SOLUTION: Help, please, answer is not 5942 :) A research group conducted an extensive survey of 2968 wage and salaried workers on issues ranging from relationships with their bosses to hous

Question 1077012: Help, please, answer is not 5942 :)
A research group conducted an extensive survey of 2968 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?" 1631 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 2.3% of the population percentage? (Hint: Use p ≈ 0.55 as a preliminary estimate. Round your answer up to the nearest whole number.)

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
We'll use this formula

where,

n = sample size

p = estimated proportion

= critical z value (the notation is pronounced "z sub c")

E = margin of error desired

Note: Refer to this page. Scroll to the very bottom of the page.

--------------------------------

In this case, we have...

n = unknown for now

p = 0.55 is the estimate proportion (note: if no estimates are given, assume p = 0.5)

roughly for a 95% confidence interval (see this table. Look for the 95% confidence interval, at the bottom of the table, and look at the value just above it)

is the margin of error in decimal form (2.3% = 2.3/100 = 0.023)

------------------------------------------------------------------------------------

Plug in the values discussed above to get...

Round UP to the nearest whole number

So the min sample size needed is 1798.

We always round up to make sure we clear the hurdle needed. It doesn't matter what the decimal portion is (and how close it is to either endpoint).

If we rounded to 1797, then we'd be short of our goal. I recommend plugging n = 1797 into the margin of error formula and you should see E being more than 2.3%, which is the opposite of what we want.
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