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Penny for your thoughts Take a sample of size 5 from the population Penny for your thoughts Take a sample of size 5 from the population

Penny for your thoughts Take a sample of size 5 from the population - PowerPoint Presentation

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Penny for your thoughts Take a sample of size 5 from the population - PPT Presentation

Penny for your thoughts Take a sample of size 5 from the population Calculate p the proportion of pennies minted in the 2000s Calculate the mean year of the pennies as well KEEP THIS DATA WHERE YOU CAN USE IT AGAIN ID: 761961

distribution sample population sampling sample distribution sampling population deviation standard size normal proportion find year probability cards activity square

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Penny for your thoughts Take a sample of size 5 from the populationCalculate p = the proportion of pennies minted in the 2000sCalculate the mean year of the pennies as wellKEEP THIS DATA WHERE YOU CAN USE IT AGAIN! ^ Only one person can pick pennies at a time!!!!!

Ch 7 - Sampling Distribution

Vocabulary Population (review)Sample (review)ParameterStatistic*Hint: the p’s and the s’s go together!

Notation VERY IMPORTANT! and you will loose points on the exam if you use the wrong notation (as well as if you use the wrong word)Population mean: Sample mean: xPopulation Proportion: pSample Proportion: p _ ^

Identify the population, the parameter, the sample, and the statistic in each of the following settings. a) A pediatrician wants to know the 75th percentile for the distribution of heights of 10-year-old boys, so she takes a sample of 50 patients and calculates Q3 = 56 inches.b) A Pew Research Center Poll asked 1102 12- to 17-year-olds in the United States if they have a cell phone. Of the respondents, 71% said Yes.

The Airport Problem This activity will review binomial calculations and introduce the idea of sampling distribution of sample proportions using simulationYou will need your TI-83/84 calculators

At Guadalajara Airport in Mexico, passengers must claim their luggage and then proceed to Customs. In the Customs area, each passenger will press a button that activates a modified stoplight. This light has only red and green bulbs. If the green lights shows, the passenger is free to go. If the light turns red, then Customs agents will inspect the passenger’s luggage. Customs officials claim that the light has a probability of 0.30 of showing red on any press of the button.

This activity is due at the beginning of class tomorrow. You do not have to draw the graphs, if you are describing them. Remember your “SOCS!” You can look at them on your calculators.

Randomly select 5 cards from a shuffled deck (of only cards 2 - 10) and note the median value. Then replace your cards. We will repeat this “many, many” times Create a table of sample mediansCreate a dotplot of sample mediansDescribe what you see.

Sampling Distribution The sampling distribution of a statistic is the distribution of values taken by the statistic in all possible samples of the same size from a population.For example, in our card activity if we were able to take every possible sample of size 5 of cards 2 through 10 and record the outcomes, that would be the sampling distribution of the median of a deck of cards when selecting 5 cards. That would be 36 nCr 5 = 376,992 different samples! It’s very time consuming to take all possible samples, instead simulations are done to imitate the process we just did. (FATHOM)

Using Fathom to simulate choosing 500 SRSs of size 5 from the deck of cards 2 though 10 and finding the sample medians produced the following dotplot. Describe it! What is the population median?Is there any connection to the population and the sampling distribution?

Distribution, Distribution, Distribution... There are 3 different types of distributions:1. Population Distribution2. Distributions of Sample Data3. Sampling Distributions

What’s the difference between the 3 distributions? A population distribution is one graph of everything possible as a whole.What would it have been for out card activity?

A distribution of sample data is an individual graph depicting each outcome from the sample you drew from the population.What would it have been for out card activity?

Sampling distribution describes how a statistic varies in many, many samples of the population. You are no longer looking at the individual elements in the sample/population.What would it have been for out card activity?

Helpful Hint Population & Sample Distributions describe individualsSampling Distribution describes a statistic.Let’s look at page 420 for a diagram to help

More questions about our card simulation: Was that a sampling distribution?If someone claims to set up the same activity and they select a sample of size 5 and get a median of 4, is that convincing evidence that they set their deck up wrong?

