Posted by Anonymous on Monday, November 22, 2010 at 12:46pm.
A research group conducted an extensive survey of 3078 wage and salaried workers on issues ranging from relationships with their bosses to household chores. The data were gathered through hourlong telephone interviews with a nationally representative sample. In response to the question, "What does success mean to you?" 1482 responded, "Personal satisfaction from doing a good job." Let p be the population proportion of all wage and salaried workers who would respond the same way to the stated question. How large a sample is needed if we wish to be 95% confident that the sample percentage of those equating success with personal satisfaction is within 1.4% of the population percentage? (Hint: Use p ≈ 0.48 as a preliminary estimate. Round your answer up to the nearest whole number.)

Statistics  MathGuru, Tuesday, November 23, 2010 at 8:34am
Try this formula:
n = [(zvalue)^2 * p * q]/E^2
...where zvalue is 1.96 which represents 95% confidence using a ztable, p = 0.48, q = 1  p, and E = .014 (1.4% in decimal form).
I'll let you calculate it from here. Round your answer to the next highest whole number.

Statistics  Anonymous, Sunday, November 29, 2015 at 11:16pm
0.495 < p < 0.525
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