SOLUTION: Any Help with this question is much appreciated! A recent article in The Wall Street Journal entitled� As Identity Theft Moves Online, Crime Rings Mimic Big Business� states tha

Question 1119303: Any Help with this question is much appreciated!
A recent article in The Wall Street Journal entitled� As Identity Theft Moves Online, Crime Rings Mimic Big Business� states that 39% of the consumer scam complaints by American consumers are about identity theft. Suppose a random sample of 90 complaints is obtained. Of these complaints, 40 were regarding identity theft. Based on these sample data, what conclusion should be reached about the statement made in the Wall Street Journal? (Test using  (alpha) = 0.10).
Thank You!
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
claim:
39% of the consumer scam complaints by american consumers are about identity theft.

out of a sample of 90, 40 were regarded as identity theft.

test using alpha .10.

i believe a two tail test since claim is not equal to and not greater or less than.

proportion of claim is .39.
proportion of sample is .40.

mean of sample proportion is .4

standard error is sqrt(p*q/n) = sqrt(.39*.61/90) = .0514133575.

the population proportion is used in this calculation.

z-score of sample = (.4 - .39) / .0514133575 = .1945019831.

critical z-score = plus or minus 1.644853626

results of approximately z-score of .2 appear to be well within the confidence interval limits of approximately -1.64 to 1.65.

null hypothesis can't be refuted as the results are not significant.

if you're using alpha as a criteria.

alpha is .05 on either end of the normal distribution curve.

alpha for a z-score of .2 is .5792596878 on the low side and .4207403122 on the high side.

this is not lower than .05 on either end which means the results are not significant.

here's a reference on one sample proportion test.

http://www.statisticslectures.com/topics/onesamplezproportions/

in this reference example, the results of the test were significant.

the standard error of the test was sqrt(.9*.1/100) = .03.

.9 was the population proportion.

the z-score was (.82-.9)/.3 = -2.67.

this resulted in an alpha on the low side of .0038 which was lower than the critical alpha of .025.

the critical alpha of .025 was associated with a critical z-score of -1.96.

the samle z-score of -2.67 was greater than this in a negative direction, which meant the results of the test were significant.

either way the results were significant, whether you used critical alpha as a crieteria or critical z-score as the criteria.

in your problem, the results were not significant.

you could not, therefore, refute the claim that 39% of the claims ere based on identify theft, and had to accept as being reasonable based on sample test your performed.

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