MM150Unit 9 Seminar

Statistics II

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9.1

Measures of Central Tendency

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Averages• Several different types of averages can be

calculated for a given set of data.

• All averages, in general, are called measures of central tendency.

• The three most common measures of central tendency are mean, median, mode, and midrange.

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Mean – To find the arithmetic mean, or mean, sum the data scores and then divide by the number of data scores.

Example: Find the mean of the data scores 5, 6, 2, 9, 8

5 + 6 + 2 + 9 + 8 = 30 = 6 5 5

Median – To find the median, put the data scores in ascending or descending order and then find the middle data score. If there are an even number of data scores, after ranking the scores, find the mean of the middle two.

Example: Find the median of the data scores 5, 7, 2, 9, 8

Put the scores in ascending order: 2, 5, 7, 8, 9

Example: Find the median of the data scores 4, 7, 2, 9Put the scores in ascending order: 2, 4, 7, 9Find the mean of 4 and 7: (4 + 7)/2 = 11/2 = 5.5

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Mode – The mode is the data score that occurs most frequently.

Example: Find the mode of the data scores 6, 4, 9, 8, 6, 5

It may help to put the scores in ascending order: 4, 5, 6, 6, 8, 9

You can see that the data score 6 occurs most often.

•You can have data sets that don’t have a mode (each score occurs once) and you can have data sets that are bimodal – which means they have 2 modes.

Midrange – The midrange is the value halfway between the greatest and least data score. To find it, take the mean of the greatest and least data score.

Example: Find the midrange of the data scores 6, 4, 9, 8, 6, 5

It may help to put the scores is ascending order: 4, 5, 6, 6, 8, 9

The midrange is (4 + 9)/2 = 13/2 = 6.5

*Please read on page 362 of your text when each is the ‘better’ average.

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EVERYONE: page 366 #21

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EVERYONE: page 366 #21

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1, 2, 2, 3, 3, 3, 4, 5, 5, 7

Mean: 1 + 2 + 2 + 3 + 3 + 3 + 4 + 5 + 5 + 7 = 3.5

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Median: 1, 2, 2, 3, 3, 3, 4, 5, 5, 7 3 weeks

(3 + 3)/2 = 3

Mode: 1, 2, 2, 3, 3, 3, 4, 5, 5, 7 3 weeks

Midrange: 1 + 7 = 4 weeks 2

Mean ExampleTodd is taking a math class where his end of term grade is based on 4 exams, each having the same number of points and weighted the same. He scored 98, 82, and 87 on the first three exams in his class. What does he need to score on the 4th exam to get at least a 90% for the final grade? The instructor uses mean as the average.

98 + 82 + 87 + x = 90 4

267 + x = 90 4

267 + x = 360

x = 93

Todd must score a 93 or higher on the test.

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Measures of Position• There are 2 measures of position, percentiles and quartiles.

• They are used to make comparisons for a large amount of data.

• Percentile – There are 99 percentiles that divide the data up into 100 equal parts.

• When you are reported to be in the 98th percentile, this does not mean your score is a 98%. This means that you outperformed about 98% of the population. 9

Measures of Position (con’t)

Quartile – Quartiles divide data into 4 equal parts, called quartiles. The first quartile is at 25%, the second at 50%, and the third at 75%.

Q1, Q2, Q3 divide ranked scores into four

equal parts25% 25% 25% 25%

Q3Q2Q1(minimum) (maximum)

(median)

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ExampleDetermine Q1, Q2, and Q3 of the data below:

15, 10, 19, 18, 11, 15, 13, 18, 19, 17,19, 15, 16, 13, 15, 16, 13, 12, 14

1.First put the data in ascending order

10, 11, 12, 13, 13, 13, 14, 15, 15, 15, 15, 16, 16, 17, 18, 18, 19, 19, 19

2.Find the median, 15 is Q2

3.Find the median of the lower half for Q1, 13

4.Find the median of the upper half for Q3, 1811

9.2

Measures of Dispersion

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2 Measures of DispersionRange – The range is the difference between the greatest and least data score.

Example: Find the range of the data scores

55, 59, 51, 64, 60

Put the data scores in ascending order 51, 55, 59,

60, 64

The range is 64 – 51 = 13

Standard deviation – The standard deviation tells us how much the data differ from the mean.

