Question 1204948: In a recent year, the Better Business Bureau settled 75% of complaints they received. You have been hired by the Bureau to investigate complaints this year involving computer stores. Assume the population proportion of complaints settled for the computer stores is the 0.75. Suppose your sample size is 152. What is the probability that the sample proportion will be within 4 percentage points of the population proportion?
my work:
.75=x 152=n
.04*.75=.03=Z
x/sqrt n = .75/sqrt 152 = 0.7097
.75+.03=.78
.75-.03=.72
(.78-.75)/0.7097=0.0423
(.72-.75)/0.7097=-0.0423
difference = 0.0337
the issue is that I know I have done something wrong because the answer key says that I am supposed to have gotten 0.7456 and I do not understand what I am doing wrong or which steps that I have mixed up or missed so that I am getting something so different.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
p = population proportion
phat = sample proportion
n = sample size
The claim is that p = 0.75
4 percentage points within this value represents the interval from 0.71 to 0.79 (because 0.75-0.04 = 0.71 and 0.75+0.04 = 0.79)
We're tasked to compute P(0.71 < phat < 0.79)
When phat = 0.71, we get this z score
z = (phat - p)/( sqrt(p*(1-p)/n) )
z = (0.71 - 0.75)/( sqrt(0.75*(1-0.75)/152) )
z = -1.138888
z = -1.14
When phat = 0.79, we get this z score
z = (phat - p)/( sqrt(p*(1-p)/n) )
z = (0.79 - 0.75)/( sqrt(0.75*(1-0.75)/152) )
z = 1.138888
z = 1.14
Side note: sqrt(p*(1-p)/n) represents the standard error (SE)
refer to page 2 of this formula sheet
https://apcentral.collegeboard.org/media/pdf/statistics-formula-sheet-and-tables-2020.pdf
P(0.71 < phat < 0.79) is roughly equivalent to P(-1.14 < z < 1.14) when p = 0.75 and n = 152.
Next I'll use a Z table as shown here
https://www.ztable.net/
This table should be found in the back of your stats textbook.
On exams your teacher will likely hand out the table.
In the table, locate the row that starts with -1.1
Place a piece of paper under it so you keep track of this row.
If working a digital copy, then open your favorite paint program to mark this row.
Next, highlight the column that has 0.04 at the top.
The intersection of this row and column yields the value 0.12714
It means P(z < -1.14) = 0.12714 approximately
The area under the standard normal Z curve to the left of -1.14 is roughly 0.12714
Follow similar steps to find that P(z < 1.14) = 0.87286
Then we subtract the items to get the area between
P(a < Z < b) = P(Z < b) - P(Z < a)
P(-1.14 < Z < 1.14) = P(Z < 1.14) - P(Z < -1.14)
P(-1.14 < Z < 1.14) = 0.87286 - 0.12714
P(-1.14 < Z < 1.14) = 0.74572
This leads back to P(0.71 < phat < 0.79) = 0.74572 which rounds to 0.7457
The discrepancy of 0.7457 and 0.7456 is likely due to some rounding error.
It's possible that I may have made a mistake somewhere.
Perhaps your table uses slightly different values.
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If you want to use a calculator then you have plenty of options to pick from.
On TI83 and TI84 calculators, use the normalCDF function
https://www.statology.org/normal-probabilities-ti-84-calculator/
If your TI calculator input is normalCDF(-1.1389,1.1389,0,1) then the result is approximately 0.745 255 0333 which rounds to 0.7453
If your TI calculator input is normalCDF(-1.138888,1.138888,0,1) then the result is approximately 0.745 250 0276 which rounds to 0.7453
Play around with the accuracy of the 1.138888 values to see how the area changes. The change shouldn't be too much.
Another calculator to use is this online one
https://davidmlane.com/hyperstat/z_table.html
Yet another option is to use the spreadsheet function NormDist
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