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Sampling distributions BPS chapter 10 © 2006 W. H. Freeman and Company.

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Presentation on theme: "Sampling distributions BPS chapter 10 © 2006 W. H. Freeman and Company."— Presentation transcript:

1 Sampling distributions BPS chapter 10 © 2006 W. H. Freeman and Company

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

3 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 x bar 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

4 The Central Limit Theorem is valid as long as we are sampling many small random events, even if the events have different distributions (as long as no one random event has an overwhelming influence). What should this mean to you? It explains why so many variables are normally distributed. Further properties So height is very much like our sample mean. The “individuals” are genes and environmental factors. Your height is a mean. Now we have a better idea of why the density curve for height has this shape. Example: Height seems to be determined by a large number of genetic and environmental factors, like nutrition.

5 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.

6 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?

7 Confidence intervals: The basics BPS chapter 13 © 2006 W.H. Freeman and Company

8 Uncertainty and confidence Although the sample mean,, is a unique number for any particular sample, if you pick a different sample, you will probably get a different sample mean. In fact, you could get many different values for the sample mean, and virtually none of them would actually equal the true population mean, . x

9 But the sample distribution is narrower than the population distribution, by a factor of √n. Thus, the estimates gained from our samples are always relatively close to the population parameter µ. n Sample means, n subjects  Population, x individual subjects If the population is normally distributed N(µ,σ), so will the sampling distribution N(µ,σ/√n).

10 Red dot: mean value of individual sample Ninety-five percent of all sample means will be within roughly 2 standard deviations (2*  /√n) of the population parameter  Because distances are symmetrical, this implies that the population parameter  must be within roughly 2 standard deviations from the sample average, in 95% of all samples. This reasoning is the essence of statistical inference.

11 The weight of single eggs of the brown variety is normally distributed N(65g,5g). Think of a carton of 12 brown eggs as an SRS of size 12. You buy a carton of 12 white eggs instead. The box weighs 770g. The average egg weight from that SRS is thus = 64.2g.  Knowing that the standard deviation of egg weight is 5g, what can you infer about the mean µ of the white egg population? There is a 95% chance that the population mean µ is roughly within ± 2  /√n of, or 64.2g ± 2.88g. population sample  What is the distribution of the sample means ? Normal (mean , standard deviation  /√n) = N(65g,1.44g).  Find the middle 95% of the sample means distribution. Roughly ± 2 standard deviations from the mean, or 65g ± 2.88g.


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