BASIC STATISTICS For the HEALTH SCIENCES Fifth Edition

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BASIC STATISTICS For the HEALTH SCIENCES Fifth Edition

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BASIC STATISTICS For the HEALTH SCIENCES Fifth Edition. By Kuzma. “It’s all Greek to me!!!!”. = Summation. x x x x x x. 5 4 6 5 2 8. . . . . . . . . x i. = 30. CHAPTER 4. Summarizing Data . OUTLINE. 4.1 MEASURES OF CENTRAL TENDENCY

BASIC STATISTICS For the HEALTH SCIENCES Fifth Edition

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BASIC STATISTICSFor the HEALTH SCIENCESFifth Edition

By

Kuzma

= Summation

x

x

x

x

x

x

5

4

6

5

2

8

xi

= 30

Summarizing Data

4.1MEASURES OF CENTRAL TENDENCY

Explains why the selection of an appropriate sample has an imprint bearing the reliability of inferences about a population

4.2MEASURES AND VARIATION

Describes several measure of variation or variability including the standard deviation

4.3COEFFICIENT OF VARIATION

Defines the coefficient of variation, useful in comparing levels of variation

4.4MEASURING AND INTERPRETING SKEWNESS

Explains how to measure skewness and how to determine if a distribution is symmetrical or skewed

MEANS AND STANDARD DEVIATIOS OF POPULATIONS

Contrasts the equations for the parameters of a population to the statistics of a sample

- 1.Compute and distinguish between the uses of measures of central tendency: mean, median, and mode
- 2.Compute and lists some uses for measures of variation: range, variance, and standard deviation
- 3.Compare sets of data by computing their coefficients of variation
- 4.Be able to compute the mean and standard deviation for grouped and ungrouped data
- 5.Determine if a data set is symmetrical or skewed
- 6.Understand the distinction between the population mean and the sample mean

- A.The Mean
- 1.the arithmetic or simple mean is computed by sunning all the observations in the sample and dividing the sum by the number of observations
- 2.there are also harmonic and geometric means
- 3.The arithmetic mean may also be considered the balance point, or, fulcrum

- B.The Median
- 1.the observation that divides the distribution into equal parts
- 2.considered the most typical observation in the a distribution
- 3.that value above which there are the same number of observations below
- 4.Symbolically the mean is represented by

- C.The Mode
- 1.Observation that occurs most frequently
- 2.If all values are different, there is no mode

- D.Which Average Should You Use
- 1.Arithmetic mean is the most commonly used
- 2.Median gives the typical observation for a distribution – good for income

- A.Range
- 1.the difference in value between the highest (maximum) and lowest (minimum) observation
- Range =

- 1.the difference in value between the highest (maximum) and lowest (minimum) observation
- 2.can be computed easily but is not very useful because it considers only the extremes

- B.Mean Deviation
- 1.the average deviation of all observations from the mean
- 2.the sum of all of the absolute values divided by the number of observations
- Mean Deviation =
- or

- C.Standard Deviation
- 1.by far the most widely used measure of variation
- 2.the square root of the variance of the observations
- 3.computed by:
- -squaring each deviation from the mean
- -adding them up
- - dividing their sum by less than the sample size

or

s =

Standard Deviation = 4.43

Variance = Standard Deviation2

Variance = Standard Deviation

Number of observations or N 26

Initial HDL values 31, 41, 44, 46, 47, 47, 48, 48, 49, 52, 53, 54, 57, 58, 58, 60, 60, 62, 63, 64, 67, 69, 70, 77, 81, 90 mg/dl

Highest values90 mg/dl

Lowest value31 mg/dl

Mode47, 48, 58, 60 mg/dl

Median(57 + 58)/2 = 57.5 mg/dl

Sum of the values (xi)1496 mg/dl

Means, x 1496/26 = 57.5 mg/dl

Interquartile range64 – 48 = 16 mg/dl

(|xi - x|)

N

2

(xi - x )

2

s

=

Degrees of Freedom

N -1

2

(xi - x )

N -1

- Mean deviation =
- Variance =
- Standard deviation = s =

Number of observations or N = 26

Initial HDL values 31, 41, 44, 46, 47, 47, 48, 48, 49, 52, 53, 54, 57, 58, 58, 60, 60, 62, 63, 64, 67, 69, 70, 77, 81, 90 mg/dl

Highest values90 mg/dl

Lowest value31 mg/dl

Mode47, 48, 58, 60 mg/dl

Median(57 + 58)/2 = 57.5 mg/dl

Sum of the values (xi)1496 mg/dl

Means, x 1496/26 = 57.5 mg/dl

Interquartile range64 – 48 = 16 mg/dl

Sum of squares (TSS)4,298.46 mg/dl squared

Variance, “s” squared 171.94 mg/dl

Standard Deviation, s 171.94 mg/dl = 13.1 mg/dl

- D.Computing Central Tendency Using SPSS
- 1.To calculate mean and standard deviation:
- a.go to the menu and choose Analyze
- b.Select Descriptives
- -select variables you want to analyze

