Find the 90% confidence limits for the corresponding true proportion for the population sampled. 
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p-hat = 284/400 = 0.71
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ME = 1.645*sqrt[0.71*0.29/400] = 0.037
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90% CI: 0.71 - 0.037 &lt; p &lt; 0.71 + 0.037
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Cheers,
Stan H.
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