9.1 – Sampling Distributions. Many investigations and research projects try to draw conclusions about how the values of some variable x are distributed

9.1 – Sampling Distributions

Many investigations and research projects try to draw conclusions about how the values of some variable x are distributed in a population. Often, attention is focused on a single characteristic of that distribution. Examples include:

1. x = fat content (in grams) of a quarter-pound hamburger, with interest centered on the mean fat content μ of all such hamburgers

2. x = fuel efficiency (in miles per gallon) for a 2003 Honda Accord, with interest focused on the variability in fuel efficiency as described by σ, the standard deviation for the fuel efficiency population distribution

3. x = time to first recurrence of skin cancer for a patient treated using a particular therapy, with attention focused on p, the proportion of such individuals whose first recurrence is within 5 years of the treatment.

Parameter:

Statistic:

A number that describes the population. This number is typically unknown.

A number that describes the sample. We use this number to estimate the parameter.

Population Sample

Mean

Standard Deviation

Proportion

Standard deviation of the proportion

Parameter Statistic

x

p p̂

p

x

p̂

Sampling Distribution:

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

Ex: Take 100 samples of size n = 20.

Sampling Variability:

The variation between each groups of samples of the same size.

If I compare many different samples and the statistic is very similar in each one, then the sampling variability is low. If I compare many different samples and the statistic is very different in each one, then the sampling variability is high.

Unbiased:

When the statistic is equal to the true value of the parameter

Unbiased Estimator:

The unbiased statistic

Ex: 20 and = 20x

How sampling works:

1. Take a large number of samples from the same population.

2. Calculate the sample mean or sample proportion for each sample

3. Make a histogram of the values of the statistics

4. Examine the distribution

Facts about Samples:

If the population mean ( ) and the population standard

deviation ( ) are unknown, we can use x to estimate

and use to estimate . These estimates may or may

not be reliable.x

•

• If I chose a different sample, it would still represent the same population. A different sample almost always produces different statistics.

• A statistic can be unbiased and still have high variablility. To avoid this, increase the size of the sample. Larger samples give smaller spread.

Example #1: Classify each underlined number as a parameter or statistic. Give the appropriate notation for each.

a. Forty-two percent of today’s 15-year-old girls will get pregnant in their teens.

p =

Parameter

0.42

b. The National Center for Health Statistics reports that the mean systolic blood pressure for males 35 to 44 years of age is 128 and the standard deviation is 15. The medical director of a large company looks at the medical records of 72 executives in this age group and finds that the mean systolic blood pressure for these executives is 126.07.

Example #1: Classify each underlined number as a parameter or statistic. Give the appropriate notation for each.

=

128 and 15 are parameters 126.07 is a statistic

128

= 15

x 126.07

Example #2: Suppose you have a population in which 60% of the people approve of gambling.

a. Is 60% a parameter or a statistic? Give appropriate notation for this value.

p = 0.60Parameter,

You want to take many samples of size 10 from this population to observe how the sample proportion who approve of gambling vary in repeated samples.b. Describe the design of a simulation using the partial random digits table below to estimate the sample proportion who approve of gambling. Label how you will conduct the simulation. Then carry out five trials of your simulation. What is the average of the samples? How close is it to the 60%?

0 – 5 approve of gambling 6-9 don’tAssign:

After choosing 10 Stop:

# of people that approve of gamblingCount:

Ok to have repeat numbers, represent new person

Repeaters:

3 6 0 0 9 1 9 3 6 5 1 5 4 1 23 9 6 3 8 8 5 4 5 3 4 6 8 1 6

3 8 4 4 8 4 8 7 8 9 1 8 3 3 82 4 6 9 7 3 9 3 6 4 4 2 0 0 6

8 2 7 3 9 5 7 8 9 0 2 0 8 0 74 7 5 1 1 8 1 6 7 6 5 5 3 0 0

6 0 9 4 0 7 2 0 2 4 1 7 8 6 8 2 4 9 4 3 6 1 7 9 0 9 0 6 5 6

6 8 4 1 7 3 5 0 1 3 1 5 5 2 97 2 7 6 5 8 5 0 8 9 5 7 0 6 7

A D A A D A D A D A

1: 6/10 = 60%

2: 4/10 = 40%

3: 4/10 = 40%

4: 7/10 = 70%

5: 7/10 = 70%

.6 .4 .4 .7 .7

5

p̂

ˆ 0.56p

c. The sampling distribution of is the distribution of from all possible SRSs of size 10 from this population. What would be the mean of this distribution if this process was repeated 100 times?

p = 0.60

p̂p̂

d. If you used samples of size 20 instead of size 10, which sampling distribution would give you a better estimate of the true proportion of people who approve of gambling? Explain your answer.

20, larger the sample size means less variability

e. Make a histogram of the sample distribution. Describe the graph.

C: 60%

U: none

S: Approx. symmetrical

S: Range = 10-1 = 9

9.2 – Sample Proportions

count of "successes in sampleˆ

size of sample

Xp

n

Using proportions:

Remember Ch8?

x np (1 )x np p

Use these when you know “p”

What if you only know the proportion of a sample?

Sampling Distribution of a Sample Proportion:

ˆ ˆp p p

ˆ(1 )

pp p

n

Rule of Thumb #1:

You can only use if the population is 10X the sample size . A census should be impractical!

p̂

ˆ(1 )

pp p

n

N 10nwhen

Rule of Thumb #2:

Only use the Normal approximation of the sampling distribution of when: p̂

10np and (1 ) 10n p

Conclusion:

If p is the population proportion then,

, (1 )N np np p

If is the sample proportion then,

(1 ),p p

N pn

p̂

ONLY if 10np and (1 ) 10n p

So, to calculate a Z-score for this!

