STAT 200 Week 5 Homework Problems
Homework Problems:
1) 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.
Solution:
Random variable: x = number of consumer complaints from identity theft in Alaska
Population parameter = proportion of consumer complaints from identity theft in Alaska
Hypotheses = H o : P= 0.23 H a : P <0.23
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%? Test at the 5% level.
Solution:
x = 321
n = 1432
p = 0.23
a = 0.05
i. State the random variable and the parameter in words.
The random variable is given by: x = number of consumer complaints from identity theft
in Alaska
The parameter of interest is by: p = proportion of complaints from identity theft in Alaska
ii. State the null and alternative hypotheses and the level of significance
The hypotheses for this experiment are given by:
H o : p = 0.23 H a : p <0.23
The level of significance is a = 0.05
, iii. State and check the assumptions or a hypothesis test
a. A simple random sample of the category of 1432 complaints of identity theft in
Alaska was taken. The study says that the complaints were out of all complaints
that year, but the year could have been chosen at random. The assumption may be
met, but you can’t be sure.
b. There are 1432 complaints in this sample. The reason for the complaint does not
affect the next complaint. There are only two outcomes, either the complaint was
for identity theft or it wasn’t. The chance the one complaint was for identity theft
does not change. Therefore, the condition for the binominal distribution are
satisfied.
c. In this case p = 0.23 and n = 1432.
np = 1432 * 0.23 = 329.36 ≥ 5 and nq = 1432 * (1 – 0.23) = 1102.64 ≥ 5,
the sampling distribution for pˆ is a normal distribution.
iv. Find the sample statistic, test statistic, and p-value
The sample proportion is given by:
x = 321
n = 1432
x 321
pˆ = = ≈ 0.2242
n 1432
The test statistic is given by:
pˆ− p 0.2242−.023
z= pq = 0.23(1−0.23) ≈ -0.52
√ n √ 1432
p-value:
p-value = P (z > -0.52) = -1.645
v. Conclusion
The p-value is greater than the level of significance therefore we fail to reject H o .
vi. Interpretation
There is not enough evidence to show that the proportion of complaints due to identity
theft are less than 23% in Alaska.
3) In 2008, there were 507 children in Arizona out of 32,601 who were diagnosed with Autism
Spectrum Disorder (ASD) ("Autism and developmental," 2008). Nationally 1 in 88 children are
Homework Problems:
1) 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.
Solution:
Random variable: x = number of consumer complaints from identity theft in Alaska
Population parameter = proportion of consumer complaints from identity theft in Alaska
Hypotheses = H o : P= 0.23 H a : P <0.23
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%? Test at the 5% level.
Solution:
x = 321
n = 1432
p = 0.23
a = 0.05
i. State the random variable and the parameter in words.
The random variable is given by: x = number of consumer complaints from identity theft
in Alaska
The parameter of interest is by: p = proportion of complaints from identity theft in Alaska
ii. State the null and alternative hypotheses and the level of significance
The hypotheses for this experiment are given by:
H o : p = 0.23 H a : p <0.23
The level of significance is a = 0.05
, iii. State and check the assumptions or a hypothesis test
a. A simple random sample of the category of 1432 complaints of identity theft in
Alaska was taken. The study says that the complaints were out of all complaints
that year, but the year could have been chosen at random. The assumption may be
met, but you can’t be sure.
b. There are 1432 complaints in this sample. The reason for the complaint does not
affect the next complaint. There are only two outcomes, either the complaint was
for identity theft or it wasn’t. The chance the one complaint was for identity theft
does not change. Therefore, the condition for the binominal distribution are
satisfied.
c. In this case p = 0.23 and n = 1432.
np = 1432 * 0.23 = 329.36 ≥ 5 and nq = 1432 * (1 – 0.23) = 1102.64 ≥ 5,
the sampling distribution for pˆ is a normal distribution.
iv. Find the sample statistic, test statistic, and p-value
The sample proportion is given by:
x = 321
n = 1432
x 321
pˆ = = ≈ 0.2242
n 1432
The test statistic is given by:
pˆ− p 0.2242−.023
z= pq = 0.23(1−0.23) ≈ -0.52
√ n √ 1432
p-value:
p-value = P (z > -0.52) = -1.645
v. Conclusion
The p-value is greater than the level of significance therefore we fail to reject H o .
vi. Interpretation
There is not enough evidence to show that the proportion of complaints due to identity
theft are less than 23% in Alaska.
3) In 2008, there were 507 children in Arizona out of 32,601 who were diagnosed with Autism
Spectrum Disorder (ASD) ("Autism and developmental," 2008). Nationally 1 in 88 children are
