Biostatistics

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Biostatistics

• Unit 5 Samples

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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.A) Distribution of the sample mean
  B) Distribution of the difference between two
  means
• C) Distribution of the sample proportion D)
  Distribution of the difference between two
  proportions

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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

• 1.  The distribution of will be normal2.
   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 or 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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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 64 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
  .9332.  This is subtracted from 1 to give the
  probability P (z gt 1.5) .0668

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Example

• (5) Complete the answerThe probability that x
  is greater than 53 is .0668.

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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 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 resultsPopulation
  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 informationn1 40, 
346, 2800 n2  35, 
300,  3250
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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
  .9998.  This is subtracted from 1 to give the
  probability        P (z gt 3.6) .0002

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Solution

• (5) Complete the answer    The probability
  that  is as large as given is
  .0002.

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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 .194     Find P( lt .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 tableA
  value of z -1.36 gives an area of .0869 which
  is the probability        P (z lt -1.36) .0869

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Solution

• (5) Complete the answer
• The probability that lt .15 is .0869.

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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
•  n2 90,  p2 .40
•   .09Find P(
  .09)

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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
  .8790 which is subtracted from 1 to give the
  probability        P (z gt 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