Griffiths D. Head First Statistics

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Some books tell you to

divide by n – 1 for a

sample, and some tell

you to divide by n.

This is because different

books make different assumptions about

what you’re using your sample for. If

you’re using the sample to estimate the

population variance, then you need to

divide by n – 1. You only need to divide

by n if you want to calculate the variance

of that exact set of values.

If you’re taking a statistics exam, check

the approach that your exam board takes.

Which formula’s which?

Sometimes it can be tricky deciding whether you should

divide by n for the variance, or whether you should divide by

n – 1. The golden rule to remember is that dividing by n gives

you the actual variance for the set of data that you

have.

If you have the data for the entire population, then dividing

by n gives you the actual variance of the population. You

need to use the formula for σ

and divide by n.

If you have a sample of data from the population, then

chances are you’ll want to use this to estimate the variance of

the population. This means that you need to use the formula

for s

and divide by n – 1.

Here’s a reminder of the data from the Mighty Gumball sample.

What do you estimate the population variance to be?

61.9 62.6 63.3 64.8 65.1 66.4 67.1 67.2 68.7 69.9

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Here’s a reminder of the data from the Mighty Gumball sample.

What do you estimate the population variance to be?

61.9 62.6 63.3 64.8 65.1 66.4 67.1 67.2 68.7 69.9

We can estimate the population variance by calculating s

= s

= (x - x)

n - 1

= (-3.8)

+ (-3.1)

+ (-2.4)

+ (-0.9)

+ (-0.6)

+ (0.7)

+ (1.4)

+ (1.5)

+ (3)

+ (4.2)

= 14.44 + 9.61 + 5.76 + 0.81 + 0.36 + 0.49 + 1.96 + 2.25 + 9 + 17.64

= 62.32/9

= 6.92 (to 2 decimal places)

Why do I divide by n – 1 for the

sample variance? Why can’t I divide by

A: You divide by n – 1 for a sample

because most of the time, you use your

sample data to estimate the variance of the

population. Dividing by n – 1 gives you a

slightly more accurate result than dividing by

n. This is because the variance of values in

the sample is likely to be slightly lower than

the population variance.

Is there some mathematical basis

for this?

A: Yes there is. It’s something that we’re

going to touch upon at the end of the chapter,

but hold onto that thought; it’s a good one.

How do I remember which symbols

are used for the population, and which

are used for the sample?

A: In general, Greek letters are used for

the population, and normal Roman letters

are used for the mean and variance for the

sample.

Is there a point estimator for the

standard deviation in the same way that

there is for the variance? How do I find it?

A: If you need to estimate the standard

deviation, start by calculating the estimator

for the variance. The estimator for the

standard deviation is the square root of this.

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estimating populations and samples

Mighty Gumball has done more sampling

The Mighty Gumball CEO is so inspired by the results of the taste test that

he’s asked for another sampling exercise that he can use for his television

advertisements. This time, the CEO wants to be able to say how popular

Mighty Gumball’s candy is compared with that of their main rival.

The Mighty Gumball staff have asked a random sample of people whether

they prefer gumballs produced by Mighty Gumball or whether they prefer

those of their main rival. They’re hoping they can use the results to predict

what proportion of the population is likely to prefer Mighty Gumball.

Mighty Gumball has found that in a sample of 40 people, 32 of them prefer

their gumballs. The other 8 prefer those of their rival.

How would you find the proportion of people in the sample who prefer Mighty

Gumball’s candy? What distribution do you think this follows? How do you think

you could apply this to the population?

Ewww! These

gumballs are gross.

I’ll choose Mighty

Gumball over any

other manufacturer.

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It’s a question of proportion

For the latest Mighty Gumball sample, the thing the CEO is interested in is

whether or not each person prefers Mighty Gumball confectionery to that of

their chief rival. In other words, every person who prefers Mighty Gumball

candy can be classified as a success.

So how do we use the sample data to predict the proportion of successes in

the population?

