Chapter 9 found online and modified slightly!

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# Chapter 9 found online and modified slightly! - PowerPoint PPT Presentation

Chapter 9 found online and modified slightly!. Sampling Distributions. Parameter. A number that describes the population Symbols we will use for parameters include m - mean s – standard deviation p – proportion (p) a – y-intercept of LSRL b – slope of LSRL. Statistic.

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### Chapter 9found online and modified slightly!

Sampling Distributions

Parameter
• A number that describes the population
• Symbols we will use for parameters include

m - mean

s – standard deviation

p – proportion (p)

a – y-intercept of LSRL

b – slope of LSRL

Statistic
• A number that that can be computed from sample data without making use of any unknown parameter
• Symbols we will use for statistics include

x – mean

s– standard deviation

p– proportion

a– y-intercept of LSRL

b– slope of LSRL

The sampling distribution of a statistic is the distribution of values taken by the statistic in all possible samples of the same size from the same population.

Consider the population of 5 fish in my pond – the length of fish (in inches):

2, 7, 10, 11, 14

What is the mean and standard deviation of this population?

mx = 8.8

sx = 4.0694

2, 7, 10, 11, 14

Let’s take samples of size 2 (n = 2) from this population:

How many samples of size 2 are possible?

2,7

2,10

2,11

2,14

7,10

7,11

7,14

10,11

10,14

11,14

5C2 = 10

mx = 8.8 Same as the population mean!

Find all 10 of these samples and record the sample means.

What is the mean of the sample means?

sx = Can’t be calculated, because our population is not at least 10 times

as large as our population size.

2, 7, 10, 11, 14

Repeat this procedure with sample size n = 3. How many samples of size 3 are possible?

2,7,10

2,7,11

2,7,14

2,10,11

2,10,14

2,11,14

7,10,11

7,10,14

7,11,14

10,11,14

5C3 = 10

mx = 8.8 Same as our population mean.

What is the mean and standard deviation of the sample means?

Find all of these samples and record the sample means.

sx = Can’t be calculated, because our population is not at least 10

times as large as our population size.

mx= m

sx

What do you notice?
• The mean of the sampling distribution EQUALS the mean of the population.
• As the sample size increases, the standard deviation of the sampling distribution decreases.

as n

We will see this occur in our “fishing activity”

A statistic used to estimate a parameter is unbiasedif the mean of its sampling distribution is equal to the true value of the parameter being estimated.

mx= m

s

sx =

n

General Properties

Rule 1:

Rule 2:

This rule is approximately correct as long as population size is at least 10 times larger than the sample size.

General Properties

Rule 3:

When the population distribution is normal, the sampling distribution of x is also normalfor any sample size n.

General Properties

Rule 4: Central Limit Theorem

When n is sufficiently large, the sampling distribution of x is well approximated by a normal curve, even when the population distribution is not itself normal.

How large is “sufficiently large” anyway?

CLT can safely be applied if n exceeds 30.

EX) 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

normalcdf (23.5, e99, 22.8, 1.1)

lower b, upper b, mean, s.d.

P(X > 23.5) =

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)? Explain

What normal curve are you now working with?

normalcdf (23.5, e99, 22.8, 1.1/√(5))

.0774

Suppose a team of biologists has been studying the Pinedale children’s fishing pond. Let x represent the length of a single trout taken at random from the pond. This group of biologists has determined that the length has a normal distribution with mean of 10.2 inches and standard deviation of 1.4 inches. 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

normalcdf (8, 12, 10.2, 1.4)

P(8< x <12) = .9978

x = 11.23 inches

What is the probability that the mean length of five trout taken at random is between 8 and 12 inches long?

What sample mean would be at the 95th percentile? (Assume n = 5)

Do you expect the probability to be more or less than the answer to part (a)? Explain

Normalcdf (8, 12, 10.2, 1.4/√(5))

invnorm (.95, 10.2, 1.4/√(5))

P(x >12.1) = .0062

A soft-drink bottler claims that, on average, cans contain 12 oz of soda. Let x denote the actual volume of soda in a randomly selected can. Suppose that 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 will exceed 12.1 oz?

normalcdf (12.1, e99, 12, .16/√(16))

A hot dog manufacturer asserts that one of its brands of hot dogs has a average fat content of 18 grams per hot dog with standard deviation of 1 gram. Consumers of this brand would probably not be disturbed if the mean was 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 that the sample mean is greater than 18.4 grams?

normalcdf (18.4, e99, 18, 1/√(36))

P(x >18.4) = .0082

No. The manufacturer’s claim was that its hotdogs had an average fat content of 18 grams.

We found that the probability (fat content was greater than 18.4 grams) = .0082 which is less than 1%. This means that approximately 99% of their hotdogs have fat content less than 18.4 grams.