The guessing game
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The Guessing Game. The entire business of Statistics is dedicated to the purpose of trying to guess the value of some population parameter. In what follows the population parameter (the target parameter ) will be either a mean or a proportion p .

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The Guessing Game

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The guessing game

The Guessing Game

The entire business of Statistics is dedicated to the purpose of trying to guess the value of some population parameter. In what follows the population parameter (the target parameter) will be either a mean or a proportion p.

All that we have at our disposal is n values taken at random from the population

x1, x2, …, xn

and their average and their variance:


The guessing game

Not much to go on, but when n is large (admittedly we don’t know what that means,

but n ≥ 30 will make us feel better) we have the wonderful

Central Limit Theorem

that tells us that the distribution of the sample mean (the average) is approximately normal


The guessing game

Trouble is, we don’t knowand we don’t know . We just have the following picture:

and the one number we know, is somewhere on the horizontal line. Actually


The guessing game

“somewhere”is a cop-out, we know exactly where is, we just don’t know where the red curve is relative to ! Could be like this


The guessing game

or like this:

(in both figures the blue dot is )


The guessing game

Looks somewhat hopeless, but there are some statements we can make for sure.

The next slides show four of them!

Hopefully you’ll catch on.Replace the blue dot (that represents ) with the REDDOT (that represents the standardization of )


The guessing game

80% chance the red dot is inside the blue bars.(why?)


The guessing game

85% chance the red dot is inside the blue bars.(why?)


The guessing game

90% chance the red dot is inside the blue bars.(why?)


The guessing game

95% chance the red dot is inside the blue bars.(why?)


The theory

The Theory

What went on in each of the previous four slides?

Let’s see. We picked a percentage of area

  • 80% - first slide

  • 85% - second

  • 90% - third

  • 95% - fourth

    From the chosen percentage we got

    (via the standard normal tables)


The guessing game

symmetric z-scores

  • 80%gave us-1.2851.285

  • 85%gave us-1.4391.439

  • 90%gave us-1.6491.649

  • 95%gave us-1.9601.960

    In fact, if you give me any

    positive area ≤ 1(call it1 - )

    I can find the corresponding

    symmetricz-scores

    by looking forthe area value


The guessing game

(figure it out from the figure!)

The two z-scores you get are written as

±Z/2

and the number 1 - iscalled

confidence coefficient if in decimal form

confidence level if in percent form


Confidence intervals

Confidence Intervals

Why are we using the word “confidence”?

Confidence in what?

Of course, we hope it is confidence in our prediction! In fact we want the confidence level to be just the probability that our prediction is correct.

Trouble is ….

We haven’t predicted anything !!

We just have established that


The guessing game

For any confidence coefficient 1 -we can find z-values and

so that

P(red dot between and ) = 1 -

Recall that the red dot stands for the standardized value


The guessing game

So we obtain the statement

that a little 7-th grade algebra transforms into

This is translated into English as:

is inside the interval


The guessing game

… with probability 1 -

Or, in slightly different (and more pompous sounding) words

“We are (1 - )% confident that is inside the interval

We call this interval the (1 - )%

confidence interval.

THAT’S OUR PREDICTION !


The guessing game

One last step: what do we use for sigma ?

If we know it (sometimes we do) …

HALLELUYA !

If not, we approximate sigma using the sample standard deviation

wheresis the (computed) sample standard deviation.


An example

An example

The numbers shown have been obtained as timeT(in seconds) elapsed from the time thecage door is openedto the time ofexit from the cagefor40 lab miceinseparate cages; (20 of the mice have been given a tranquilizer, the other 20 a placebo, but this is for another problem later.)

Construct the following

confidence intervals

for the mean ofT


The question

The Question:

  • 90% confidence interval

  • 95% confidence interval

  • 99% confidence interval

  • 30% confidence interval

  • 10% confidence interval


The guessing game

The data:

3.5 2.2 1.4 3.6 3.5 2.6 2.7 2.1 1.9 4.1

2.7 2.8 2.3 1.9 1.3 3.3 2.8 2.6 2.1 3.8

4.3 4.4 2.8 2.0 3.3 4.1 1.4 3.1 2.8 3.0

4.1 4.2 3.8 3.9 4.1 3.4 3.1 1.3 4.5 3.2

The sample mean and standard deviation are:

therefore


The z scores

The z-scores

For each we compute z’s (from my “stats” program or from the table)

  • 90% 1.645

  • 95% 1.960

  • 99% 2.579

  • 30% 0.385

  • 10% 0.126


The answers

The answers

Using the formula

we get the intervals

  • 90% [2.760, 3.239]

  • 95% [2.715, 3.285]

  • 99% [2.625, 3.375]

  • 30% [2.944, 3.056]

  • 10% [2.982, 3.018]

    Note that the higher the confidence the wider the interval. Is this reasonable?


Or who s gonna win the elections

orwho’s gonna win the elections?

Quite often one needs to estimate what

proportion p of a population prefers option A over option B.

One takes a “large enough” random sample of the population, counts how many prefer A, divides by the size n of the sample and gets a number,

denoted by(a statistic!) .


The guessing game

Of course is a random variable, and it turns out that it is an unbiased estimator ofp, that is

E( ) = p

If we knew the standard deviation of

we could construct confidence intervals forpas we did for the parameter .

(If n is big enogh the Central Limit Theorem still holds)

We can show that= pq/n (remember that q = 1 - p), but this is tautological (we don’t know p !)

However, if n is large enough, we can use

for p and proceed as with .


The guessing game

In other words, we use

instead of

and

instead of

and get the interval


An example1

An Example

What proportion p of Notre Dame students know a language other than English?

In a random sample of 1,500 Notre Dame students, 855 stated they knew some language other than English.

Develop a 98% confidence interval for p based on this sample.


The guessing game

We have = 0.02

Therefore= 2.33 (why?)

Now=855/1500 = 0.57 and therefore we approximate with 0.0128(why?)

We get the 98% confidence interval as

(0.57 – 2.33x0.128, 0.57 + 2.33x0.128)

That is

(0.54, 0.60)


The formulas

THE FORMULAS

  • To estimate with confidence 1 -


The guessing game

  • To estimate with confidence 1 -


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