1. Objectives (BPS chapter 11) Sampling distributions  Parameter versus statistic  The law of large numbers  What is a sampling distribution?  The sampling distribution of  The central limit theorem  Statistical process control

2. Reminder: Parameter versus statistic  Sample: the part of the population we actually examine and for which we do have data.  A statistic is a number describing a characteristic of a sample. We often use a statistic to estimate an unknown population parameter.  Population: the entire group of individuals in which we are interested but can’t usually assess directly.  A parameter is a number describing a characteristic of the population. Parameters are usually unknown. Population Sample

3. The law of large numbers Law of large numbers: As the number of randomly-drawn observations (n) in a sample increases, the mean of the sample ( ) gets closer and closer to the population mean  (quantitative variable). the sample proportion ( ) gets closer and closer to the population proportion p (categorical variable).

4. What is a sampling distribution? The sampling distribution of a statistic is the distribution of all possible values taken by the statistic when all possible samples of a fixed size n are taken from the population. It is a theoretical idea—we do not actually build it. The sampling distribution of a statistic is the probability distribution of that statistic. Note: When sampling randomly from a given population,  the law of large numbers describes what happens when the sample size n is gradually increased.  The sampling distribution describes what happens when we take all possible random samples of a fixed size n.

5. Sampling distribution of (the sample mean) We take many random samples of a given size n from a population with mean  and standard deviation  Some sample means will be above the population mean  and some will be below, making up the sampling distribution. Sampling distribution of “x bar” Histogram of some sample averages

6. Sampling distribution of  √n√n For any population with mean  and standard deviation  :  The mean, or center of the sampling distribution of, is equal to the population mean .  The standard deviation of the sampling distribution is  /√n, where n is the sample size.

7.  Mean of a sampling distribution of : There is no tendency for a sample mean to fall systematically above or below  even if the distribution of the raw data is skewed. Thus, the mean of the sampling distribution of is an unbiased estimate of the population mean  —it will be “correct on average” in many samples.  Standard deviation of a sampling distribution of : The standard deviation of the sampling distribution measures how much the sample statistic varies from sample to sample. It is smaller than the standard deviation of the population by a factor of √n.  Averages are less variable than individual observations.

8. For normally distributed populations When a variable in a population is normally distributed, then the sampling distribution of for all possible samples of size n is also normally distributed. If the population is N(  ), then the sample means distribution is N(  /√n). Population Sample means

9. IQ scores: population vs. sample In a large population of adults, the mean IQ is 112 with standard deviation 20. Suppose 200 adults are randomly selected for a market research campaign.  The distribution of the sample mean IQ is A) exactly normal, mean 112, standard deviation 20. B) approximately normal, mean 112, standard deviation 20. C) approximately normal, mean 112, standard deviation 1.414. D) approximately normal, mean 112, standard deviation 0.1. C) approximately normal, mean 112, standard deviation 1.414. Population distribution: N (  = 112;  = 20) Sampling distribution for n = 200 is N (  = 112;  /√n = 1.414)

10. Application Hypokalemia is diagnosed when blood potassium levels are low, below 3.5mEq/dl. Let’s assume that we know a patient whose measured potassium levels vary daily according to a normal distribution N(  = 3.8,  = 0.2). If only one measurement is made, what's the probability that this patient will be misdiagnosed hypokalemic? z =  1.5, P(z <  1.5) = 0.0668 ≈ 7% If instead measurements are taken on four separate days, what is the probability of such a misdiagnosis? z =  3, P(z <  1.5) = 0.0013 ≈ 0.1% Note: Make sure to standardize (z) using the standard deviation for the sampling distribution.

11. Practical note  Large samples are not always attainable.  Sometimes the cost, difficulty, or preciousness of what is studied limits drastically any possible sample size.  Blood samples/biopsies: no more than a handful of repetitions acceptable. Often we even make do with just one.  Opinion polls have a limited sample size due to time and cost of operation. During election times, though, sample sizes are increased for better accuracy.  Not all variables are normally distributed.  Income is typically strongly skewed for example.  Is still a good estimator of  then?

12. The central limit theorem Central Limit Theorem: When randomly sampling from any population with mean  and standard deviation , when n is large enough, the sampling distribution of is approximately normal: N(  /√n). Population with strongly skewed distribution Sampling distribution of for n = 2 observations Sampling distribution of for n = 10 observations Sampling distribution of for n = 25 observations

13. Income distribution Let’s consider the very large database of individual incomes from the Bureau of Labor Statistics as our population. It is strongly right-skewed.  We take 1000 SRSs of 100 incomes, calculate the sample mean for each, and make a histogram of these 1000 means.  We also take 1000 SRSs of 25 incomes, calculate the sample mean for each, and make a histogram of these 1000 means. Which histogram corresponds to the samples of size 100? 25?

14. In many cases, n = 25 isn’t a huge sample. Thus, even for strange population distributions we can assume a normal sampling distribution of the mean, and work with it to solve problems. How large a sample size? It depends on the population distribution. More observations are required if the population distribution is far from normal.  A sample size of 25 is generally enough to obtain a normal sampling distribution from a strong skewness or even mild outliers.  A sample size of 40 will typically be good enough to overcome extreme skewness and outliers.

15. Statistical process control Industrial processes tend to have normally distributed variability, in part as a consequence of the central limit theorem applying to the sum of many small influential factors. Random samples taken over time can thus be used to easily verify that a given process is not getting out of “control.” What is statistical control? A variable that continues to be described by the same distribution when observed over time is said to be in statistical control, or simply in control.

16. Process-monitoring What are the required conditions? We measure a quantitative variable x that has a normal distribution. The process has been operating in control for a long period, so that we know the process mean µ and the process standard deviation σ that describe the distribution of x as long as the process remains in control. An control chart displays the average of samples of size n taken at regular intervals from such a process. It is a way to monitor the process and alert us when it has been disturbed so that it is now out of control. This is a signal to find and correct the cause of the disturbance.

17. control charts For a process with known mean µ standard deviation σ, we calculate the mean of samples of constant size n taken at regular intervals.  Plot (vertical axis) against time (horizontal axis).  Draw a horizontal center line at µ.  Draw two horizontal control limits at µ ± 3σ/√n (UCL and LCL).

18. A machine tool cuts circular pieces. A sample of four pieces is taken hourly, giving these average measurements (in 0.0001 inches from the specified diameter). Because measurements are made from the specified diameter, we have a given target µ = 0 for the process mean. The process standard deviation σ = 0.31. What is going on? Sample 1 −0.14 2 0.09 3 0.17 4 0.08 5 −0.17 6 0.36 7 0.30 8 0.19 9 0.48 10 0.29 11 0.48 12 0.55 13 0.50 14 0.37 15 0.69 16 0.47 17 0.56 18 0.78 19 0.75 20 0.49 21 0.79 The process mean has drifted. Maybe the cutting blade is getting dull, or a screw got a bit loose. For the chart, the center line is 0 and the control limits are ±3σ/√4 = ± 0.465. x x x x x x