CHAPTER 7 THE SAMPLING DISTRIBUTION OF THE MEAN. 7.1 Sampling Error; The need for Sampling Distributions

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Arron Harvey
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1 CHAPTER 7 THE SAMPLING DISTRIBUTION OF THE MEAN 7.1 Sampling Error; The need for Sampling Distributions Sampling Error the error resulting from using a sample characteristic (statistic) to estimate a population characteristic (parameter). Eample 1 (Household Income): Try to estimate the population mean income,, of all U.S. households by the sample mean income,, of the 60,000 households surveyed. In 1998, it was reported to be $51,855 ( ) in CPR (Current Population Reports). Question: How accurate is our sample mean (estimate) likely to be? What is the probability that a sample mean from SRS of 60,000 households will estimate the population mean households income with an error of no more than $1000? To answer these questions we need to eamine the sampling distribution of. Sampling Distribution: The distribution of a statistic, or the distribution of all possible observations of the statistic for samples of a given size. Eample 2. Weight of certain breed of dogs. Below we have unrealistic small population of 5 dogs and their weight in pound. dog A B C D E weight The population mean height is μ= =42+ =52 pounds. σ=6.57 pounds N 5 Possible samples of size two and its means are summarized in the following table. Samples A, B A, C A, D A, E B, C B, D B, E C, D C, E D, E Weights 42,48 42,52 42,58 42,60 48,52 48,58 48,60 52,58 52,60 58, Any sample of size 2 we will take is going to be one of the above 10 possible samples, so probability of obtaining each value of is 1/10. How confident of our sample mean of size two is going to estimate our population mean within 2 pounds of the population mean weight? In other words what is P(50 54)? Since there are 5 samples ({A,D}, {A,E}, {B,C}, {B,D} and {B, E} ) which lie within 2 pounds of the population mean 52, P(50 54)=50 %

2 We can also compute mean and standard deviation of all μ =52 and σ =4.02 values, they are: Notice that mean of all values is the same as the population mean and standard deviation is smaller than the population standard deviation. Now we repeat our eample for samples of size 4 Possible samples of size four and its means are summarized in the following table. Samples A, B, C,D A,B,C,E A,B,D,E A,C,D,E B,C,D,E Weights 42,48, 52, 58 42,48, 52, 60 42,48,58,60 42,52,58,60 48,52,58, Any sample of size 4 we will take is going to be one of the above 5 possible samples, so probability of obtaining each value of is 1/5. This time P(50 54) = 4/5=80% We can also compute mean and standard deviation of all μ =52 and σ =1.64 values, they are: In conclusion we can clearly see that mean the distribution of Ȳ remains the same as a population mean, regardless of the sample size. Standard deviation of that distribution decreases as n increases. Sampling Distribution of the sample mean - the distribution of the variable (i.e., of all possible sample means) for a given variable. Sample size and Sampling Error As sample size increases, the more sample means cluster around the population mean, and the sampling error of estimating, by is smaller. What we do in practice? Large and unknown population -> Obtaining the sampling distribution is not feasible. But we can find approimate sampling distribution of the sample mean for any underlying population distributions, which we discuss in 7.2 and The Mean and Standard deviation of We use the sampling distribution of the sample mean to make inferences about a population mean based on the mean of a sample from the population. Bur generally we do not know the eact distribution of the sample mean (sampling distribution).

3 Under certain conditions, we can approimate the sampling distribution of the sample mean ( ) by the normal distribution. Normal distribution is determined by its mean and standard deviation. So let s denote its mean is and its standard deviation is. Mean of the variable For samples of size n, the mean of the variable equals the mean of the variable under consideration: (the mean of all possible sample mean equals the population mean). Standard Deviation of the variable For samples of size n, the standard deviation of the variable equals the standard deviation of the variable under consideration divided by the square root of the sample size: (the standard deviation of all possible sample means equals the population standard deviation divided by the square root of the sample size) n Sample Size and Sampling Error 1. The larger the sample size, the smaller the standard deviation of. 2. The smaller the standard deviation of, the more closely its possible values cluster around the mean of. 3. The mean of is the same as the population mean: NOTE: The standard deviation of determines the amount of sampling error to be epected when a population mean is estimated by s ample mean. So often it is referred to as the standard error of the sample mean. Standard error (SE) of a statistic standard deviation of a statistic 7.3 The Sampling Distribution of the Mean Sampling Distribution of the Mean for a Normally Distributed Variable If the variable of a population is normally distributed with mean and standard deviation, then, for any sample of size n 1, the variable is also normally distributed with mean and standard deviation. n

