New Year Resolutions: Some Startling Statistics About American Consumers
By Jeanette Joy Fisher
Before you whip out your credit card, stop and think. America is often called a consumer society, and it's true. We're also a nation of credit card debt, which is evidenced by some of these startling facts.
We all know that acquiring overwhelming debt is stressful from a financial standpoint. However, it might surprise you to know that more than 70 percent of divorces in America are brought about due to financial problems.
One thing Americans are NOT is savers. We simply don't put away enough money to protect us in the case of financial emergency. In fact, the average American consumer carries so much debt load that they're a mere three paychecks away from having to declare bankruptcy.
Financial emergencies arise all the time, of course, but Americans continue to build up credit card debt, without thought of the possible consequences. That's why nearly 1,500,000 people in the United States are forced to file for bankruptcy every year. Another 1,500,000 people will turn to the various consumer credit counseling organizations for help in order to try to avert bankruptcy. That's a staggering number, but it's dwarfed in comparison to the 37,000,000 people who endeavor to work out plans with their creditors on their own. That's 40 million people a year who are in enough financial trouble to take drastic action!
How do people get in such horrible financial condition? One way is through the use of credit cards. They may be convenient, but they certainly make consumers pay more for the merchandise they buy. For instance, a person paying for a purchase with a credit card will pay, on average, more than 130 percent for that item than if they had simply paid cash for it.
A large majority of consumer only pay the minimum payment on their charge cards, but those payments are generally 90 percent interest, with only 10 percent going toward reducing the principal. A staggering 71% of all credit card holders pay only the minimum payment, and they do it without a thought to the consequences. If they can easily afford the monthly payment, they just keep paying it, without realizing how expensive their purchases ultimately have become.
If you are one of the millions of Americans who are only paying the minimum amount on your charge cards, start paying MORE each month. Even if it's only a little more. Reducing the principal balance will shorten your overall payment schedule--sometimes by hundreds or even thousands of dollars.
The statistics are mind-boggling, and with interest rates headed even higher, you must educate yourself on the wise use of credit, pay cash whenever possible, and try to live within your means, even if it means doing without some things. It's the only way you can avoid becoming one of those sad financial statistics yourself. Include monitoring your credit card use in your New Year's Resolutions.
Copyright © Jeanette J. Fisher
What percentage of Americans have credit card debt, I need statistics on credit debt.?
please give me the links for the websites….thanks in advance
47% of Americans pay their credit cards in full each month.
Google: Statistics on credit card debt…
Are credit cards having a positive or negative effect when it comes to personal debt and finances?
I am working on a paper for my college class and would like to get some feedback about the use of credit cards by American consumers. My assignment is to argue that Americans should not use credit cards. I have found several sites full of statistics but I would like to expand my sources of information for my paper. I appreciate and and all constructive feedback on the issue. Relevant links would also be helpful. Thank you for your help!
Stat test. Pleaase someone help me!?
The monthly credit card debts for individual accounts are normally distributed with a standard deviation of $25. A researcher wishes to estimate the mean monthly credit card debt for all individual accounts. Find the sample size needed to assure with a 95.44% confidence that the sample mean will not differ from the population mean by more than 5 units.
A. 54
B. 100
C. 104
D. 124
A public health researcher wished to estimate the proportion of the adult population of the US that has high blood pressure. How large a sample is needed in order to be 95% confident that the sample proportion will not differ from the true proportion by no more than 0.005?
A. 1,639
B. 38,416
C. 41,871
D. 158,118
Thirty randomly selected students took the calculus final. If the sample mean was 84 and the standard deviation was 12.8, construct a 99% confidence interval for the mean score of all students.
A. (79.58 , 90.42)
B. (80.03 , 87.97)
C. (77.56 , 90.44)
D. 78.25 , 89.75)
A simple random sample of students is selected, and the students are asked how much time is spent preparing for a test. The time (in hours) are as follows: 1.3, 7.2, 4.2, 12.5, 6.6, 2.5, 5.5 Based on these results, a confidence interval for the population mean is found to be µ = (1.3 , 10.1). Find the degree of confidence. Hint: find s.
A. 98%
B. 96%
C. 94%
D. 92%
Find the critical values that correspond to a 95% confidence level with a sample of size ten.