According to the National Center for Health Statistics, the distribution of heights for 16-year-old females is modeled well by a Normal density curve with mean of 64 inches and standard deviation of 2.5 inches. Make a graph of the population distribution.Sketch a graph of at the distribution of sample data for an SRS of size 20.

Exit Slip - you must complete p420 CYU #1 - 3 ON YOUR OWN BEFORE OU LEAVE!

Penny for your thoughts ^ Take a sample of size 10 from the populationCalculate p = the proportion of pennies minted in the 2000s Calculate the mean year of the pennies as well KEEP THIS DATA WHERE YOU CAN USE IT AGAIN!

Biased or Unbiased? So which sample statistics are biased and which are unbiased? To find out, let’s collect some quantitative data:1. On the piece of paper given to you write how many hours of sleep you got last night. 2. Each of you will randomly select a sample of 4 cards.3. You will need to record the following information: the four numbers, the sample mean, and the sample range. 4. Replace the cards. 5. Pass the bag to the next person then record your sample mean and sample range on the corresponding dotplots on the board. 6. Once everyone plots their data, we will analyze it and compare it to the population mean and population range.

Definitions Biased Estimator - the mean of a sampling distribution is not close to the population’s parameter being estimatedUnbiased Estimator - the mean of a sampling distribution is equal to the value of the parameter being estimated

Biased and Unbiased Estimators This is different than the sampling process being biased. When using an estimator (i.e. a measure of center or spread) we are assuming the sampling process is not biased.The actual statistic we are finding can be biased or unbiased as well.

Variability of a Statistic “the spread of the sampling distribution”What will help decrease the variability?Will it decrease bias? To answer these questions, let’s read p424 - 426

Penny for your thoughts ^ Take a sample of size 20 from the populationCalculate p = the proportion of pennies minted in the 2000s Calculate the mean year of the pennies as well KEEP THIS DATA WHERE YOU CAN USE IT AGAIN!

7.2 - Sample Proportions Remember your notation:population parameter is psample proportion is p ^

Penny Activity Let’s look at the population dotplot. Describe what you see.

Penny Activity Let’s create the proportion dotplots!

Compare the 4 graphs Shapes?Centers?Spreads?

More Notations We have a population, we take a sample, and find some proportion.If we want to investigate those sample proportions we can find the mean and standard deviation of the sampling distribution of the sample proportions.mean of sample proportions: standard deviation of sample proportions: ^ ^

Describe the Sampling Distribution of p SHAPE: sometimes it can be approximated by the Normal curve. It depends on the sample size n and the population proportion p.CENTER: = p because p is an unbiased estimator of p. SPREAD: gets smaller as n gets larger. The value of depends on both n and p . ^ ^ ^ ^ p. 436 shows a good, small proof of why these are true ^

Sampling Distribution of a Sample Proportion Choose an SRS of size n from a population of size N with proportion p of successes. Let p be the sample proportion of successes. Then: the mean of the sampling distribution of p is = p the standard deviation of the sampling distribution of p is AS LONG AS THE 10% CONDITION IS SATISFIED: if and , the Normal conditions are satisfied and the sampling distribution of p is approximately Normal. ^ ^ ^ ^ ^ ^ Formulas are on your formula sheet

In a Gallop Poll of 1785 random adults, 44% said that attend a religious service last week. Suppose the actual adult population that attends a religious service is 40%. a) What is the mean of the sampling distribution of p-hat?b) Find the standard deviation & check the 10% conditionc) Is the sampling distribution of p-hat approximately Normal?d) Find probability of obtaining a sample of 1785 adults with 44% or more attended a service? Do you have doubts in this poll?e) What would the sample size need to be to reduce the standard deviation of the sampling distribution by 1/3?

The superintendent of a large school district wants to know what proportion of middle school students in her district are planning on attending a four-year college or university. Suppose that 80% of all middle school students in her district are planning on attending a four-year college or university. What is the probability that an SRS of size 125 will give a result within 7 percentage points of the true value? We will use the 4-step method to solve this problem.

