SOLUTION: 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
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Question 1076716: 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.)
You can put this solution on YOUR website! p = .55
q = 1 - p = .45
alpha = .05
divide that by 2 and you get an alpha of .025 on each end of the normal distribution curve.
your critical z-score becomes plus or minus 1.95996 which can be rounded to 1.96
your mean population proportion is assumed to be equal to .55 which is equal to p.
your standard error is equal to sqrt(p*q/n) which is equal to sqrt(.55*.45/n) = sqrt(.2475/n).
s is equal to the standard error.
n is equal to the sample size required.
your formula is:
s = sqrt(.2475/n)
since the normal distribution is symmetric about the mean, you just need to work with the high z-score and you will automatically get results applicable to the low z-score.
high z-score = 1.96
low z-score = -1.96
your mean proportion is equal to .55
plus or minus 2.3% of that is equal to plus or minus .01265
your high mean proportion is equal to .56265
your low mean proportion is equal to .55 - .01265 = .53735
that's the range your looking for at a two sided confidence interval of 95% which give you an alpha of .025 at each end of the normal distribution curve.
working with the high z-score, the formula would be:
1.96 = (.56265 - .55) / x
simplify to get 1.96 = .01265 / s
solve for s to get s = .01265 / 1.96
this results in s = .006454 rounded to 6 decimal digits.
since s is equal to sqrt(.45*.55)/n) which is equal to sqrt(.2475/n), you get:
sqrt(.2475/n) = .006454
square both sides to get .2475/n = 4.1654116 * 10^-5.
multiply both sides by n and divide both sides by 4.1654116 * 10^-5 and solve for n to get n = .2475 / (4.1654116 * 10^-5)
result is n = 5941.789... which can be rounded to 5942.
results indicate you would need a sample size of 5942 in order to get a sample mean proportion that would include the population mean proportion within a 95% confidence interval.
to test this out, assume your sample size is 5942.
your mean proportion is .55
your standard error is equal to sqrt(p*q/n) which is equal to sqrt(.45*.55/5942) which is equal to .006454 rounded to 6 decimal places.
your high critical z-score is 1.96
your high raw score becomes equal to (x-m)/s which is equal to (x-.55)/.006454.
your formula becomes 1.96 = (x-.55)/.006454
multiply both sides by .006454 and then add .55 to both sides to get:
1.96 * .006454 + .55 = x
solve for x to get x = .56265 rounded to 4 decimal digits.
do the same for the low z-score.
the formula for that is -1.96 = (x-.55)/.006454
multiply both sides by .006454 and add .55 to both sides to get:
-1.96 * .006454 + .55 = x
solve for x to get x = .53735 rounded to 4 decimal digits.
your acceptable range at 95% confidence interval is .53735 to .56265
your low score and you high score are both with 2.3% of the assumed population mean proportion of .55.
.55 - .53735 = .01265
.36265 - .55 = .01265
.01265 = .55 * .023 which is 2.3% plus or minus from the .55 mean proportion.
this can be shown graphically using the following normal distribution calculator.
