SOLUTION: According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, 23% of all complaints in 2007 were for identity theft. In that year, Alaska ha

Question 1184879: According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, 23% of all complaints in 2007 were for identity theft. In that year, Alaska had 321 complaints of identity theft out of 1,432 consumer complaints ("Consumer fraud and," 2008). Does this data provide enough evidence to show that Alaska had a lower proportion of identity theft than 23%? State the random variable, population parameter, and hypotheses. show the process
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
population mean assumed to be 23% = .23
alaska had 321 complaints out of 1432.
the proportion for alaska is 321/1432 = .22416.
this is the sample mean.
the sample size is 1432.

the standard error is equal to sqrt(population mean * (1 - population mean) / sample size) = sqrt(.23 * (1-.23) / 1432) = .0111208.

z-score is (x - m) / s
x is the sample mean.
m is the population mean
s is the standard error.

z-score = (.22416 - .23) / .0111208 = -.52514.

the area to the left of that z-score is .2997.

the critical z-score at .05 two tailed significance level is equal to -1.96.

the absolute value of the test test z-score is not greater than the absolute value of the critical z-score.

this means the results are not significant and there is not enough evidence to show that the alaska proportion is different from the general population proportion.

also, the critical p-value is .025 and the test p-value is .2997.
this is another indication that the test results are not significant.

the critical z-score and the critical p-value will always agree.
either one is sufficient to determine significance of the test.

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