- 1.To calculate mean and standard deviation:

- A.Defined as the ratio of the standard deviation to the absolute value of the mean, expressed as a percentage
- CV =

- A.Skewed – a distribution that has a marked or pronounced asymmetry
- B.Use SPSS to measure skewness
- 1.Open dialog box titled Options
- 2.Select Skewness
- 3.Two columns under skewness
- a.Statistic
- b.Std Error

- 4. The Statistics column gives the actual measure of skewness, or the standard error of skewness

- A.Population Mean
- 1.defined as the sum of the values divided by the number of observations for the entire population (N)
- 2.it is the sum of the squared deviations from the population mean, divided by N
- 3.The sample mean (x) is an estimate of the population mean and is the sum of values in the sample divided by n, the number of observations in the sample alone
- 4.The population variance is the is the sum of the squared deviations from the population mean divided by N, whereas the sample variance is the sum of the squared deviations from the sample mean

In describing data by use of a summary measure, it is important to select the measure of central tendency that most accurately represents the data. A key factor is to determine of the data are symmetrical or skewed. Data are most commonly represented by two summary measures – one to indicate central tendency and one to indicate variation. The most commonly used pair is the arithmetic mean and the standard deviation.

Probability

5.1WHAT IS PROBABILITY

Discusses the concept of probability as a measure of the likelihood of occurrence of a particular event

5.2COMPLEMENTARY EVENTS

Demonstrates how to calculate probability when the events are complementary

5.3PROBABILITY RULES

Solves problems involving the probability of compound events by use of the addition rule or the multiplication rule, or conditional probability

5.4COUNTING RULES

Explains how to compute the number of possible ways an event can occur by use of permutations and combinations

5.5PROBABILITY DISTRIBUTIONS

Illustrates the concept of a probability distribution, which lists the probabilities associated with the various outcomes of a variable

5.6LOTTERY PROBABILITY AND SAMPLING

Uses lottery probabilities to illustrate sampling with and without replacement

5.7BINOMIAL DISTRIBUTION

Describes a common distribution having only two possible outcomes on each trial

- 1.Define probability and compute it in a given situation
- 2.State the basic properties of probability
- 3.Select and apply the appropriate probability rule for a given situation
- 4.Distinguish between mutually exclusive events and independent events
- 5.Distinguish between permutations and combinations, and be able to compute them for various events
- 6.Explain what a probability distribution is, and state its major use
- 7.State the probabilities by using a binomial distribution
- 8.Interpret the symbols in the binomial term

- A.Applies exclusively to a future event
- B.Probability statements are numeric, defined in the range 0 to 1, never more and never less
- C.Defined as the ratio of the number of ways the specified event can occur to the total number of equally likely events that can occur
- D.P(E), can be defined as the proportion of times a favorable event will occur in a long series of repeated trials:
- P(E) = =

=

- Probabilities are always stated in terms between 1 and 0. (Rule 1)
- The sum of all probabilities must equal 100%
- Three basic rules
- The independence rule
- The product rule
- The addition rule

Event A is mutually exclusive of event B, if event A occurs, than event B cannot occur and conversely, if event B occurs, than event A cannot occur.

Rule 2: P(A) = 1 – P(A)

A

A

A

A

A

A

A

A

- The first law of probability: (rule 5)
- The probability of the occurrence of any two or more mutually exclusive events is equal to the sum of their separate probabilities.

- If a coin is flipped two times in a row and heads shows on the first flip is there a better chance that heads will show on the second flip?
- P{H|H}= {H|T}

Coin 1Coin 2

HH

HT

TH

TT

- A.Event is the complement of event A
- B.P(A) = sum of probabilities of outcomes in A
- C.P = sum of probabilities of outcomes in
- And
- Therefore

A

A

A

A

A

A

A

A not B

AB

(48 cards)

(4 cards)

Total 52 cards

P (B | A) = P (AB) / P(A)

The vertical line “|” between B and A indicates that the event to the right of the vertical line (event A) is a condition potentially influencing the probability of the event to the left (event B)

- Suppose that in the case of drawing cards, the first card drawn is not replaced.
- If the first card draw is an ace what is the probability that the second card with be an ace?
- What is the probability the second will be a deuce?

