In 2010 the Federal Trade Commission began a three-year study of consumer buying
ID: 3019987 • Letter: I
Question
In 2010 the Federal Trade Commission began a three-year study of consumer buying and credit habits. Much of the statistical work centered on hypothesis testing of levels of debt and the percentage of debtors who faced “excessive liabilities.” At the outset of the study in the first quarter of 2010, the FTC concluded that “residents in all states still carry an average of more than $10,000 in debt.” In 2013 a sample of 12,510 residents reported a mean level of $10,158. At the 5% level of significance does it appear the FTC’s original contention that the average debt exceeded $10,000 might still be true? Assume a standard deviation of $8,500. The average U.S. college graduate begins his or her post-college days with ”slightly less than” $2,000 in credit card debt. Twenty-five graduates had a mean debt $1,975 with a standard deviation of $700. Test that hypothesis at the 1% level of significance. A state-by-state study revealed that Delaware tops the list with an average debt for student of more than $20,000 spread across seven unsecured creditors. A sample of 1,700 residents in Delaware had a mean level of debt of $20,257 and a standard deviation of $5,700. At the 5% level of significance, does it seem that $20,000 is the prevailing amount of debt? . Credit card debt (as opposed to total debt) showed a different pattern. State authorities in Wisconsin claimed that their residents had the lowest average credit card debt of $5,062 with a median household income of $49,001. In a survey conducted by college students in Wisconsin found that 500 residents of Wisconsin averaged $4,912 credit card debt with a standard deviation of $1,200. Using a level of significance of 4%, do these findings support the state authorities’ hypothesis? These same residents posted a mean household income of $52,300 and a standard deviation of $25,000. Has this value changed if the state authorities’ claim is correct? Set alpha at 3%? Under certain conditions some credit card holders can request a reduction in their annual percentage rates charged by the banks who issue the cards. A survey conducted by the U.S. Public Interest Research Group found” that of those who contacted their credit card issuers, 56 percent received a lower APR. On average, the percentage went from 16 percent to 10.47 percent.” From 550 debtors who petitioned their issuers, 279 were granted a reduction. Does it seem the public interest research group is correct at the 3% level of significance? The reduction the 279 received averaged 4.9% and a standard deviation of 1.5%. If = 1% can the group retain their assertion regarding the mean reduction? The ease with which individuals can accumulate consumer debt beyond their means to repay has precipitated a growth industry in debt consolidation and credit counseling services that offer debt management plans (DMP). The most common benefit of a DMP as advertised by most agencies is the consolidation of multiple monthly payments into one monthly payment, which is usually less than the sum of the individual payments previously paid by the customer. This is because issuers of credit cards will usually accept a lower monthly payment from a customer in a DMP than if the customers were paying the account on their own. Some DMPs advertise that payments can be cut by 50%, although a reduction of 10-20% is more common. Of 1000 people who took part in a DMP, the average reduction in monthly payments was 18.5% and a standard deviation of 1.2%. If alpha is set at 1% can it be concluded that the mean reduction was less than 20%? After a thorough reading of the case, test ALL potential hypotheses using both approaches to check your work. Also, calculate the associated p-values. What conclusions can you draw regarding debt patterns in the U.S.?
Explanation / Answer
a) In 2010 the Federal Trade Commission began a three-year study of consumer buying and credit habits.
At the outset of the study in the first quarter of 2010, the FTC concluded that “residents in all states still carry an average of more than $10,000 in debt.”
In 2013 a sample of 12,510 residents reported a mean level of $10,158. At the 5% level of significance does it appear the FTC’s original contention that the average debt exceeded $10,000 might still be true?
Assume a standard deviation of $8,500.
Here we have to test the hypothesis that,
H0 : mu = $10000 Vs H1 : mu > $10000
where, mu is the population mean.
sample size (n) = 12510
mean level (Xbar) = $10,158
sd = $8500
Population standard deviation is known therefore we use Z-test.
The test statistic for testing hypothesis is,
Z = (Xbar - mu) / ( sd/sqrt(n) )
Z = (10158 - 10000) / (8500 / sqrt(12510))
Z = 2.0791
alpha = 5% = 0.05
There are two approaches first is P-value approach and second is critical value approach.
P-value and critical value we can find by using EXCEL.
syntax :
P-value :
=1 - NORMSDIST(z)
where, z is test statistic value.
Critical value :
=NORMSINV(probability)
where, probability = 1 - alpha (because test is right tailed)
P-value = 0.019
Critical value = 1.645
We see that, P-value < alpha and Test statistic (Z) > critical value
So reject H0 at 5% level of significance.
Conclusion : It appear the FTC’s original contention that the average debt exceeded $10,000.
b) The average U.S. college graduate begins his or her post-college days with ”slightly less than” $2,000 in credit card debt. Twenty-five graduates had a mean debt $1,975 with a standard deviation of $700. Test that hypothesis at the 1% level of significance.
Here we have to test the hypothesis that,
H0 : mu = $2,000 Vs H1 : mu < $2,000
where, mu is the population mean.
sample size (n) = 25
mean level (Xbar) = $1,975
sd = $700
Population standard deviation is known therefore we use Z-test.
The test statistic for testing hypothesis is,
Z = (Xbar - mu) / ( sd/sqrt(n) )
Z = (1975 - 2000) / (700 / sqrt(25))
Z = -0.1786
alpha = 1% = 0.01
P-value and critical value we can find by using EXCEL.
syntax :
P-value :
=NORMSDIST(z) (the test is left tailed)
where, z is test statistic value.
Critical value :
=NORMSINV(probability)
where, probability = alpha (because test is right tailed)
P-value = 0.429
Critical value = -2.33
We see that, P-value > alpha and Test statistic (Z) > critical value
So accept H0 at 1% level of significance.
