## Hope there will be no lottery winners this New Year’s Eve

According to a report in npr.org called Road Fatalities Dip Thanks To Safer Cars, Economy, an array of factors are making the road safer. According to a study by the Department of Transportation, the overall number of fatality on American roads has dropped dramatically, fallen by over 20% in the last few years. Two likely reasons for this dramatic drop are safer cars and a slower economy. However, even with the over 20% drop in fatality on the road, there is still one death every 15 minutes on the road.

I always think of dying from a crash involving a drunk driver is a lottery. It is a negative lottery for sure since no one would want to win it. In a previous post (The lottery of drunk driving fatality), I discussed the statistic of one drunk driving fatality every 45 minutes. By comparison, the number of deaths on the roads due to all causes is three times higher than just deaths from drunk driving (in the lottery analogy it is three times more likely to win)! I hope in this holiday season, no one will win this negative lottery.

Be safe on the road. Between drinking and driving, only do one of them!

Now the quantitative stuff. As reported in Road Fatalities Dip Thanks To Safer Cars, Economy, there were almost 44,000 road-related deaths in 2005. In 2009, there were about 34,000 deaths. This is a 22% decrease. There are two ways to see this.

One is to calculate the number of reduction in deaths, which is $44000-34000=10000$. Then divide $10000$ by $44000$. We have:

$\displaystyle \frac{10000}{44000}=0.2273$, which is 22.73%.

Another way to derive the 22.73% is to calculate the following ratio:

$\displaystyle \frac{34000}{44000}=0.7727$

Then subtract one from this ratio and obtain $0.7727-1=-0.2273$, which indicates a 22.73% decrease in road-related deaths.

The 2009 figure for the number of road-related deaths is 34,000. This comes out to be one death every 15 minutes. To derive this rate, we need to calculate the total number of minutes in a year. There are 365 x 24 x 60 = 525,600 minutes in a year. Then divide 525,600 by 44,000 to obtain 15.46 minutes. Then round the answer to 15 minutes.

We can get a perspective of this calculation by looking at an example of taking an exam. For example, if you have two hours (120 minutes) to take an exam and the exam has 10 problems, then on average you have 12 minutes to work one problem. Thus if you can work one problem per 12 minutes, you can expect to finish the exam in the allotted time.

Back to the calculation at hand, there are 525,600 minutes in a year and there are 34,000 events. Thus on average there are 15 minutes allotted for each event.

$\displaystyle \frac{365 \times 24 \times 60}{34000}=15.46=15$

The hope is that the denominator in the above ratio will keep getting smaller in the years to come. From 2005 to 2009, the denominator shrank from 44,000 to 34,000. I have a thought. Supose that in the next 5 years (2009 to 2013), there will be the same percent decrease in the road-related deaths as in the 5-year period from 2005 to 2009. What will be the value of the denominator? In other words, according to the same trend line, what will be the number of road-related deaths in 2013?

The answer to the above question is obtained by reducing the 34,000 deaths in 2009 by 22.73%. Try the following:

$\displaystyle 34000 \times (1-0.2273) = 34000 \times 0.7727=26271.8$

If the same trend that played out between 2005 and 2009 holds, the projection for 2013 would be about 26,000. Whether this is a realistic projection or not, I do not know. I will leave this to the experts who study traffic fatality. Let’s hope that the improvement will be as least no worse than this projection.

## Putting a value on higher education

This post is a continuation of the previous post Putting a value on college education. Here we look at the earnings of various levels of higher education relative to “below high school” and relative to high school graduates. The data are obtained from a report in College Board that is called Education Pays 2010. This reports talks about the benefits of the value of higher education to both individuals and society.

In Figure 1.1 of this report, I found information on the median earnings and tax payments of full-time year round workers ages 25 and older for various levels of education in 2008. The following table shows the median earnings.