- Probability of a Ace on first draw
4/52 = 1/13 = .0769

A

A

A

A

A

A

A

A

A

A not B

AB

(48 cards)

(4 cards)

Total 52 cards

- Probability of a Ace on first draw
4/52 = 1/13 = .0769

- Probability of an ace on second draw without replacement
3/51 = .0588

A

A

A

A

A

A

A

A

A

A not B

AB

(48 cards)

(3 cards)

Total 51 cards

- Probability of a Ace on first draw
4/52 = 1/13 = .0769

- Probability of an ace on second draw without replacement
3/51 = .0588

- Probability of a deuce on second draws without replacement
4/51 = .0788

A

2

2

2

2

2

2

2

2

A

A

A not B

AB

(48 cards)

(4 cards)

Total 51 cards

A

A

A

A

A

A

P(AB) = P(A) x P(B | A)

- The second law of probability: rule 6
- The probability of the simultaneous or successive occurrence of two or more independent events is equal to the product of their separate probabilities.

- What is the probability of drawing an ace three times in a row with replacement?

1/13

x 1/13

x 1/13

= 1/2197

Conditional probability:

P (B | A) = P (AB) / P(A)

and

Multiplication theorem

P (AB) = P(A | B) P (B)

Bayes’ Rules

P (B | A) = P (A | B) P(B) / P(A)

- All the different possible probabilities must add up to 1.0 (100%)
- A probability must always be < 1.0
- p + (1 – p) = 1.0
- P(A or B) = P (A) + P (B) – P(AB)

A

B

not

A B

A not B

AB

Total 52 cards

A

A

B

not

A B

A not B

AB

AB

B

not

A B

A not B

AB

Total 52 cards

Total 52 cards

The probability of drawing an ace or a club.

P(A or B) = P (A) + P (B) – P(AB)

P ( Ace or Club) = 4/52 + 13/52 – 1/52 = 16/52

- B.The Addition Rule:
- The probability that event A or event B (or both) will occur equals the sum of the probabilities of each individual event minus the probability of both

- A.Rule 1: Number of Ways
If event A can occur in n1, distinct ways and event B can occur in n2 ways,

then the events consisting of A and B can occur in n1 – n2 ways

- B.Rule 2: Permutations
- 1.In determining the number of ways in which you can arrange a group of objects, you must first know whether the order of arrangement plays a role.
- 2.A permutation is a selection of r objects from a group of n objects, taking the order of selection into account

- C.Rule 3: Combinations
- A selection of a subgroup of distinct objects, with order not being important

- A.Defined as a complete list of all possible outcomes, together with the probability of each
- B.Random Variable – variable which can assume any number of values
- C.A probability distribution is a list of the probabilities associated with the values of the random variable obtained in an experiment

- A.Sampling without replacement –
- once a number has been drawn (of subject selected) that number may not be drawn again

- The same number (or subject) may be picked multiple times

- A.A binomial distribution serves as a model for outcomes limited to two choices (e.g. sick or well, dead or alive…)
- B.Expansion of the binomial term , where
- 1.is the probability of successful outcome
- 2.= 1 - is the probability of an unsuccessful outcome and
- 3.is the number of trials or attempts

- C.The binomial expansion is applicable provided that
- 1.Each trial has only two possible outcomes – success or failure
- 2.The outcome of each trial is independent of the outcomes of any other trial
- 3.The probability of success, pm is constant from trial to trial

1/4

1/4

1/2

1/4

1/4

4/4 = 1

If two coins are tossed at once,

four possibilities may occur.

Coin 1Coin 2

HH

HT

TH

TT

If three coins are tossed at once,

eight possibilities may occur.

- D.In general, the probability of an event consisting of r successes out of n trials is
- where
n = the number of trials in an experiment

r = the number of successes

n- r = the number of failures

p = the probability of success

q = 1 – p, the probability of failure

- where

Factorials

- The expression
- is a term from the binomial expansion

- The entire expansion lists the terms for r successes and (n – r) failures from the binomial distribution
=

=

A table would be complex.

Formula?

(p + q)2 = pn + n/1 p(n-1)q + n(n-1)/(1)(2) p (n-2) q2 + …+ qn

There is an easier way!

(Pascal’s Triangle)

n Binomial Coefficient Denominator of p

To complete 7 trials add a column on each end of the row insert the value 1 in those cells then add two adjacent numbers in line 6 and place that value

In the cell on line 7 between the two values in the row below.

53 White balls

5 balls selected

(53!)/(5!)(48!)=1/2,869,685

42 red balls

Powerball selected

1/47

Jackpot: 1/2,869,685* 1/47

= 1/120,526,770

Probability measures the likelihood that a particular event will or will not occur. In a long series of trials, probability is the ration of the number of favorable outcomes to the total number of equally likely outcomes. Permutations and combinations are useful in determining the number of outcomes. If compound events are involved, we need to select and apply the addition rule or the multiplication rule to compute probabilities. The outcome of an experiment together with its respective probabilities constitutes a probability distribution. One very common probability distribution is the binomial distribution. It presents the probabilities of various numbers of successes in trials where there are only two possible outcomes to each trial.