stat 200 week 5 homework problems stat200 week 5 homework problems v1 questions amp answers new
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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.7.1.6
The population proportion=0.23
n=number of complaints=1432
Alaska had complaints=321=x
So,
sample proportion=nx=1432321 =0.2242
H0: p=0.23 against H1: p<0.23
Z statistic,
Z=np^(1−p^)p^−0.23 ~N(0,1)
Obs(Z)=(-)0.52
H0: p=0.23 against H1: p<0.23
p value=0.30 (Excel Formula: NORM.S.DIST(0.52,1))
As,p value> α (level of significance). Hence there is not enough evidence to
support the claim that, Alaska had a lower proportion of identity theft than
23% .
Appropriate level of significance = 0.05
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.
H0 : p = 0.23
μ=299,710.5 km/sec μ<299,710.5 lm/sec
H1 : p < 0.23
Z critical value = -1.645 at 5% level
Sample proportion
'p' = 321/1432 = 0.2242
Z stat = (p - P) / sqrt(PQ/n)
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