Predicting population proportion

If we use X to represent the number of successes in the population, then X

follows a binomial distribution with parameters n and p. n is the number of

people in the population, and p is the proportion of successes.

In the same way that our best estimate of the population mean is the

mean of the sample, our best guess for the proportion of successes in the

population has to be the proportion of successes in the sample. This means

that if we can find the proportion of people in the sample who prefer Mighty

Gumball’s treats, we’ll have a good estimate for the proportion of people who

prefer Mighty Gumball in the general population.

We can find the proportion of successes in the sample by taking the total

number of people who prefer Mighty Gumball, and then dividing by

the total number of people in the sample. If we use p

to represent the

proportion of successes in the sample, then we can estimate the proportion of

successes in the population using

p = p

In other words, we can use the proportion of successes in the sample as a

point estimator for the proportion of successes in the population. In the case

of the company’s latest sample, 32 out of 40 people prefer Mighty Gumball

confectionery, which means that p

= 0.8. Therefore, the point estimator for

the proportion of successes in the population is also 0.8.

where

= number of successes

number in sample

Proportion of successes in the

sample

Point estimator for the proportion

of successes in the population

point estimators for proportions

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Probability and proportion are related

There’s actually a very close relationship between probability and

proportion.

Imagine you have a population for which you want to find the

proportion of successes. To calculate this proportion, you take the

number of successes, and divide by the size of the population.

Now suppose you want to calculate the probability of choosing a

success from the population at random. To derive this probability,

you take the number of successes in the population, and divide

by the size of the population. In other words, you derive the

probability of getting a success in exactly the same way

as you derive the proportion of successes.

We use the letter p to represent the probability of success in the

population, but we could easily use p to represent proportion

instead—they have the same value.

p = probability = proportion

So am I right in thinking that

probability and proportion are

related? They’re both represented by

p, and they sound like they’re similar.

Mighty Gumball takes another sample of their super-long-lasting

gumballs, and finds that in the sample, 10 out of 40 people prefer

the pink gumballs to all other colors. What proportion of people

prefer pink gumballs in the population? What’s the probability of

choosing someone from the population who doesn’t prefer pink

gumballs?

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Mighty Gumball takes another sample of their super-long-lasting

gumballs, and finds that in the sample, 10 out of 40 people prefer

the pink gumballs to all other colors. What proportion of people

prefer pink gumballs in the population? What’s the probability of

choosing someone from the population who doesn’t prefer pink

gumballs?

We can estimate the population proportion with the sample proportion. This gives us

p = p

= 10/40

= 0.25

The probability of choosing someone from the population who doesn’t prefer pink gumballs is

P(Preference not Pink) = 1 - p

= 1 - 0.25

= 0.75

So is proportion the same thing as

probability?

A: The proportion is the number of

successes in your population, divided by

the size of your population. This is the

same calculation you would use to calculate

probability for a binomial distribution.

Does proportion just apply to the

binomial distribution? What about other

probability distributions?

A: Out of all the probability distributions

we’ve covered, the only one which has

any bearing on proportion is the binomial

distribution. It’s specific to the sorts of

problems you have with this distribution.

Is the proportion of the sample the

same as the proportion of the population?

A: The proportion of the sample can be

used as a point estimator for the proportion

of the population. It’s effectively a best

guess as to what the value of the population

proportion is.

Is that still the case if the sample is

biased in some way? How do I estimate

proportion from a biased sample?

A: The key here is to make sure that

your sample is unbiased, as this is what you

base your estimate on. If your sample is

biased, this means that you will come with

an inaccurate estimate for the population

proportion. This is the case with other point

estimators too.

So how do I make sure my sample

is unbiased?

A: Going through the points we

raised in the previous chapter is a good

way of making sure your sample is as

representative as possible. The hard work

you put in to preparing your sample is

worth it because it means that your point

estimators are a more accurate reflection of

the population itself.

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This is awesome!

