Chapter 7 Introduction to Sampling Distributions

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Chapter 7 Introduction to Sampling Distributions. Hmm…déjà vu?. Statistic is a numerical descriptive measure of a sample Parameter is a numerical descriptive measure of a population. So…let’s do this. What are the symbols for Statistic mean, variance, standard deviation?. Something new. - PowerPoint PPT Presentation

Transcript of Chapter 7 Introduction to Sampling Distributions

Chapter 7 Introduction to

Sampling Distributions

Hmm…déjà vu?

0Statistic is a numerical descriptive measure of a sample

0Parameter is a numerical descriptive measure of a population

So…let’s do this

0What are the symbols for Statistic mean, variance, standard deviation?

Something new

0Proportion

Note:

0We are going from raw data distribution to a sampling distribution

Note #2:

0We often do not have access to all the measurements of an entire population because of constraints on time, money, or effort. So we must use measurements from a sample.

Type of inferences

01) Estimation: In this type of inference, we estimate the value of a population parameter

02) Testing: In this type of inference, we formulate a decision about the value of a population parameter

03) Regression: In this type of inference, we make predictions or forecasts about the value of a statistical variable

Sampling distribution

0 It is a probability distribution of a sample statistic based on all possible simple random samples of the same size from the same population

Group Work:

01) What is population parameter? Give an example02) What is sample statistic? Give an example03) What is a sampling distribution?

Answer

01) Population parameter is a numerical descriptive measure of a population

02) A sample statistic or statistic is a numerical descriptive measure of a sample

03) A sampling distribution is a probability distribution for the sample statistic we are using

Read page 295-298

Homework Practice

0Pg 298-299 #1-9

Central Limit Theorem

Central Limit Theorem

0For a Normal Probability Distribution:0Let x be a random variable with a normal distribution

whose mean is and whose standard deviation is . Let be the sample mean corresponding to random samples of size n taken from the x distribution. Then the following are true:0 A) The distribution is a normal distribution0 B) The mean of the distribution is 0 C) The standard deviation of the distribution is

Note:

0 We conclude from the previous theorem that when x has a normal distribution, the distribution will be normal for any sample size n. Furthermore, we can convert the distribution to the standard normal z distribution by using these formulas:

0 Where n is the sample size0 is the mean of the distribution, and0 is the standard deviation of the x distribution

Example

0 Suppose a team of biologist has been studying the height in human. Let x be the height of a single person. The group has determined that x has a normal distribution with mean and standard deviation feet .

0 A) What is the probability that a single person taken at random will be in between 4.7 and 6.5 feet tall?

0 B) What is the probability that the mean length of 5 people taken at random is between 4.7 and 6.5 feet tall?

Answer

0A) 0 )

0B) =

Group Work

0 Suppose a team of feet analysts has been studying the size of man’s foot for particular area. Let x represent the size of the foot. They have determined that the size of the foot has a normal distribution with and standard deviation inches.

0 A) What’s the probability of the foot of a single person taken at random will be in between 6 and 8 inches?

0 B) What’s the probability that the mean size of 8 people taken at random will be in between 7 and 9 inches?

Standard error

0Standard error is the standard deviation of a sampling distribution. For the sampling distribution,

0Standard error =

Using Central limit theorem to convert the distribution to

the standard normal distribution

0Where n is the sample size ,0 is the mean of the x distribution, and0 is the standard deviation of the x distribution

Group Work: Central Limit Theorem

0 A) Suppose x has a normal distribution with population mean 18 and standard deviation 3. If you draw random samples of size 5 from the x distribution and x bar represents the sample mean, what can you say about the x bar distribution? How could you standardize the x bar distribution?

0 B) Suppose you know that the x distribution has population mean 75 and standard deviation 12 but you have no info as to whether or not the x distribution is normal. If you draw samples of size 30 from the x distribution and x bar represents sample mean, what can you say about the x bar distribution? How could you standardize the x bar distribution?

0 C) Suppose you didn’t know that x had a normal distribution. Would you be justified in saying that the x bar distribution is approximately normal if the sample size were n=8?

Answer

0A) Since you are given it to be normal, the x bar distribution also will be normal even though sample size is much less than 30.

0B) Since sample size is large enough, the x bar distribution will be an approximately normal distribution.

0C) No, sample size is too small. Need to be 30 or more

Note:

0A sample statistic is unbiased if the mean of its sampling distribution equals the values of the parameter being estimated

0The spread of the sampling distribution indicates the variability of the statistic. The spread is affected by the sampling method and the sample size. Statistics from larger random samples have spread that are smaller.