State We want to find the probability that the percentage of middle school students that plan to attend a 4-year college or university falls between 73% and 87%or in symbols: P(0.73 < < 0.87)  

Plan = 0.80.Since the school district is large, we’ll assume the 10% condition is satisfied and there are more than 1250 students. (10*125 = 1250). So, = 0.036 ^ ^ We can consider the distribution of p to be approximately Normal since the following are true: np = 125(.8) = 100 > 10 n(1-p)= 125(.2) = 25 > 10

Do P(0.73 0.87) = normalcdf(0.73, 0.87, 0.80, 0.036) = 0.948If you want full credit on the exam, you must have clearly said everything in the “Plan” step and these calculations will receive full credit. ^ Sketching a Normal curve will help. You can also use Table A to find the answer. Remember to standardize (z-score) first!

Conclude About 95% of all SRSs of size 125 will give a sample proportion within 7 percentage points of the true proportion of middle school students who want to attend a four-year college or university.

7.3 - Sample Means

More Notation! Mean of the sampling distribution: Standard deviation of the sampling distribution of the sample means:All of the notations in this chapter are very important and very similar. You will loose credit for using the wrong notations on the exam. So if you can’t remember it’s always best to write out what you are finding rather than try to use a notation. _ _

The mean of the sampling distribution of is   Suppose that x-bar is the mean of an SRS of size n drawn from a large population with mean and standard deviation , it does not matter what shape the population has.   The standard deviation of the sampling distribution of is as long as the 10% condition is satisfied!   These formulas are on your formula sheet    

If you are asked to find the sampling distribution of x, these means to state if it is Normal and find the mean and standard deviation. Hint: if the population itself is approximately Normal, then so is the sampling distribution of x.Hint: Please read carefully! Make sure you know if you are using the population standard deviation or the sample means standard deviation before you standardize or use your normalcdf on your calculator. _ _

A grinding machine in an auto parts plant prepares axels with a target diameter mu = 40.125 mm. The machine has some variability, so the standard deviation of the diameters is sigma = 0.002 mm. The machine operator inspects a random sample of 4 axles each hour for quality control purposes and records the sample mean diameter x. a) Assuming the process is working properly, what are the mean and standard deviation of the sampling distribution of x? b) Can you find the probability that x is within + .05 mm if you are choosing an SRS of 100 axels? Explain c) In order for you to pass this inspection the standard deviation of the sampling distribution of x needs to be 0.0005 mm. How many axels would you have to sample? _ - _ _

The composite scores of individual students on the ACT in 2009 followed a Normal distribution with mean 21.1 and standard deviation 5.1. a) What is the probability that a single student randomly chosen from all those taking the test scores 23 or higher? Show your work.b) Now take an SRS of 50 students who took the test. What is the probability that the mean score x of these students is 23 or higher?

What if the population shape is not Normal? http://onlinestatbook.com/stat_sim/sampling_dist/index.html

Central Limit Theorem Draw an SRS of size n from any population with mean and finite standard deviation CLT - when n is large, the sampling distribution of the sample means x is approximately Normal. _ NOTE: this is of the sample means, not just any sample!!!

How large is large? In order for the Normal conditions to apply for the sample means, and the population is not Normal the CLT will apply in most cases if

The number of flaws per square yard in a type of carpet material varies with the mean 1.6 flaws per square yard and standard deviation 1.2 flaws per square yard. The population distribution cannot be Normal, because a count takes only whole-number values. An inspector studies 200 square yards of material, records the number of flaws found in each square yard, and calculates x, the mean number of flaws per square yard inspected. Find the probability that the mean number of flaws exceeds 2 per square yard. _

State What’s the probability that the mean number of flaws per square yard of carpet is more than 2?

Plan The mean of the sampling distribution of the sample means is 10% condition is met since there’s more than 2000 square yards of carpet, soSince the sample size is large, 200 > 30, we can safely use the Normal distribution as an approximation for the sampling distribution of x _ _ _

Do Draw the curve,find P(x > 2) = normalcdf(2, 100, 1.6, 0.085) = 0**identify what these numbers are or:z = (2 - 1.6)/.085 = 4.705P(z > 4.705) = 0

Conclude There is virtually no chance that the average number of flaws per yard in the sample will be greater than 2.