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 hour-long 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 = [(z-value)^2 * p * q]/E^2

...where z-value is 1.96 which represents 95% confidence using a z-table, 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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