$\displaystyle \begin{pmatrix} \text{Education Level}&\text{Median Earnings in Dollars} \\{\text{ }}&\text{ }&\text{ } \\\text{Below High School}&\text{24,300} \\\text{High School Graduate}&\text{33,800} \\\text{Some college but no degree}&\text{39,700} \\\text{Associate Degree}&\text{42,000} \\\text{Bachelor's Degree}&\text{55,700} \\\text{Master's Degree}&\text{67,300} \\\text{Doctoral Degree}&\text{91,900} \\\text{Professional Degree}&\text{100,000}\end{pmatrix}$

To help us compare, I convert the median earnings into ratios using “below high school” as baseline.

$\displaystyle \begin{pmatrix} \text{Education Level}&\text{Ratio}&\text{Percent Increase} \\{\text{ }}&\text{ }&\text{ } \\\text{Below High School}&\text{1.0000}&\text{0 percent} \\\text{High School Graduate}&\text{1.3909}&\text{39.09 percent} \\\text{Some college but no degree}&\text{1.6337}&\text{63.37 percent} \\\text{Associate Degree}&\text{1.7284}&\text{72.84 percent} \\\text{Bachelor's Degree}&\text{2.2922}&\text{129.22 percent} \\\text{Master's Degree}&\text{2.7695}&\text{176.95 percent} \\\text{Doctoral Degree}&\text{3.7819}&\text{278.19 percent} \\\text{Professional Degree}&\text{4.1152}&\text{311.52 percent}\end{pmatrix}$

For a full explanation on how the math is done, see Putting a value on college education. As an example, I show the calculation for the bachelor’s degree level. The following is the ratio of bachelor’s degree to the below high school level.

$\displaystyle \frac{55700}{24300}=2.2922$

Thus, the median earning of college graduates is 2.2922 times the median earning of the below high school level, or equivalently 229.22% of the below high school level. Subtracting one from this ratio gives the percent increase going from below high school to bachelor’s degree. This means that the additional amount a college graduate can expect to earn is 129% of the median earning of the below high school level!

In general, the ratio column in the above table gives information about the median earning at each level as a percentage of the baseline (below high school). The column for percent increase gives the percentage increase going from the baseline to each level, that is, the additional earning power at each level expressed as a percentage. It is clear that the higher the educational level, the higher the additional earning power.

Now the same comparison but using the high school level as the baseline.

$\displaystyle \begin{pmatrix} \text{Education Level}&\text{Ratio}&\text{Percent Increase} \\{\text{ }}&\text{ }&\text{ } \\\text{Below High School}&\text{0.7189}&\text{-28.11 percent} \\\text{High School Graduate}&\text{1.0000}&\text{0 percent} \\\text{Some college but no degree}&\text{1.1746}&\text{17.46 percent} \\\text{Associate Degree}&\text{1.2426}&\text{24.26 percent} \\\text{Bachelor's Degree}&\text{1.6479}&\text{64.79 percent} \\\text{Master's Degree}&\text{1.9911}&\text{99.11 percent} \\\text{Doctoral Degree}&\text{2.7189}&\text{171.89 percent} \\\text{Professional Degree}&\text{2.9586}&\text{195.86 percent}\end{pmatrix}$

With high school graduates as baseline, high school dropouts only earn about 71.89% of the median earning of the high school graduates. The additional earning power of the below high school level is -28.11%, which would actually be a decrease in earning power. Once again, a similar clear pattern here: learning has dollar value. The higher the educational level, the higher the additional earning power. On average, a college graduate can expect to earn 65% (64.79%) more than a high school graduate. The additional earning power for a professional degree is even greater, 195.86%, which is three times the addtional earning power for a college dregree.

## Putting a value on college education

Recently I read a piece in Time.com about high school dropout rates. According to this article, the economic outlook for high school dropouts is bleak. The statistics cited by the article are the average earnings and unemployment rates. In 2009, an average college degree holder earned $1,015 per week and an averaged high school dropout earned just$454 per week. The unemployment rate for the college degree holders is 5.2% and 14.6% for high school dropouts. The figures speak volume. They not only put a dollar value on college education. These numbers also give me an opportunity to talk about the concept of ratio, which is a handy way to compare two quantities.

