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Biostatistics Unit 5 Samples Needs to be completed. 12/24/13

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**Sampling distributions**

Sampling distributions are important in the understanding of statistical inference. Probability distributions permit us to answer questions about sampling and they provide the foundation for statistical inference procedures.

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Definition The sampling distribution of a statistic is the distribution of all possible values of the statistic, computed from samples of the same size randomly drawn from the same population. When sampling a discrete, finite population, a sampling distribution can be constructed. Note that this construction is difficult with a large population and impossible with an infinite population.

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**Construction of sampling distributions**

1. From a population of size N, randomly draw all possible samples of size n. 2. Compute the statistic of interest for each sample. 3. Create a frequency distribution of the statistic.

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**Properties of sampling distributions**

We are interested in the mean, standard deviation, and appearance of the graph (functional form) of a sampling distribution.

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**Types of sampling distributions**

We will study the following types of sampling distributions. Distribution of the sample mean Distribution of the difference between two means Distribution of the sample proportion Distribution of the difference between two proportions

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**(A) Sampling distribution of**

Given a finite population with mean (m) and variance (s2). When sampling from a normally distributed population, it can be shown that the distribution of the sample mean will have the following properties.

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**Properties of the sampling distribution**

The distribution of will be normal. The mean , of the distribution of the values of will be the same as the mean of the population from which the samples were drawn; = m. 3. The variance, , of the distribution of will be equal to the variance of the population divided by the sample size; =

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Standard error The square root of the variance of the sampling distribution is called the standard error of the mean which is also called the standard error.

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**Nonnormally distributed populations**

When the sampling is done from a nonnormally distributed population, the central limit theorem is used.

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**The central limit theorem**

Given a population of any nonnormal functional form with mean (m) and variance (s2) , the sampling distribution of , computed from samples of size n from this population will have mean, m, and variance, s2/n, and will be approximately normally distributed when the sample is large (30 or higher).

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**The central limit theorem**

Note that the standard deviation of the sampling distribution is used in calculations of z scores and is equal to:

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**Sampling distribution of the mean and Central Limit Theorem**

We do in class together

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**Data A small apartment building has 3 apartments.**

How many people live in each apartment? Apartment People A B C

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**Find m and s Use the TI to obtain the values for the population.**

The values are: m = s =

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Form samples of size 2 Samples Sample mean A, A A, B A, C B, A B, B B, C C, A C, B C, C We need to form all samples of size 2, using replacement since the population is very small. Then we find the sample mean for each sample of 2 apartments.

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Find m and s Use the TI to obtain the values for the means of the samples. The values are: m = s =

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**Results Mean of Sample means**

Mean of population equals mean of the sample means

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**Results of Standard deviation of the sample means**

S.D. equals the population standard deviation divided by the square root of the sample size

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**Distribution of the sample means**

If the population is normally distributed, then the sample means will be normally distributed. If the population is not normally distributed, then the sample means will be normally distributed if the sample size is at least 30.

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**Important Consequence**

If we take samples of size n from some population, under the previous conditions, then we can determine the probability of the sample means fulfilling some condition. We use:

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Example #1 The heights of kindergarten children are approximately normally distributed with a mean of 39 and a standard deviation of 2. If one child is randomly selected, what is the probability that the child is taller than 41 inches? This is 1 child – Not the Central Limit Theorem!

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Example #2 Suppose we have a class of 30 kindergarten children. What is the probability that the mean height of these children exceeds 41 inches? This is the Central Limit Theorem as it is asking about the probability of a sample mean!

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Conclusion It is not unusual for one child, selected at random from a kindergarten class, to be taller than 41 inches. It is highly unlikely that the mean height for 30 kindergarten students exceeds 41 inches.

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An analogy It would not be unusual for a student to get an A on a statistics test. It would be unusual if the class average for a statistics class was an A!

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**Demonstration that Central Limit Theorem Really Works (1)**

We start with a dwelling that has 3 apartments. Here is the list of occupancies. Apt A = 3 Apt B = 4 Apt C = 2 This is the entire population. It is entered into a list on the TI-83

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**Demonstration that Central Limit Theorem Really Works (2)**

We calculate 1-Var Stats to obtain the population parameters for this population. Mean: m = 3 Standard Deviation: s = Note: we do not use s = 3 because this is the entire population, not a sample.

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**Demonstration that Central Limit Theorem Really Works (3)**

Knowing the population parameters of m and s, we now determine them using a sampling distribution. We can find the population parameters because it is a very small population. Normally, populations are too large to determine m and s directly from the population.