ˆ

(1 )

p pZ

p pn

statistic parameterStandardized test statistic:

standard deviation of statistic

Or ˆ

ˆ

p

p

pZ

Example #1Suppose you are going to roll a fair six-sided die 60 times and record , the proportion of times that a 1 or a 2 is showing.

a. Where should the distribution of the 60 -values be centered?

2

6p

1

3

b. What is the standard deviation of the sampling distribution of , the proportion of all rolls of the die that show a 1 or a 2 out of the 60 rolls ?

.33(1 .33)

60

ˆ

(1 )p

p p

n

Rule of Thumb #1: Population is 10X sample size

p̂

0.60858

c. Describe the shape of the sampling distribution of Justify your answer.

p̂

10np and (1 ) 10n p

160 10

3

20 10

160 1 10

3

40 10

Approximately Normal. 0.5,0.60858N

Rule of Thumb #2:

Example #2According to government data, 22% of American children under the age of 6 live in households with incomes less than the official poverty level. A study of learning in early childhood chooses an SRS of 300 children. What is the probability that more than 20% of the sample are from poverty households?

Rule of Thumb #1: N 10n

N 10(300)

N 3000

Population is 10X sample size, ok to use standard deviation


0.22p


10np and (1 ) 10n p 300 0.22 10

66 10

300 1 0.22 10 234 10

Approximately Normal.

Rule of Thumb #2:

(1 ),p p

N pn

0.22p

.22(1 .22)

300

ˆ

(1 )p

p p

n

0.0239

0.22,0.0239N


ˆ

(1 )

p pZ

p pn

0.20 0.22

0.0239

0.8362

0.22

= 0.0239

0.20

P(Z – 0.8362) = 1 – P(Z – 0.8362)

0.22

= 0.0239

0.20

P(Z – 0.8362) =

= 1 – 0.2005

Or: normalcdf(0.20, 1000000, 0.22, 0.0239) = 0.7985

1 – P(Z – 0.8362)

= 0.7995

b. How large a sample would be needed to guarantee that the standard deviation of is no more than 0.01? Explain.

n

ppp

)1(ˆ

n

)22.1(22.01.0

0.17160.0001

n

1716n

0.0001 0.1716n

9.3 – Sample Means

x

Sample Means Distribution:

nx

How do you determine normality?

• If sample distribution is drawn from a Normal population, sample distribution is Normal, no matter how big n is

• If sample distribution is drawn from a Skewed population, sample distribution is Skewed, if n is small.

Central Limit Theorem: (CLT)

• No matter what the population distribution looks like, if n 30, then the sample distribution is approximately normal.

To calculate z-scores:

xZ

n

statistic parameterStandardized test statistic:

standard deviation of statistic

Or x

x

Z

Example #1A 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 = 0.16 oz. Sixteen cans are to be selected, and the soda volume will be determined for each one.

a. Describe the shape of the sample distribution

Because the population is approx normal, so is the sample distribution

b. Calculate the sample mean and standard deviation


x

x

12

n

0.16

16 0.04


c. Determine the probability the sample mean soda volume is between 11.9 oz and 12.1 oz of the company’s claim.

12 12.111.9

= 0.04x

Z

n

12.1 12

0.04

2.5

xZ

n

11.9 12

0.04

2.5

P( -2.5 < Z < 2.5) =

= P(Z < 2.5) – P(Z< -2.5)

P(Z < 2.5) – P(Z< -2.5)

P( -2.5 < Z < 2.5) =

= 0.9938 – 0.0062

P(Z < 2.5) – P(Z< -2.5)

= 0.9876

Or: normalcdf(11.9, 12.2, 12, 0.04) = 0.9876

Example #2The weights of newborn children in the United States vary according to the normal distribution with mean 7.5 pounds and standard deviation 1.25 pounds. The government classifies a newborn as having low birth weight if the weight is less than 5.5 pounds.

a. What is the probability that a baby chosen at random weighs less than 5.5 pounds at birth?

=7.5

= 1.25

5.5

P(Z < -1.6) =

xZ

5.5 7.5

1.25

1.6

P(Z < -1.6) = 0.0548

Or: normalcdf(-1000000, 5.5, 7.5, 1.25) = 0.0548

Example #2The weights of newborn children in the United States vary according to the normal distribution with mean 7.5 pounds and standard deviation 1.25 pounds. The government classifies a newborn as having low birth weight if the weight is less than 5.5 pounds.b. You choose forty babies at random and compute their mean weight. What are the mean and standard deviation of the mean weight of the three babies?

Distribution approx normal because population is, also n 30

x x 7.5n

1.25

400.1976

Example #2The weights of newborn children in the United States vary according to the normal distribution with mean 7.5 pounds and standard deviation 1.25 pounds. The government classifies a newborn as having low birth weight if the weight is less than 5.5 pounds.c. What is the probability that the forty babies average birth weight is less than 5.5 pounds?

=7.5

= 0.1976

5.5

P(Z < -10.12) =

xZ

n

5.5 7.5

0.1976

10.12

P(Z < -10.12) = 0

Or: normalcdf(-1000000, 5.5, 7.5, 0.1976) = 0

Example #2The weights of newborn children in the United States vary according to the normal distribution with mean 7.5 pounds and standard deviation 1.25 pounds. The government classifies a newborn as having low birth weight if the weight is less than 5.5 pounds.d. Would your answers to a, b, or c be affected if the distribution of birth weights in the population were distinctly nonnormal?

Yes, you couldn’t use the normal approximation for part a.

Part b and c are fine because n 30, and by the CLT, the distribution is approximately normal

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9.1 – Sampling Distributions. Many investigations and research projects try to draw conclusions about how the values of some variable x are distributed