SOLUTION: In a recent year, the Better Business Bureau settled 75% of complaints they received. (Source: USA Today, March 2, 2009) You have been hired by the Bureau to investigate complaints
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-> SOLUTION: In a recent year, the Better Business Bureau settled 75% of complaints they received. (Source: USA Today, March 2, 2009) You have been hired by the Bureau to investigate complaints
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Question 1119490: In a recent year, the Better Business Bureau settled 75% of complaints they received. (Source: USA Today, March 2, 2009) You have been hired by the Bureau to investigate complaints this year involving computer stores. You plan to select a random sample of complaints to estimate the proportion of complaints the Bureau is able to settle. Assume the population proportion of complaints settled for the computer stores is the 0.75, as mentioned above. Suppose your sample size is 139. What is the probability that the sample proportion will be within 3 percent of the population proportion?
Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations.
Answer = (Enter your answer as a number accurate to 4 decimal places.) Answer by Theo(13342) (Show Source):
your standard error is equal to square root of (.75 * .25 / 139).
your standard error is therefore equal to .036727658.
the formula for z-score is:
z = (x-m)/s
z is the z-score.
x is the raw score.
m is the mean.
s is the standard error.
you want the sample proportion to be within 3% of the population proportion.
.03 * .75 = .0225.
.75 - .0225 = .7275.
.75 + .0225 = .7725.
you want your population proportion to be between .7275 and .7725.
when the sample proportion is .7275, the z-score associated with that would be:
z = (x-m)/s becomes z = (.7275 - .75) / .036727658 = -.6126173357.
when the sample proportion is .7725, the z-score associated with that would be:
z = (x-m)/s becomes z = (.7725 - .75) / .036727658 = .6126173357.
round those z-score to 4 decimal digits and you get:
low z = -.6126
high z = .6126.
the area of the normal distribution curve to the left of a z-score of -.6126 is equal to .2700703716.
the area of the normal distribution curve to the left of a z-score of .6126 is equal to .7299296284.
the area in between is the smaller area subtracted from the larger area.
this is equal to .4598592567.
that's the probability that the sample proportion will be within 3% of the population proportion.
visually this looks like this:
the calculator does some rounding so the display shows you that the area between those 2 z-scores is equal to .4599 which is agreeable to the more detailed result i got using the TI-84 Plus calculator.
the calculator display above used raw scores.
the display below shows the same calculations using z-scores.
