Sampling Distributions of Proportions

Toss a penny 20 times and count the number of heads Calculate the proportion (percent) of heads & mark it on the dot plot on the board What shape do you think the dot plot should have?

Proportions Parameter: A number that describes a population  p Statistic: A number computed from sample data

Why not always take a census? Not always accurate Difficult/impossible to do Very expensive

Sampling Distribution All the values a variable can be Sampling Distribution The values a statistic can be in all possible samples of the same size from the same population  our dotplot is a partial sampling distribution

Suppose we have a population of six people: Alice, Ben, Charles, Denise, Edward, & Frank What is the true proportion of females? How many samples of size two are possible? 1/3 6C2 =15

Find the 15 possible samples and the proportion of females in each sample. B & F 0 C & D .5 C & E 0 C & F 0 D & E .5 D & F .5 E & F 0 A & B .5 A & C .5 A & D 1 A & E .5 A & F .5 B & C 0 B & D .5 B & E 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.

Unbiased Estimator A statistic is an unbiased estimator of a parameter if the mean of its sampling distribution equals the true parameter  p is an unbiased estimator of p

Formulas: On the green sheet!

Does the standard deviation formula work for our example? NO! Condition: We MUST have a sample size that is less than 10% of the population! WHY? Correction factor: Multiply by Each sample contains more than 10% of our population, so we can no longer assume independence of people in our samples!

Conditions for Proportions Population is at least 10n (independence) Large enough n so that sampling dist. is approx. normal np > 10 & n(1 – p) > 10

Remember back to the binomial distribution Why does the second condition ensure an approximate normal distribution? Remember back to the binomial distribution Suppose n = 10 & p = 0.1 A histogram of this distribution is strongly skewed right! np > 10 & n(1 – p) > 10 ensures the binomial distribution can be approximated by a normal distribution Now use n = 100 & p = 0.1 (So np > 10). While the histogram is still skewed right, look what happens to the tail!

P(p-hat > .1) = normalcdf(.10, 1E99, .07, .01804) = .0482 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 sample who may not make payments on time? Are the conditions met? What is the probability that over 10% of these clients will not make payments on time? Yes: 1. Banks give out at least 10(200) = 2000 loans 2. np = 200(.07) = 14 > 10 n(1 – p) = 200(.93) = 186 > 10 P(p-hat > .1) = normalcdf(.10, 1E99, .07, .01804) = .0482

Suppose a student tossed a coin 200 times and claimed to find only 42% heads. Do you believe them? Conditions: 1. Coin flips are already independent 2. np = 200(.5) = 100 > 10; n(1-p) = 200(.5) = 100 > 10 There's only a 1.18% chance of this happening, so I do not believe the student did this.

About 60% of the students at GBHS wear contacts About 60% of the students at GBHS wear contacts. In a sample of 100 students, what is the probability that more than 65% of them wear contacts? Conditions: 1. There are more than 10(100) = 1000 students at GBHS 2. np = 100(.6) = 60 > 10; n(1-p) = 100(.4) = 40 > 10 mp-hat = .6 & sp-hat = .049 normalcdf(.65, 1E99, .6, .049) = .1538

What is the mean and standard deviation of this population? Sampling Distributions – Means  Population: length of fish (in inches) in my pond. Distribution: 2, 7, 10, 11, 14 What is the mean and standard deviation of this population? mx = 8.8 sx = 4.0694

Let’s take samples of size 2 (n = 2) from this population. How many samples of size 2 are possible? 5C2 = 10 mx = 8.8 Find all 10 of these samples and record the sample means. What is the mean and standard deviation of the sample means? sx = 2.4919

Repeat with sample size n = 3. How many samples of size 3 are possible? 5C3 = 10 mx = 8.8 What is the mean and standard deviation of the sample means? Find all of these samples and record the sample means. sx = 1.66132

What do you notice? mx = m sx as n Sampling distribution mean EQUALS population mean As sample size increases, std. dev. of the sampling distribution decreases mx = m sx as n

General Properties mx = m  x is an unbiased estimator of μ s sx = n Rule 1: Rule 2: Standard Error of the Mean  x is an unbiased estimator of μ s n sx =

General Properties Rule 3: When population is normal, sampling distribution of x is also normal for any n

General Properties Rule 4: Central Limit Theorem When n is sufficiently large, the sampling distribution of x is approx. normal, even if the population is not  CLT can be safely applied when n > 30

1) The army reports that the distribution of head circumference among soldiers is approximately normal with mean 22.8 inches and standard deviation of 1.1 inches. a) What is the probability that a randomly selected soldier’s head will have a circumference that is greater than 23.5 inches? P(X > 23.5) = .2623

What normal curve are you now working with? b) What is the probability that a random sample of five soldiers will have an average head circumference that is greater than 23.5 inches? Do you expect the probability to be more or less than the answer to part (a)? Why? What normal curve are you now working with? P(X > 23.5) = .0774

2) A team of biologists has been studying a fishing pond 2) A team of biologists has been studying a fishing pond. Let X = the length of a single trout taken at random from the pond. The biologists have determined that X has a normal distribution with mean 10.2 in. and standard deviation 1.4 in. a) What is the probability that a single trout taken at random from the pond is between 8 and 12 inches long? P(8 < X < 12) = .8427

P(8< x <12) = .9978 x = 11.23 inches b) What is the probability that the mean length of five trout taken at random is between 8 and 12 inches long? c) What sample mean would be at the 95th percentile (when n = 5)? Do you expect the probability to be more or less than the answer to part (a)? Why? P(8< x <12) = .9978 x = 11.23 inches

3) A soft-drink bottler claims that, on average, cans contain 12 oz of pop. Let X = actual volume of pop in a randomly selected can. Suppose X is normally distributed with s = .16 oz. Sixteen cans are randomly selected and a mean of 12.1 oz is calculated. What is the probability that the mean of 16 cans exceeds 12.1 oz? P(x >12.1) = .0062

4) Koegels asserts that its “Viennas” brand has an average fat content of 18 grams per hot dog with standard deviation 1 gram. Consumers of this brand would probably not be disturbed if the mean were less than 18 grams, but would be unhappy if it exceeded 18 grams. An independent testing organization is asked to analyze a random sample of 36 hot dogs. Suppose the resulting sample mean is 18.4 grams. What is the probability of getting a sample mean of 18.4 grams or higher? P(x > 18.4) = .0082

Does this result indicate that the Koegels claim is incorrect? Yes – if it were true, it would be almost impossible to get our results by chance.