STAT 200 Week 5 Homework Problems
7.1.2
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
...
STAT 200 Week 5 Homework Problems
7.1.2
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.
x = number of consumer complaints from identity theft in Alaska
Population parameter = proportion of consumer complaints from identity theft in Alaska
Hypothesis = HO : P=0.23 HA : P<0.23
7.1.6
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 type I and
type II errors in this case, consequences of each error type for this situation, and the appropriate alpha
level to use.
Skipped problem per direction by professor.
7.2.4
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%? Test at the 5% level.
x = 321
n = 1432
HO : p=0.23
HA : p<0.23
α = 0.05
q = 1 – 0.23 = 0.77
p-hat = 321 / 1432 = 0.224162011
z = (0.224162011 – 0.591) / √((0.591 * 0.77)/321) = -0.36683798 / 0.03765187314599194 = -9.74
p-value = normalcdf(-1e99,-9.74,0,1)
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