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The Guessing Game

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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:

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

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

“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

or like this:

(in both figures the blue dot is )

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 )

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

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

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

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

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)

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

(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

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

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

So we obtain the statement

that a little 7-th grade algebra transforms into

This is translated into English as:

is inside the interval

… 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 !

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.

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

- 90% confidence interval
- 95% confidence interval
- 99% confidence interval
- 30% confidence interval
- 10% confidence interval

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

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

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?

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!) .

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 .

In other words, we use

instead of

and

instead of

and get the interval

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.

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)

- To estimate with confidence 1 -

- To estimate with confidence 1 -