A. left kigh square = 1.700 & right kigh square = 17.023
B. left kigh square = 17.023 & right kigh square = 1.700
C. left kigh square = 2.700 & right kigh square = 19.023
D. left kigh square = 19.023 & right kigh square = 2.700
An estimate of the population proportion from a study on the incidence of flu in a certain time period is 0.253. In a new study, the researchers want a margin of error of 0.005 with a 99% confidence level. Find the minimum sample size you should use to assure that your proportion point estimate will be within the required margin of error around the population proportion p.
A. 2,511
B. 29,041
C. 45,113
D. 50,126
According to a recent poll, 53% of Americans would vote for the incumbent president. If a random sample of 100 people results in 45% who would vote for the incumbent, test the claim that the actual percentage is 53%. Use a 0.10 significance level. Find the claim.
A. p <> 0.53
B. p < = 0.53
C. p >= 0.53
D. p = 0.53
Given the sample statistics, determine if you should use the t distribution, normal distribution, or neither to construct a confidence interval for an estimate of µ. From a sample of 22 observations, the sample mean is 65.7 , standard deviation is 3.6 , sigma is unknown , and the population appears to be skewed.
A. t distribution
B. normal distribution
C. neither
D. not enough information
You want to be 99% confident that the sample standard deviation s is within 5% of the population standard deviation. Find the appropriate minimum sample size.
A. 923
B. 1335
C. 2553
D. 2638
Suppose that you wish to estimate a population mean. Which of the following does the necessary sample size depend upon:
A) The desired degree of confidence
B) The desired margin of error
C) The population mean
D) The population size
E) The population standard deviation
A. A,B,E
B. A,B,D
C. A,B,D,E
D. All of the above
The following confidence interval is obtained for a population proportion, p: (0.528 , 0.570). Use these confidence interval limits to find the margin of error, E.
A. 0.011
B. 0.021
C. 0.031
D. 0.041
< = means less than or equal to
>= means greater than or equal to
<> means not equal to
ANSWER: Sample Size = 100
Why???
SMALL SAMPLE, LEVEL OF CONFIDENCE, NORMAL POPULATION DISTRIBUTION
Margin of Error (half of confidence interval) = 5
The margin of error is defined as the “radius” (or half the width) of a confidence interval for a particular statistic.
Level of Confidence = 95.44
?: population standard deviation = 25
(‘z critical value’) from Look-up Table for 95.44% = 2
significant digits = 2
Margin of Error = (‘z critical value’) * ?/SQRT(n)
n = Sample Size
Algebraic solution for n:
n =[('z critical value') * ?/Margin of Error]^2
= [ (2 * 25)/5 ]^2
Sample Size = 100
ANSWER: SAMPLE SIZE = 38416
Why???
CHOOSING THE SAMPLE SIZE, POPULATION PROPORTION
?: POPULATION PROPORTION = 50
(50% – assumption)
B: SPECIFIED BOUND ON THE ERROR = 0.005
significant digits = 2
Level of Confidence = 95
‘z critical value’ from Look-up Table for 95% = 1.96
FORMULA FOR COMPUTATION OF SAMPLE SIZE:
B = ‘z critical value’ * SQRT [ ? * (1 - ? )/n]
Algebraically solve for SAMPLE SIZE n :
n = ? * (1 – ?) * [ 'z critical value' / B ]^2 = 0.5 * (1 – 0.5) * [ 1.96 / 0.005 ]^2
SAMPLE SIZE = 38416
The Table for Standard Normal Distribution is organized as a cummulative ‘area’ from the LEFT corresponding to a ‘z critical value’. The Standard Normal Distribution is also symmetric (called a ‘Bell Curve’) which means its an interpretive procedure to Look-Up the ‘area’ from the Table. For Level of Confidence = 95 the Look-Up ‘z critical value’ = -1.96 which corresponds to the ‘area’ from the LEFT outside of the Level of Confidence.
And since the Table for Standard Normal Distribution is symmetric, the ‘area’ from the LEFT is one-half outside of the Level of Confidence. There is also another one-half from the RIGHT as well. For the purpose of solving this question, the Look-Up ‘z critical value’ from the LEFT is satisfactory.
Alternatively; use Excel NORMSINV(probability) = NORMSINV ( 0.5*( 1 – 95 / 100 )))