The numbers I want to focus on are the two weekly earnings of $1,015 and$454. How much larger is $1,015 over$454? In terms of absolute amount, the weekly earning of $1,015 is$561 more than $454 since $1015-454=561$. But the absolute amount does not tell the whole story. We can also look at the ratio of$1,015 over $454, which is obtained by$1,015 divided by $454. $\displaystyle (1) \ \ \ \ \ \ \ \ \ \ \frac{1015}{454}=2.236$ What to make of the ratio 2.236? First, it says that the average earning of a college graduate is 2.236 times of the average earning of a high school dropout. For each one dollar in income for a high school dropout, the average college graduate makes two dollars and twenty four cents. So on average a college degree holder makes over two times more than someone with less than a high school education. In this ratio, the baseline (or denominator) is the average weekly earning of a high school dropout. We can also state the ratio 2.236 in terms of percent. Multiplying it by 100, we get 223.6. So the average earning of a college graduate is 223.6% of the average earning of a high school dropout. This is the same information as in the above paragraph, just that the scale is a little different. For each one hundred dollars in income for a high school dropout, the average college graduate makes two hundred and twenty three dollars and sixty cents. The average college grad’s weekly paycheck is$561 more than the weekly paycheck of an average high school dropout. The additional amount of $561 as a percentage of the paycheck for the average high school dropout is: $\displaystyle (2) \ \ \ \ \ \ \ \ \ \ \frac{561}{454}=1.236$ Note that the answer in (2) is just one less than the ratio in (1). Thus the ratio (1) above has another interpretation. Subtracting one from the ratio (1) gives the percent increase going from$454 to $1,015. The increase is 1.236 (or 123.6% after multiplying by 100). For example, if someone is currently making$454 a week and if his income is to increase to $1,015 a week, this represents a 123.6% raise! This is the average premium of a college education, i.e, the additional amount a college grad can expect to make. We can also flip the ratio in (1) and obtain the following ratio. $\displaystyle (3) \ \ \ \ \ \ \ \ \ \ \frac{454}{1015}=0.4473$ What to make of this ratio? Now the baseline is the college graduate. It tells us that the average earning of a high school dropout is only 0.4473 (or 44.73%) of the average earning of a college grad. For each one dollar earned by a college grad, the high school dropout only makes about 44.73 cents. For each one hundred dollars earned by a college grad, the high school dropout makes about fourty four dollars and seventy three cents. If we subtract one from the ratio in (3), we get the percentage change going from$1,015 to $454 (percentage decrease in this case). Note that $0.4473-1=-0.5527$. Multiplying this by 100, we get negative 55.27%. This says that going from$1,015 to $454 represents a 55.27% decrease in income. So if someone is currently making$1,015 a week, and if his income is to decrease to $454 a week, this represents a 55.27% decrease in income. To summarize, when we compare two quantities $P$ and $Q$ with $Q$ as a baseline, we can calculate the ratio $\displaystyle R=\frac{P}{Q}$. The ratio $R$ gives information about the quantity $P$ as a percentage of the baseline quantity $Q$. The quantity $R-1$ is the percentage change going from the baseline quantity $Q$ to the quantity $P$. If $R-1$ is a positive number, then it is a percentage increase when going from $Q$ to $P$. Otherwise it is a percentage decrease. In our income example in this post, $R-1$ can be interpreted as the additional earning of college graduates expressed as a percentage. ## The stealth price increases at the grocery store One particular snack containing almonds, cashews and cranberries that I buy at a local grocery store is now in a reduced size of 14 ounces per pack (previously a 16-ounce pack). I am getting 2 ounces less with the same price. This is a stealth price increase. In this tough economic time, this must be a favorite way for manufacturers to raise prices. Breakfast cereal boxes seem to be getting smaller. It will be an interesting exercise to find out what the percentage price increases are for these items with reduced sizes. For example, what is the percentage price increase for the snack pack that I buy (going from 16-ounce to 14-ounce)? Since the amount I pay at the cash register is the same whether it is 16-ounce or 14-ounce, I must look at the unit price (price per ounce). The old unit price is$0.28 (=$4.49/16) and the new unit price is$0.32 (=$4.49/14). Thus they are charging me 4 cents more per ounce by giving me two ounces less almonds and cashews. The percentage increase in price is 14.29% (0.04/0.28=0.1429). As a percentage, this is a hefty price increase! The following is a general formula for finding percentage price increase. $\displaystyle (1) \ \ \ \ \ \ \ \ \ \ \ \text{Percentage Price Increase}=\frac{\text{Dollar Price Increase}}{\text{Old Dollar Price}}$ There is another way to compute the percentage increase. Observe that 0.32/0.28=1.1429. Thus the ratio of the new price to the old price contains the information of the percentage increase. All we need to do is to subtract 1 from this ratio. We have the following formula, which is equivalent to the above formula. $\displaystyle (2) \ \ \ \ \ \ \ \ \ \ \ \text{Percentage Price Increase}=\frac{\text{New Price}}{\text{Old Price}}-1$ It turns out that in the situation of price increase by reducing weight while charging the same price, calculating the percentage increase does not require the price amount. In our example, the percentage price increase does not depend on the price being$4.49. All we need to do is to compute the ratio of the old weight to the reduced weight. This ratio contains the information of the percentage price increase. Simply subtract 1 from this ratio, we obtain the percentage increase. In our example, 16/14=1.1429.

