Can we predict the future? In a way, yes. For example, it's Friday and if you commute, you'd probably predict it will take you longer to drive home tonight than it does on other nights. If I toss a coin ten times, you guess that I should get 5 heads.
A problem with our predictions is there is always the chance for error So even though we know tonight's commute will take longer, it's tough to say how much longer. Even though we expect 5 heads, we might get 4 or 6 or some other number in 10 tosses.
A better word for prediction in the context of Stat 10 is Expectation and Expected Value.
Definition.The EXPECTED VALUE in Chapter 17 is the number of draws from a box times the average of that box. The draws must be random with replacement for this to work.
Associate it with the idea of a "most likely outcome"
A basic example: a coin toss -- it has 2 outcomes. Head or Tails. Suppose we're interested in the count of heads in some number of tosses. We could assign a value of 1 if a toss comes up heads and a value of 0 if it comes up tails (because when we sum it up, it's just like a count of heads). We expect 50% for each outcome (i.e. half heads, half tails). The average of the box is .50 or 1/2 or 50% (i.e 0 + 1 divided by 2 -- a simple average of its "tickets").
In a situation of 10 tosses (draws), you wind up with an expected value of 5 (10 times 1/2). Or think of this situation as 5 heads is the most likely outcome.
Another Example: Suppose you are a sales executive and market research information suggests thatthese are your sales estimates for this month.
UNITS SOLD 300 500 750
PROBABILITY 1/3 1/3 1/3
The expected number sold (average) is 516.67 units. That is (300x 1/3) + (500 x 1/3) + (750 x 1/3) = 516.67 and multiplied by 1 (a single draw).
The expected number sold in 12 months would be
516.67 * 12 = 6,200 (approximately)
Possible interpretations: in a given month (assuming all months are the same) you'd expect to sell a little over 516.67 units. In 12 months, you'd expect to sell about 6200 units.
Actual Outcome (observed value) from some number of draws is
= expected value + chance error
where chance error is just some amount above or below the expected value.
Think about tossing a coin ten times. If I toss it ten times and get 9 heads, you might think I'm extremely lucky or I'm cheating.
If I toss a coin ten times and get 6 heads, you probably wouldn't think I was extremely lucky or that I was cheating. 6 seems reasonable, 9 doesn't. This is where the chance error component enters. What you are sensing, intuitively is the size of the chance error in a coin toss. For a coin box, the SD of the box is also 0.5 .
The standard error is a measure of the chance error. An outcome (sum) from some number of draws will be around an expected value but it can (and will be) off by chance error . The error should be close in magnitude to the standard error.
Formula: standard error = square root (number of draws) X the standard deviation (SD) of the box. So for a coin box, in 10 tosses, the standard error = ROOT(10) * 0.5 = 1.6 (approximately)
Remember, Standard deviation is a measure of spread. What the formula suggests is that the more draws you make, the larger the standard error.. Example: 4 draws, the multiplier is 2 (root 4) , 9 draws it is 3 (root 9) , 25 draws it is 5 (root 25) and so forth.
Note: the standard error is not the same as the standard deviation. The SD is calculated for lists, but the SE is for some kind of chance or random process, like a lottery, like drawing tickets from a box, like tossing a coin. The SD is part of an SE though.
Since Sunday is Halloween. I have a treat for you.
All that is required is that you (a) calculate the expected value of the box and (b) calculate the SE. based on the number of draws and the SD (standard deviation) of the box
Then, you can calculate standard units with a familiar formula:
Z = (observed value - expected value)
--------------------------------
SE
Example. Let's go back to the 9 heads in 10 tosses of a coin idea. The SE for the coin toss situation is SQRT(10) x SQRT(.5 x .5) (read 17.5 for the specifics). Let's see how likely it is to get 9 heads in 10 tosses.
SE = 1.5811 and Z = ((9 -5) / 1.5811) = 2.52 or about 2.55. The area between + and - 2.55 is 98.92% which leaves 1% total outside of the area. So the chance of getting 9 heads or better is about 1/2 of a percent. The chance of getting 6 heads or more is about 25%
Your intuitive sense works well. The combination of the expected value, standard error, and normal curve validates your suspicions.
This same method can be used to figure out chances in all kinds of situations.