The central limit theorem states that if you have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement Show
For the random samples we take from the population, we can compute the mean of the sample means: and the standard deviation of the sample means: Before illustrating the use of the Central Limit Theorem (CLT) we will first illustrate the result. In order for the result of the CLT to hold, the sample must be sufficiently large (n > 30). Again, there are two exceptions to this. If the population is normal, then the result holds for samples of any size (i..e, the sampling distribution of the sample means will be approximately normal even for samples of size less than 30). Central Limit Theorem with a Normal PopulationThe figure below illustrates a normally distributed characteristic, X, in a population in which the population mean is 75 with a standard deviation of 8. If we take simple random samples (with replacement) The mean of the sample means is 75 and the standard deviation of the sample means is 2.5, with the standard deviation of the sample means computed as follows: If we were to take samples of n=5 instead of n=10, we would get a similar distribution, but the variation among the sample means would be larger. In fact, when we did this we got a sample mean = 75 and a sample standard deviation = 3.6. Central Limit Theorem with a Dichotomous OutcomeNow suppose we measure a characteristic, X, in a population and that this characteristic is dichotomous (e.g., success of a medical procedure: yes or no) with 30% of the population classified as a success (i.e., p=0.30) as shown below. The Central Limit Theorem applies even to binomial populations like this provided that the minimum of np and n(1-p) is at least 5, where "n" refers to the sample size, and "p" is the probability of "success" on any given trial. In this case, we will take samples of n=20 with replacement, so min(np, n(1-p)) = min(20(0.3), 20(0.7)) = min(6, 14) = 6. Therefore, the criterion is met. We saw previously that the population mean and standard deviation for a binomial distribution are: Mean binomial
probability: Standard deviation:
The distribution of sample means based on samples of size n=20 is shown below. The mean of the sample means is and the standard deviation of the sample means is: Now, instead of taking samples of n=20, suppose we take simple random samples (with replacement) of size n=10. Note that in this scenario we do not meet the sample size requirement for the Central Limit Theorem (i.e., min(np, n(1-p)) = min(10(0.3), 10(0.7)) = min(3, 7) = 3).The distribution of sample means based on samples of size n=10 is shown on the right, and you can see that it is not quite normally distributed. The sample size must be larger in order for the distribution to approach normality. Central Limit Theorem with a Skewed DistributionThe Poisson distribution is another probability model that is useful for modeling discrete variables such as the number of events occurring during a given time interval. For example, suppose you typically receive about 4 spam emails per day, but the number varies from day to day. Today you happened to receive 5 spam emails. What is the probability of that happening, given that the typical rate is 4 per day? The Poisson probability is: Mean = μ Standard deviation =
The mean for the distribution is μ (the average or typical rate), "X" is the actual number of events that occur ("successes"), and "e" is the constant approximately equal to 2.71828. So, in the example above Now let's consider another Poisson distribution. with μ=3 and σ=1.73. The distribution is shown in the figure below.
This population is not normally distributed, but the Central Limit Theorem will apply if n > 30. In fact, if we take samples of size n=30, we obtain samples distributed as shown in the first graph below with a mean of 3 and standard deviation = 0.32. In contrast, with small samples of n=10, we obtain samples distributed as shown in the lower graph. Note that n=10 does not meet the criterion for the Central Limit Theorem, and the small samples on the right give a distribution that is not quite normal. Also note that the sample standard deviation (also called the "standard error return to top | previous page | next page How is SS defined formula calculated?We do this by squaring the deviations and then taking the square root of the sum of the squared deviations. The equation that we just used (SS = S (X - m)2) is refered to as the definitional formula for the Sum of Squares.
What is the formula for calculating population deviation scores?The formula reads: sigma (standard deviation of a population) equals the square root of the sum of all the squared deviation scores of the population (raw scores minus mu or the mean of the population) divided by capital N or the number of scores in the population.
What is the formula for calculating variance?Steps for calculating the variance. Step 1: Find the mean. To find the mean, add up all the scores, then divide them by the number of scores. ... . Step 2: Find each score's deviation from the mean. ... . Step 3: Square each deviation from the mean. ... . Step 4: Find the sum of squares. ... . Step 5: Divide the sum of squares by n – 1 or N.. What does it mean for a sample to have a standard deviation of s 5?Answer and Explanation: A sample with a standard deviation equal to 5 indicates that, on average, the distance between each data point in an entire dataset is different from the mean of the dataset by a value of 5.
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