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**Chapter 7 Sampling Distributions**

7.1 What is a Sampling Distribution? 7.2 Sample Proportions 7.3 Sample Means

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**Section 7.3 Sample Means After this section, you should be able to…**

Learning Objectives After this section, you should be able to… FIND the mean and standard deviation of the sampling distribution of a sample mean CALCULATE probabilities involving a sample mean when the population distribution is Normal EXPLAIN how the shape of the sampling distribution of sample means is related to the shape of the population distribution APPLY the central limit theorem to help find probabilities involving a sample mean

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**Sample means are among the most common statistics.**

Sample proportions arise most often when we are interested in categorical variables. When we record quantitative variables we are interested in other statistics such as the median or mean or standard deviation of the variable. Sample means are among the most common statistics. Sample Means

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Example: Making Money: P- 443 Consider the mean household earnings for samples of size 100. Compare the population distribution on the left with the sampling distribution on the right. What do you notice about the shape, center, and spread of each?

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**With different SRS of size 100 households, x-bar will be different but still close to $69,750.**

Although the distribution of individual earnings is skewed and very spread out, the distribution of sample mean is roughly symmetric and much less spread out. (Averages are less variable than individual observations)

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**The Sampling Distribution of**

When we choose many SRSs from a population, the sampling distribution of the sample mean is centered at the population mean µ and is less spread out than the population distribution. Here are the facts. as long as the 10% condition is satisfied: n ≤ (1/10)N. Mean and Standard Deviation of the Sampling Distribution of Sample Means

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**Sampling from a Normal Population**

Sample Means

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**9.3: Sample Means Sampling Distribution of a Sample mean (x-bar)**

Standard deviation (σ = ) Only use if N > 10n If an SRS of size n is taken from a population that is Normally distributed, then the sampling distribution is also Normal.

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**Behavior of x-bar ( x-bar is Mean of sample drawn from large population with mean μ and sd σ)**

X-bar is unbiased estimator of pop mean μ Values of x-bar are less spread out for larger samples The sd decreases at the rate of √n , so take the sample 4 times as large to cut the sd of x-bar in half Use σ / √n for sd of x-bar when population is at least 10 times as large as the sample.

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**Sampling Distribution of x- normally distributed population**

/10 Population distribution: N( , )

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**The Central Limit Theorem**

Most population distributions are not Normal. What is the shape of the sampling distribution of sample means when the population distribution isn’t Normal? It is a remarkable fact that as the sample size increases, the distribution of sample means changes its shape: it looks less like that of the population and more like a Normal distribution! When the sample is large enough, the distribution of sample means is very close to Normal, no matter what shape the population distribution has, as long as the population has a finite standard deviation.

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Definition: . Note: How large a sample size n is needed for the sampling distribution to be close to Normal depends on the shape of the population distribution. More observations are required if the population distribution is far from Normal.

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**The Central Limit Theorem**

Consider the strange population distribution from the Rice University sampling distribution applet.( P- 450) Describe the shape of the sampling distributions as n increases. What do you notice?

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**Normal Condition for Sample Means**

. Normal Condition for Sample Means

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**Example: Servicing Air Conditioners**

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 air conditioners. You have budgeted 1.1 hours per unit. Will this be enough? Since the 10% condition is met (there are more than 10(70)=700 air conditioners in the population), the sampling distribution of the mean time spent working on the 70 units has The sampling distribution of the mean time spent working is approximately N(1, 0.12) since n = 70 ≥ 30. We need to find P(mean time > 1.1 hours) If you budget 1.1 hours per unit, there is a 20% chance the technicians will not complete the work within the budgeted time.

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Section 7.3 Sample Means Summary In this section, we learned that…

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Section 7.3 Sample Means Summary In this section, we learned that…

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**Looking Ahead… In the next Chapter…**

We’ll learn how to estimate population parameters with confidence, using sample statistics. We’ll learn about Confidence Intervals Estimating Population Proportions Estimating Population Means In the next Chapter… Looking Ahead…

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Try P- 454 # 50,52,56,60,64

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50.

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52.

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54.

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56.

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60.

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64. State:What is the probability that the mean number of flaws per square yard of carpet is more than 2? Plan: The sampling distribution of the sample mean number of flaws per square yard has mean (10% condition OK since there are more than 2000 square yards of material). Since the sample size is so large(200>30), we can safely use the Normal distribution as an approximation for the sampling distribution of x-bar. Do: Conclude: There is virtually no chance that the average number of flaws per yard in the sample will be found to be greater than 2.

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