Sampling Sampling Distributions of Distributions of

ProportionsProportions

• Toss a penny 20 times and record the number of heads.

• Calculate the proportion of heads & mark it on the dot plot on the board.

What shape do you think the dot plot will have?

The dotplot is a partial graph of the sampling distribution of all sample proportions of sample

size 20. If I found all the possible sample proportions – this would

be approximately normal!

Sampling DistributionSampling Distribution

• Is the distribution of possible values of a statistic from all possible samples of the same size from the same population

• In the case of the pennies, it’s the distribution of all possible sample proportions (p)

We will use:p for the population

proportionand

p-hat for the sample proportion

Suppose we have a population of six Suppose we have a population of six people: Alice, Ben, Charles, Denise, people: Alice, Ben, Charles, Denise, Edward, & FrankEdward, & Frank

What is the proportion of females?

What is the parameter of interest in this population?

 

Draw samples of two from this population.Draw samples of two from this population.

How many different samples are possible?

gendergender

66CC22 =15 =15

1/31/3

Find the 15 different samples that are Find the 15 different samples that are possible & find the sample proportion of the possible & find the sample proportion of the

number of females in each sample.number of females in each sample.

Alice & Ben .5Alice & Charles .5Alice & Denise 1Alice & Edward .5Alice & Frank .5Ben & Charles 0Ben & Denise .5Ben & Edward 0

Ben & Frank 0

Charles & Denise .5

Charles & Edward 0

Charles & Frank 0

Denise & Edward.5

Denise & Frank .5

Edward & Frank 0

Find the mean & standard deviation of all p-hats.Find the mean & standard deviation of all p-hats.

29814.0σ&31

μ ˆˆ pp

How does the mean of the sampling distribution (p-hat) compare to the population

parameter (p)?p-hat = p

Formulas:Formulas:

npp

p

p

p

1σ

μ

ˆ

ˆ

These are found on the formula chart!

Does the standard deviation of the Does the standard deviation of the sampling distribution equal the sampling distribution equal the

equation?equation? NO -

29814.031

23

231

σˆ p

WHY?WHY?We are sampling more than 10% of our population! If we use the correction factor, we will see that we are correct.

29814.0

1626

23

231

σˆ

p

Correction factor – multiply by

1

NnN

So – in order to calculate the standard deviation of the sampling distribution, we

MUST be sure that our sample size is less than 10%

of the population!

Assumptions (Rules of Thumb)Assumptions (Rules of Thumb)

• Sample size must be less than 10% of the population (independence)

• Sample size must be large enough to insure a normal approximation can be used.

np np >> 10 & n (1 – p) 10 & n (1 – p) >> 10 10

Why does the second assumption insure Why does the second assumption insure an approximate normal distribution?an approximate normal distribution?

Suppose n = 10 & p = 0.1 (probability of a success), a histogram of this distribution is strongly skewed right!

Remember back to binomial distributions

Now use n = 100 & p = 0.1 (Now np > 10!) While the histogram is still strongly skewed right – look what happens to the tail!

np > 10 & n(1-p) > 10 insures that the sample size is large enough to

have a normal approximation!

Based on past experience, a bank believes that 7% of the people who receive loans will not make payments on time. The bank recently approved 200 loans.

What are the mean and standard deviation of the proportion of clients in this group who may not make payments on time?

Are assumptions met?

What is the probability that over 10% of these clients will not make payments on time?

01804.

20093.07.

σ

07.μ

ˆ

ˆ

p

p

Yes – np = 200(.07) = 14n(1 - p) = 200(.93) = 186

Ncdf(.10, 1E99, .07, .01804) = .0482

Suppose one student tossed a coin 200 times and found only 42% heads. Do you believe that this is likely to happen?

0118.200

)5(.5.,5,.42,.

ncdf

No – since there is approximately a 1% chance of

this happening, I do not believe the student did this.

np = 200(.42) = 84 & n(1-p) = 200(.58) = 116Since both > 10, I can use a normal curve!

Find & using the formulas.

Assume that 30% of the students at SHS wear contacts. In a sample of 100 students, what is the probability that more than 35% of them wear contacts?

Check assumptions!p-hat = .3 & p-hat = .045826

np = 100(.3) = 30 & n(1-p) =100(.7) = 70

Ncdf(.35, 1E99, .3, .045826) = .1376