Read 304 and 305

Homework Practice

0Pg 306-308 #1-18 eoe

Sampling Distributions for Proportions

Think Back to Section 6.4

0We dealt with normal approximation to the binomial.

0How is this related to sampling distribution for proportions?

0Well in many important situations, we prefer to work with the proportion of successes r/n rather than the actual number of successes r in binomial experiments.

Sampling distribution for the proportion

0 Given0 n= number of binomial trials (fixed constant)0 r= number of successes0 p=probability of success on each trial0 q=1-p= probability of failure on each trial

0 If np>5 and nq>5, then the random variable can be approximated by a normal random variable (x) with mean and standard deviation

Standard Error

0The standard error for the distribution is the standard deviation

Where do these formula comes from?

0Remember is as known as expected value or average.

0So (r is number of success)

How to Make continuity corrections to intervals

0 If r/n is the right endpoint of a interval, we add 0.5/n to get the corresponding right endpoint of the x interval

0 If r/n is the left endpoint of a interval, we subtract 0.5/n to get the corresponding left endpoint of the x interval

Example:

0Suppose n=30 and we have a interval from 15/30=0.5 and 25/30=.83 Use the continuity correction to convert this interval to an x interval

Answer

00.5/30 = 0.02 (approx)

0So x interval is .5-.02 and .83+.02 which is .48 to .85

0 interval: .5 to .83

0 x interval: .48 to .85

Group Work

0Suppose n=50 and interval is .64 and 1.58. Find the x interval

Word Problem

0 Annual cancer rate in L.A. is 209 per 1000 people. Suppose we take 40 random people.

0 A) What is the probability p that someone will get cancer and what’s the probability q that they won’t get cancer?

0 B) Do you think we can approximate with a normal distribution? Explain

0 C) What are the mean and standard deviation for ?

0 D) What is the probability that between 10% and 20% of the people will be cancer victim? Interpret the result

Answer0 A)

0 B) 0 Since both are greater than 5, we can approximate with a normal distribution

0 C)

0 D) Since probability is we need to convert into x distribution

0 Continuity correction=0.5/n=0.5/40=0.0125, so we subtract .01 and add .01 from the interval

0 So , then convert this into z value.

Group work

0 In the OC, the general ethnic profile is about 47% minority and 53% Caucasian. Suppose a company recently hired 78 people. However, if 25% of the new employees are minorities then there is a problem. What is the probability that at most 25% of the new fires will be minorities if the selection process is unbiased and reflect the ethnic profile? (Follow the last example’s footstep)

Answer

0 0 Since both are greater than 5, so normal approximation is appropriate

0 Continuity correction 0.5/n = 0.5/78 = .006

0 Since it is to the right endpoint, you add .006

0 P(

Remember Control Chart?

0Sketch how a control chart looks like

P-Chart

0 It is just like the control chart

How to make a P-Chart?

0 1. Estimate

0 2. The center like is assigned to be 0 3. Control limits are located at and

0 4. Interpretation: out-of-control signals0 A) any point beyond a 3 standard deviation level0 B) 9 consecutive points on one side of the center line0 C) at least 2 out of 3 consecutive points beyond a 2 standard deviation level

0 Everything is in control if no out-of-control signals occurs

Example situation:

0Civics and Economics is taught in each semester. The course is required to graduate from high school, so it always fills up to its maximum of 60 students. The principal asked the class to provide the control chart for the proportion of A’s given in the course each semester for the past 14 semesters. Make the chart and interpret the result.

Example:

Semester 1 2 3 4 5 6 7

r=#of A’s 9 12 8 15 6 7 13

=r/60 .15 .20 .13 .25 .1 .12 .22

Semester 8 9 10 11 12 13 14

r=#of A’s 7 11 9 8 21 11 10

=r/60 .12 .18 .15 .13 .35 .18 .17

Answer

0 (this is pooled proportion of success)0 n=60

0 Control Limits are: .077 and .273 (2 standard deviation)0 .028 and .322 (3 standard deviation)

0 Then graph it! Y-axis is proportion, and x-axis is sample number

Group Work

0Mr. Liu went on a streak of asking 30 women out a month for 11 months. Complete the table and make a control chart for Mr. Liu and interpret his “game”.

month 1 2 3 4 5 6 7

r=says yes 10 1 3 15 12 20 5

=r/30

month 8 9 10 11

r=says yes 2 8 7 23

=r/30

Homework Practice

0P317-320 #1-13 eoo