The Central Limit TheoremIn Note 6.5 "Example 1" in Section 6.1 "The Mean and Standard Deviation of the Sample Mean" we constructed the probability distribution of the sample mean for samples of size two drawn from the population of four rowers. The probability distribution is: Show
x- 152154156158160162164P(x-)116 216316416316216116 Figure 6.1 "Distribution of a Population and a Sample Mean" shows a side-by-side comparison of a histogram for the original population and a histogram for this distribution. Whereas the distribution of the population is uniform, the sampling distribution of the mean has a shape approaching the shape of the familiar bell curve. This phenomenon of the sampling distribution of the mean taking on a bell shape even though the population distribution is not bell-shaped happens in general. Here is a somewhat more realistic example. Figure 6.1 Distribution of a Population and a Sample Mean Suppose we take samples of size 1, 5, 10, or 20 from a population that consists entirely of the numbers 0 and 1, half the population 0, half 1, so that the population mean is 0.5. The sampling distributions are: n = 1: x-01P(x-)0.50.5 n = 5: x-00.20.40.6 0.81P(x-)0.030.160.310.310.160.03 n = 10: x-00.1 0.20.30.40.50.60.70.80.91P(x-) 0.000.010.040.120.210.250.210.120.040.010.00 n = 20: x-00.050.100.15 0.200.250.300.350.400.450.50P(x-)0.00 0.000.000.000.000.010.040.070.120.160.18 x-0.550.600.650.700.750.800.85 0.900.951P(x-)0.160.120.070.040.01 0.000.000.000.000.00 Histograms illustrating these distributions are shown in Figure 6.2 "Distributions of the Sample Mean". Figure 6.2 Distributions of the Sample Mean As n increases the sampling distribution of X- evolves in an interesting way: the probabilities on the lower and the upper ends shrink and the probabilities in the middle become larger in relation to them. If we were to continue to increase n then the shape of the sampling distribution would become smoother and more bell-shaped. What we are seeing in these examples does not depend on the particular population distributions involved. In general, one may start with any distribution and the sampling distribution of the sample mean will increasingly resemble the bell-shaped normal curve as the sample size increases. This is the content of the Central Limit Theorem. The Central Limit TheoremFor samples of size 30 or more, the sample mean is approximately normally distributed, with mean μX-=μ and standard deviation σ X-=σ/n, where n is the sample size. The larger the sample size, the better the approximation. The Central Limit Theorem is illustrated for several common population distributions in Figure 6.3 "Distribution of Populations and Sample Means". Figure 6.3 Distribution of Populations and Sample Means The dashed vertical lines in the figures locate the population mean. Regardless of the distribution of the population, as the sample size is increased the shape of the sampling distribution of the sample mean becomes increasingly bell-shaped, centered on the population mean. Typically by the time the sample size is 30 the distribution of the sample mean is practically the same as a normal distribution. The importance of the Central Limit Theorem is that it allows us to make probability statements about the sample mean, specifically in relation to its value in comparison to the population mean, as we will see in the examples. But to use the result properly we must first realize that there are two separate random variables (and therefore two probability distributions) at play:
Example 3Let X- be the mean of a random sample of size 50 drawn from a population with mean 112 and standard deviation 40.
Solution
Note that if in Note 6.11 "Example 3" we had been asked to compute the probability that the value of a single randomly selected element of the population exceeds 113, that is, to compute the number P(X > 113), we would not have been able to do so, since we do not know the distribution of X, but only that its mean is 112 and its standard deviation is 40. By contrast we could compute P(X->113 ) even without complete knowledge of the distribution of X because the Central Limit Theorem guarantees that X- is approximately normal. Example 4The numerical population of grade point averages at a college has mean 2.61 and standard deviation 0.5. If a random sample of size 100 is taken from the population, what is the probability that the sample mean will be between 2.51 and 2.71? Solution The sample mean X- has mean μX-=μ=2.61 and standard deviation σX-=σ/n=0.5/10=0.05, so P(2.51<X-<2.71)=P(2.51−μX- σX-<Z<2.71−μX-σX-)=P(2.51−2.610.05<Z<2.71−2.610.05) =P(−2<Z<2)=P(Z<2)−P(Z <−2)=0.9772−0.0228=0.9544 Normally Distributed PopulationsThe Central Limit Theorem says that no matter what the distribution of the population is, as long as the sample is “large,” meaning of size 30 or more, the sample mean is approximately normally distributed. If the population is normal to begin with then the sample mean also has a normal distribution, regardless of the sample size. For samples of any size drawn from a normally distributed population, the sample mean is normally distributed, with mean μX-=μ and standard deviation σX-=σ/n, where n is the sample size. The effect of increasing the sample size is shown in Figure 6.4 "Distribution of Sample Means for a Normal Population". Figure 6.4 Distribution of Sample Means for a Normal Population Example 5A prototype automotive tire has a design life of 38,500 miles with a standard deviation of 2,500 miles. Five such tires are manufactured and tested. On the assumption that the actual population mean is 38,500 miles and the actual population standard deviation is 2,500 miles, find the probability that the sample mean will be less than 36,000 miles. Assume that the distribution of lifetimes of such tires is normal. Solution For simplicity we use units of thousands of miles. Then the sample mean X- has mean μX-=μ= 38.5 and standard deviation σX-=σ/n=2.5/5=1.11803 . Since the population is normally distributed, so is X-, hence P(X-<36)=P(Z<36−μX-σX-)=P(Z<36−38.51.11803)=P(Z<−2.24)=0.0125 That is, if the tires perform as designed, there is only about a 1.25% chance that the average of a sample of this size would be so low. Example 6An automobile battery manufacturer claims that its midgrade battery has a mean life of 50 months with a standard deviation of 6 months. Suppose the distribution of battery lives of this particular brand is approximately normal.
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Additional ExercisesAnswers
How do you find the variance of the sampling distribution of the mean?The formula to find the variance of the sampling distribution of the mean is: σ2M = σ2 / N, where: σ2M = variance of the sampling distribution of the sample mean.
What is sampling distribution of sample variance?The sampling distribution of the sample variance is a chi-squared distribution with degree of freedom equals to n−1, where n is the sample size (given that the random variable of interest is normally distributed).
What is the sampling distribution of all possible sample variance S 2 when you collect sample of size n?The average of all possible sample variances, denoted as μ s 2 , is as follows. Note that the average of all possible sample variances is the same as the population variance which means the sample variance is an unbiased estimate of the population variance.
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Answer.. What is the mean of the sampling distribution of the means?The mean of the sampling distribution of the mean is the mean of the population from which the scores were sampled. Therefore, if a population has a mean μ, then the mean of the sampling distribution of the mean is also μ.
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