Conclusion : The average U.S. college graduate begins his or her post-college days with $2,000 in credit card debt.
c) A state-by-state study revealed that Delaware tops the list with an average debt for student of more than $20,000 spread across seven unsecured creditors.
A sample of 1,700 residents in Delaware had a mean level of debt of $20,257 and a standard deviation of $5,700. At the 5% level of significance, does it seem that $20,000 is the prevailing amount of debt? .
Here we have to test the hypothesis that,
H0 : mu = $20000 Vs H1 : mu > $20000
where, mu is the population mean.
sample size (n) = 1700
mean level (Xbar) = $20257
sd = $5700
Population standard deviation is known therefore we use Z-test.
The test statistic for testing hypothesis is,
Z = (Xbar - mu) / ( sd/sqrt(n) )
Z = (20257 - 20000) / (5700 / sqrt(1700))
Z = 1.8590
alpha = 5% = 0.05
P-value and critical value we can find by using EXCEL.
syntax :
P-value :
=1 - NORMSDIST(z)
where, z is test statistic value.
Critical value :
=NORMSINV(probability)
where, probability = 1 - alpha (because test is right tailed)
P-value = 0.032
Critical value = 1.645
We see that, P-value < alpha and Test statistic (Z) > critical value
So reject H0 at 5% level of significance.
Conclusion : A state-by-state study revealed that Delaware tops the list with an average debt for student of more than $20,000 spread across seven unsecured creditors.
d) Credit card debt (as opposed to total debt) showed a different pattern.
State authorities in Wisconsin claimed that their residents had the lowest average credit card debt of $5,062 with a median household income of $49,001.
In a survey conducted by college students in Wisconsin found that 500 residents of Wisconsin averaged $4,912 credit card debt with a standard deviation of $1,200. Using a level of significance of 4%, do these findings support the state authorities’ hypothesis?
Here we have to test the hypothesis that,
H0 : mu = $5062 Vs H1 : mu < $5062
where, mu is the population mean.
sample size (n) = 500
mean level (Xbar) = $4912
sd = $1200
Population standard deviation is known therefore we use Z-test.
The test statistic for testing hypothesis is,
Z = (Xbar - mu) / ( sd/sqrt(n) )
Z = (4912 - 5062) / (1200 / sqrt(500))
Z = -2.7951
alpha = 4% = 0.04
P-value and critical value we can find by using EXCEL.
syntax :
P-value :
= NORMSDIST(z)
where, z is test statistic value.
Critical value :
=NORMSINV(probability)
where, probability = alpha (because test is right tailed)
P-value = 0.000
Critical value = -1.7507
We see that, P-value < alpha and Test statistic (Z) < critical value
So reject H0 at 4% level of significance.
Conclusion : There is not sufficient evidence to say that State authorities in Wisconsin claimed that their residents had the lowest average credit card debt of $5,062.
e) Under certain conditions some credit card holders can request a reduction in their annual percentage rates charged by the banks who issue the cards.
A survey conducted by the U.S. Public Interest Research Group found” that of those who contacted their credit card issuers, 56 percent received a lower APR.
On average, the percentage went from 16 percent to 10.47 percent.
From 550 debtors who petitioned their issuers, 279 were granted a reduction. Does it seem the public interest research group is correct at the 3% level of significance?
Here we have to test the hypothesis that,
H0 : p = 56% = 0.56 Vs H1 : p > 0.56
where,p is the population proportion.
q = 1 - p = 1 - 0.56 = 0.49
sample size (n) = 550
sample proportion (p^) = x/n = 279 / 550 = 0.51
sd = sqrt ( (p*q)/n ) = sqrt((0.51*0.49)/550) = 0.0213
The test statistic for testing hypothesis is,
Z = (p^ - p) / sqrt((p*q)/n )
Z = (0.56 - 0.51) / 0.0213
Z = 2.4734
alpha = 3% = 0.03
P-value and critical value we can find by using EXCEL.
syntax :
P-value :
=1 - NORMSDIST(z)
where, z is test statistic value.
Critical value :
=NORMSINV(probability)
where, probability = 1 - alpha (because test is right tailed)
P-value = 0.01
Critical value = 1.88
We see that, P-value < alpha and Test statistic (Z) > critical value
So reject H0 at 3% level of significance.
Conclusion : There is sufficient evidence to reject null hypothesis.
f) Some DMPs advertise that payments can be cut by 50%, although a reduction of 10-20% is more common. Of 1000 people who took part in a DMP, the average reduction in monthly payments was 18.5% and a standard deviation of 1.2%. If alpha is set at 1% can it be concluded that the mean reduction was less than 20%?
Here we have to test the hypothesis that,
H0 : mu = 20% Vs H1 : mu < 20%
where, mu is the population mean.
sample size (n) = 1000
mean level (Xbar) = 18.5% = 18.5 / 100 = 0.185
sd = 1.2% = 1.2 / 100 = 0.012
Population standard deviation is known therefore we use Z-test.
The test statistic for testing hypothesis is,
Z = (Xbar - mu) / ( sd/sqrt(n) )
Z = (0.185 - 0.2) / (1000 / sqrt(1000))
Z = -39.5285
alpha = 1% = 0.01
There are two approaches first is P-value approach and second is critical value approach.
P-value and critical value we can find by using EXCEL.
syntax :
P-value :
=NORMSDIST(z)
where, z is test statistic value.
Critical value :
=NORMSINV(probability)
where, probability = alpha (because test is right tailed)
P-value = 0.000
Critical value = -2.326
We see that, P-value < alpha and Test statistic (Z) < critical value
So reject H0 at 5% level of significance.
Conclusion : The mean reduction was less than 20%.
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