1. Sampling Distributions of Proportions

2. Sampling 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 proportion
and
p-hat for the sample proportion

3. What is the proportion of females?
Suppose we have a population of six people: Alice, Ben, Charles, Denise, Edward, & Frank
What is the proportion of females?
What is the parameter of interest in this population?
 Draw samples of two from this population.
How many different samples are possible?
1/3
Proportion of females
6C2 =15

4. Find the 15 different samples that are possible & find the sample proportion of the number of females in each sample.
Ben & Frank 0
Charles & Denise .5
Charles & Edward 0
Charles & Frank 0
Denise & Edward .5
Denise & Frank .5
Edward & Frank 0
Alice & Ben .5
Alice & Charles .5
Alice & Denise 1
Alice & Edward .5
Alice & Frank .5
Ben & Charles 0
Ben & Denise .5
Ben & Edward 0
How does the mean of the sampling distribution (mp-hat) compare to the population parameter (p)?
mp-hat = p
Find the mean & standard deviation of all p-hats.

5. These are found on the formula chart!
Formulas:
These are found on the formula chart!

6. Correction factor – multiply by
Does the standard deviation of the sampling distribution equal the equation?
NO -
So – in order to calculator the standard deviation of the sampling distribution, we MUST be sure that our sample size is less than 10% of the population!
WHY?
Correction factor – multiply by
We are sampling more than 10% of our population! If we use the correction factor, we will see that we are correct.

7. Assumptions (Rules of Thumb)
Sample size must be less than 10% of the population (independence)
Sample size must be large enough to ensure a normal approximation can be used.
np > 10 & n (1 – p) > 10

8. Remember back to binomial distributions
Why does the second assumption ensure an approximate normal distribution?
Remember back to binomial distributions
Suppose n = 10 & p = 0.1 (probability of a success), a histogram of this distribution is strongly skewed right!
np > 10 & n(1-p) > 10
insures that the sample size is large enough to have a normal approximation!
Now use n = 100 & p = 0.1 (Now np > 10!) While the histogram is still strongly skewed right – look what happens to the tail!

9. 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?
Yes –
np = 200(.07) = 14
n(1 - p) = 200(.93) = 186
Ncdf(.10, 1E99, .07, ) = .0482

10. Suppose one student tossed a coin 200 times and found only 42% heads
Suppose one student tossed a coin 200 times and found only 42% heads. Do you believe that this is likely to happen?
np = 200(.5) = 100 & n(1-p) = 200(.5) = 100
Since both > 10, I can use a normal curve!
Find m & s using the formulas.
No – since there is approximately a 1% chance of this happening, I do not believe the student did this.

11. Assume that 30% of the students at BHS wear contacts
Assume that 30% of the students at BHS wear contacts. In a sample of 100 students, what is the probability that more than 35% of them wear contacts?
Check assumptions!
mp-hat = & sp-hat =
np = 100(.3) = 30 & n(1-p) =100(.7) = 70
Ncdf(.35, 1E99, .3, ) = .1376