We have a lot of

impressive statistics

we can use in our

advertising.

people

prefer

pink!

The point estimator for the population

variance is given by

= s

where s

is given by

Σ(x - x)

n - 1

The population proportion is represented

using p. It’s the proportion of successes within

the population.



The point estimator for p is given by p

where p

is the proportion of successes in the

sample.

p = p

You calculate p

by dividing the number of

successes in the sample by the size of the

sample.

= number of successes

number in sample



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Buy your gumballs here!

Remember the Statsville Cinema? They’re recently been authorized

to sell Mighty Gumball products to film-goers, and it’s a move that’s

proving popular with most of their customers.

The trouble is, not everybody’s happy.

I really like red gumballs,

and I’d rather not eat the

other colors. How many red

gumballs come in the box?

Introducing new jumbo boxes

The cinema sells mixed boxes of gumballs, and this weekend,

they’re putting on a film marathon of classic films.

The event looks like it’s going to be popular, and tickets are selling

well. The trouble is, some people get cranky if they don’t get their

fix of red gumballs.

A jumbo box of gumballs is meant for sharing, and each box

contains 100 gumballs. 25% of gumballs in the entire gumball

population are red.

I need 40 red gumballs to

make it through the movie.

Is that likely? If there

aren’t enough red gumballs in

the box, I’ll get another snack

instead.

We need to find the probability that in one

particular jumbo box, 40 or more of the

gumballs will be red.

Since there are 100 gumballs per box, that means we need

to find the probability that 40% of the gumballs in this box

are red, given that 25% of the gumball population is red.

finding sample probabilities

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So how does this relate to sampling?

So far, we’ve looked at how to put together an unbiased sample, and

how to use samples to find point estimators for population parameters.

This time around, the situation’s different. Here, we’re told what the

population parameters are, and we have to work out probabilities

for one particular jumbo box of gumballs. In other words, instead

of working out probabilities for the population, we need to work out

probabilities for the sample proportion.

Isn’t that the sort of thing

that we were doing before?

What’s the big deal?

This time, we’re looking for probabilities for a

sample, not a population.

Rather than work out the probability of getting particular frequencies

or values in a probability distribution, this time around we need to find

probabilities for the sample proportion itself. We need to figure

out the probability of getting this particular result in this particular box

of gumballs.

Before we can work out probabilities for this, we need to figure out the

probability distribution for the sample proportion. Here’s what we need

to do:

Look at all possible samples the same size as the one we’re

considering.

If we have a sample of size n, we need to consider all possible samples of

size n. There are 100 gumballs in the box, so in this case n is 100.

Look at the distribution formed by all the samples, and

find the expectation and variance for the proportion.

Every sample is different, so the proportion of red gumballs in each box

of gumballs will probably vary.

Once we know how the proportion is distributed, use it to

find probabilities.

Knowing how the proportion of successes in a sample is distributed

means we can use it to find probabilities for the proportion in a random

sample—in this case, a jumbo box of gumballs.

Let’s take a look at how to do this.

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Sample

The sampling distribution of proportions

So how do we find the distribution of the sample proportions?

Let’s start with the gumball population. We’ve been told what the

proportion of red gumballs is in the population, and we can represent

this as p. In other words, p = 0.25.

Each jumbo box of gumballs is effectively a sample of gumballs taken from the

population. Each box contains 100 gumballs, so the sample size is 100. Let’s

represent this with n.

If we use the random variable X to represent the number of red gumballs in

the sample, then X ~ B(n, p), where n = 100 and p = 0.25.

The proportion of red gumballs in the sample depends on X, the number of

red gumballs in the sample. This means that the proportion itself is a random

variable. We can write this as P

, where P

= X/n

= X

X ~ B(n, p)

We don’t know the exact number

of red gumballs in the sample, but

we know its distribution.

represents the proportion

of successes in the sample.

25% of gumballs in the

gumball population are red,

so p = 0.25.

Population of gumballs

distribution of sample proportions