$\displaystyle (3) \ \ \ \ \ \ \ \ \ \ \ \text{Percentage Price Increase}=\frac{\text{Old Weight}}{\text{Reduced Weight}}-1$

Formulas (1) and (2) are general formulas that work for all situations. Formula (3) can only be used for calculating percentage increases as a result of reducing the weight of the product (while charging the same dollar amount). Formula (3) works because the price of the product (e.g. \$4.49 in our example) is canceled out in the derivation of Formula (2).

Formula (3) can be handy for displaying scenarios (e.g. a manufacturer may want to determine the percentage increases for different sizes of reduction in weight). The following matrix displays the percentage increases for various sizes of reduction of a 16-ounces (or any other unit of measurement for weight).

Reducing a 16-unit package:

$\displaystyle \begin{pmatrix} \text{Weight Reduction}&\text{Percentage Increase} \\{\text{ }}&\text{ }&\text{ } \\\text{1}&\text{6.67 percent} \\\text{2}&\text{14.29 percent} \\\text{3}&\text{23.08 percent} \\\text{4}&\text{33.33 percent} \\\text{5}&\text{45.45 percent} \\\text{6}&\text{60.00 percent} \\\text{7}&\text{77.78 percent} \\\text{8}&\text{100.00 percent}\end{pmatrix}$

Obviously, the more the reduction in weight, the higher the percentage increase in price. I can see that a company could very well look at a matrix such as the one above to see what they can get away with. Reducing a 16-ounce package by one ounce only gives a 6.67% increase, probably not high enough from a manufacturer’s perspective. By reducing too much, the increase will no longer be stealth. So they probably would like to target a reduction that is not so conspicuous and yet one that can achieve a significant percent increase. In the case of our example of the snack packs of almonds and cashews, reducing the package by 2 ounces may not be too noticeable while acheiving a double digit percent price increase. That was a winner for the manufacturer!

If a manufacturer is bold enough to reduce the 16-ounce package to an 8-ounce package while charging the same price, the price increase would be 100%! You would to pay twice as much to get the same 16-ounce of the product! I am sure no manufacturer in their right mind would go for this.

One final observation. For a fixed amount of reduction in weight (say 2 ounces), the larger the original weight, the smaller the percentage price increase. For example, if the original weight is 24 ounces, a reduction of 2 ounces would only acheive a 9.09% increase. See the following matrix. Each percentage is calculated using Formula (3). For example, for a reduction of 2 ounces from a 24-ounce package, the percentage increase in price is 24/22 – 1 = 0.0909.

Reducing a 24-unit package:

$\displaystyle \begin{pmatrix} \text{Weight Reduction}&\text{Percentage Increase} \\{\text{ }}&\text{ }&\text{ } \\\text{1}&\text{4.35 percent} \\\text{2}&\text{9.09 percent} \\\text{3}&\text{14.29 percent} \\\text{4}&\text{20.00 percent} \\\text{5}&\text{26.32 percent} \\\text{6}&\text{33.33 percent} \\\text{7}&\text{41.18 percent} \\\text{8}&\text{50.00 percent}\end{pmatrix}$