4 The Central Limit Theorem (CLT) one of the most important theorems is statistics For a relatively large sample size, the variable is approimately normally distributed, regardless of the distribution of the variable under consideration. The approimation becomes better and better with increasing sample size. The Sampling Distribution of the Sample Mean If a variable of a population has mean and standard deviation, then for samples of size n, 1. The mean of equals the population mean : 2. The standard deviation of (standard error of ) equals the population standard deviation divided by the square root of the sample size: n 3. If is normally distributed, then so is, regardless of sample size. 4. If the sample size is large (approimately bigger than 30), then is approimately normally distributed, regardless of the distribution of. Following graphs represent distribution of IQ scores (X) in some population (a) and sampling distributions of a sample mean ( ) for n=4 (b) and n=16 (c) Notice that all three distribution curves center at 100 ( μ ) and graphs get to be narrower as sample sizes are increasing

Following graph illustrates distribution of househols sizes in the USA (clearly not normal distribution This histogram illustrates distribution of the sample mean of 1000 samples of size n=30.

a) Suppose we randomly select one individual from that population, what is the probability that his height will eceed 72 inches? P(Y>72)= P(Z>0.82)=0.2061, where Z = 72 69.7 2.8 under N(69.7, 2.

5 Following graph illustrates distribution of househols sizes in the USA (clearly not normal distribution This histogram illustrates distribution of the sample mean of 1000 samples of size n=30. Clearly nearly normal distribution Eample. Let Y be a height of males in certain population. Assume Y has approimately normal distribution with mean μ=69.7 inches and SD σ=2.8 inches. a) Suppose we randomly select one individual from that population, what is the probability that his height will eceed 72 inches? P(Y>72)= P(Z>0.82)=0.2061, where Z = under N(69.7, 2.8) to the right of 72. and probability equals to the area b) Suppose we randomly select a sample of 4 individuals from that population. What is the probability that their average height ( Ȳ ) will estimate population mean with an error of no more than 1 inch?

P(68.7 Ȳ 70.7)=P ( 0.71 Z 0.71)=0.5223, where 0.71= 68.7 69.7 and 2.8/ 4 0.71= 70.7 69.7 and probability equals to the area under N(69.7, 2.8/ 4=1.4 ) 2.

7 ), so it will be narrower than the distribution for n=4. The % of all Ȳ values within 1 inch of off the mean will be larger, so our probability will increase.

6 P(68.7 Ȳ 70.7)=P ( 0.71 Z 0.71)=0.5223, where 0.71= and 2.8/ = and probability equals to the area under N(69.7, 2.8/ 4=1.4 ) 2.8 / 4 c) Without computations, will the answer in part b change or will it remain the same if we take a sample of size 16? If n=16, distribution of Ȳ will be N(69.7, 2.8/ 16=0.7 ), so it will be narrower than the distribution for n=4. The % of all Ȳ values within 1 inch of off the mean will be larger, so our probability will increase. d) Suppose our population was not normal, but severely left skewed, what would be the answers to questions b and c? If population is not normal, we need to use Central Limit Theorem, but that requires us to have a large sample (of size at least 30). If samples are as small as 4 and 16, we can't assume that the distributiln of Ȳ will be normal or approimately normal, so we can't answer our questions in both parts. Eample (Central Limit Theorem) Based on service records from the past year, the time (in hours) that a technician requires to complete preventative maintenance on an air conditioner follows the distribution that is strongly right-skewed, and whose most likely outcomes are close to 0. The mean time is µ = 1 hour and the standard deviation is σ = 1. Your company will service an SRS of 70 conditioners. You budgeted 1.1 hour per unit. Will that be enough? The Central Limit Theorem stateds that the sampling distribution of the mean time spent working on 70 units is approimately normal with mean 1 and SD 0.12 (since n=70>30). z= =0.83 P ( >1.1)=P (Z >0.83) = = If you budgeted 1.1 hour per unit, there is over 20% chance that the technicians will not complete the work within the budgeted time.

Notice that these facts about the mean and standard deviation of X are true no matter what shape the population distribution has

Notice that these facts about the mean and standard deviation of X are true no matter what shape the population distribution has 7.3.1 The Sampling Distribution of x- bar: Mean and Standard Deviation The figure above suggests that when we choose many SRSs from a population, the sampling distribution of the sample mean is centered