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**Demonstration that Central Limit Theorem Really Works (4)**

We need to form all samples of size 2, using replacement since the population is very small. Then we find the sample mean for each sample of 2 apartments. Samples Sample mean A, A 3.0 A, B 3.5 A, C 2.5 B, A B, B 4.0 B, C C, A C, B C, C 2.0

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**Demonstration that Central Limit Theorem Really Works (5)**

We calculate 1-Var Stats to obtain the population parameters for the sampling distribution. Mean: m = 3 Standard Deviation: s =

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**Demonstration that Central Limit Theorem Really Works (6)**

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Example Given the information below, what is the probability that x is greater than 53? (1) Write the given information. m = 50 s = 16 n = x = 53

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Example (2) Sketch a normal curve.

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Example (3) Convert x to a z score.

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Example (4) Find the appropriate value(s) in the table. A value of z = 1.5 gives an area of This is subtracted from 1 to give the probability P (z > 1.5) = .0668

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Example (5) Complete the answer. The probability that x is greater than 53 is

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**(B) Distribution of the difference between two means**

It often becomes important to compare two population means. Knowledge of the sampling distribution of the difference between two means is useful in studies of this type. It is generally assumed that the two populations are normally distributed.

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**Sampling distribution of**

Plotting mean sample differences against frequency gives a normal distribution with mean equal to which is the difference between the two population means.

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Variance The variance of the distribution of the sample differences is equal to Therefore, the standard error of the differences between two means would be equal to

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Converting to a z score To convert to the standard normal distribution, we use the formula We find the z score by assuming that there is no difference between the population means.

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**Sampling from normal populations**

This procedure is valid even when Sampling from normal populations the population variances are different or when the sample sizes are different. Given two normally distributed populations with means, and , and variances, and , respectively. (continued)

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**Sampling from normal populations**

The sampling distribution of the difference, , between the means of independent samples of size n1 and n2 drawn from these populations is normally distributed with mean, , and variance,

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Example In a study of annual family expenditures for general health care, two populations were surveyed with the following results: Population 1: n1 = 40, = $346 Population 2: n2 = 35, = $300

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**Example If the variances of the populations are**

= 2800 and = 3250, what is the probability of obtaining sample results as large as those shown if there is no difference in the means of the two populations?

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**Solution (1) Write the given information n1 = 40, = $346, = 2800**

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Solution (2) Sketch a normal curve

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Solution (3) Find the z score

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Solution (4) Find the appropriate value(s) in the table A value of z = 3.6 gives an area of This is subtracted from 1 to give the probability P (z > 3.6) = .0002

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Solution (5) Complete the answer The probability that is as large as given is

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**C) Distribution of the sample proportion ( )**

While statistics such as the sample mean are derived from measured variables, the sample proportion is derived from counts or frequency data.

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**Properties of the sample proportion**

Construction of the sampling distribution of the sample proportion is done in a manner similar to that of the mean and the difference between two means. When the sample size is large, the distribution of the sample proportion is approximately normally distributed because of the central limit theorem.

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Mean and variance The mean of the distribution, , will be equal to the true population proportion, p, and the variance of the distribution, , will be equal to p(1-p)/n.

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The z-score The z-score for the sample proportion is

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Example In the mid seventies, according to a report by the National Center for Health Statistics, 19.4 percent of the adult U.S. male population was obese. What is the probability that in a simple random sample of size 150 from this population fewer than 15 percent will be obese?

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Solution (1) Write the given information n = 150 p = Find P( < .15)

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Solution (2) Sketch a normal curve

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Solution (3) Find the z score

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Solution (4) Find the appropriate value(s) in the table A value of z = gives an area of which is the probability P (z < -1.36) = .0869

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**Solution (5) Complete the answer**

The probability that < .15 is

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**D) Distribution of the difference between two proportions**

This is for situations with two population proportions. We assess the probability associated with a difference in proportions computed from samples drawn from each of these populations. The appropriate distribution is the distribution of the difference between two sample proportions.

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**Sampling distribution of**

The sampling distribution of the difference between two sample proportions is constructed in a manner similar to the difference between two means. (continued)

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**Sampling distribution of**

Independent random samples of size n1 and n2 are drawn from two populations of dichotomous variables where the proportions of observations with the character of interest in the two populations are p1 and p2 , respectively.

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**Mean and variance The distribution of the difference between two**

sample proportions, , is approximately normal. The mean is The variance is These are true when n1 and n2 are large.

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The z score The z score for the difference between two proportions is given by the formula

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Example In a certain area of a large city it is hypothesized that 40 percent of the houses are in a dilapidated condition. A random sample of 75 houses from this section and 90 houses from another section yielded difference, , of .09. If there is no difference between the two areas in the proportion of dilapidated houses, what is the probability of observing a difference this large or larger?

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**Solution (1) Write the given information n1 = 75, p1 = .40**

= .09 Find P( )

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Solution (2) Sketch a normal curve

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Solution (3) Find the z score

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Solution (4) Find the appropriate value(s) in the table A value of z = 1.17 gives an area of which is subtracted from 1 to give the probability P (z > 1.17) = .121

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**Solution (5) Complete the answer The probability of observing **

of .09 or greater is .121.

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fin

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