See the next PowerPoint slide for an example of standard deviation.

Range = (maximum value) – (minimum value)

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Standard Deviation, cont’d

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1

x xs

n

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Standard Deviation ExampleFind the standard deviation of 11, 15, 18, 9, 12

1.) Find the mean of the data scores. 11+15+18+9+12 = 65 = 13 5 52.) Make a chart with 3 columns

Data Data – Mean (Data – Mean)2

11 11- 13 = -2 4

15 15 – 13 = 2 4

18 18 – 13 = 5 25

9 9 – 13 = -4 16

12 12 – 13 = -1 1

3. 4. 5.

6.) 507.) Divide 50 by n – 1, where n is the number of data scores. So divide 50 by 4, which is 12.5

8.) Find the square root of the number found in step 7. √12.5 ≈ 3.5355 15

Everyone Example

Calculate the standard deviation for the data set:

15, 16, 20, 13

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EVERYONE solution1. Find the mean. 15 + 16 + 20 + 13 = 64 = 16 4 4

2. Make a 3-column table.

Data Data – Mean (Data – Mean)2

15 -1 1

16 0 0

20 4 16

13 -3 9

6. 26

3. 4. 5.

7. 26/3 = 8.667

8. √8.667 ≈ 2.944

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9.3

The Normal Curve18

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Types of Distributions

• Rectangular Distribution

• All observed values occur with the same frequency.

• J-shaped distribution

• The frequency is either constantly increasing or constantly decreasing.

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Types of Distributions (con’t)

• Bimodal

• Two nonadjacent values occur more frequently than any other values in the set of data.

• Skewed to left or right

• Has more of a “tail” on one side than the other.

• The greatest frequency appears on the left or the right of the curve.

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Skewed Distributions

• In figure (a) the greatest frequency appears on the left so the mode would be on the left side of the curve.

• All the data would be used to determine the mean. The values on the right side of the curve in (a) would increase the value of the mean. So the value of the mean would be farther to the right of the mode.

• The median would be between the mean and the mode.

Normal DistributionsPROPERTIES

1. The graph of a normal distribution is called a normal curve.

2. The normal curve is bell-shaped and symmetric about the mean.

3. The mean, median, and mode of a normal distribution all have the same value and all occur a the center of the distribution.

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The Empirical RuleIn any normal distribution

1.Approximately 68% of all the data lies within one standard deviation of the mean (in both directions).

2. Approximately 95% of all the data lies within two standard deviations of the mean (in both directions).

3.Approximately 99.7% of all the data lies within

three standard deviations of the mean (in both directions).

The Empirical Rule

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Empirical Rule ExampleSuppose that the weights of newborn infants are normally distributed. If approximately 2000 infants are born at Sarasota Memorial Hospital each year, determine the approximate number of infants who are expected to weigh:

a)within one standard deviation of the mean.

b)within two standard deviations of the mean.

Solution: a) By the empirical rule, 68% of the infants weigh within one standard deviation of the mean.

(.68)(2000) = 1360 infants are expected to weigh within one

standard deviation of the mean.

a) By the empirical rule, 95% of the infants weigh within two standard deviations of the mean.

(.95)(2000) = 1900 infants are expected to weigh within two

standard deviations of the mean.25

z-Scores

• z-scores determine how far, in terms of standard deviations, a given score is from the mean of the distribution.

value of the piece of data - mean

standard deviation

xz

s

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Example: z-scores

• A normal distribution has a mean of 50 and a standard deviation of 5. Find z-scores for the following values.

• a) 55 b) 60 c) 43

• a)

A score of 55 is one standard deviation above the mean.

55 50 51

5 5z

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Example: z-scores continued

• b)

A score of 60 is 2 standard deviations above the mean.

• c)

A score of 43 is 1.4 standard deviations below the mean.

60 50 102

5 5z

43 50 71.4

5 5z

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9.4

Linear Correlation and Regression

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Linear Correlation• Linear correlation is used to determine

whether there is a relationship between two quantities and, if so, how strong the relationship is.

• The linear correlation coefficient, r, is a unitless measure that describes the strength of the linear relationship between two variables.

• If the value is positive, as one variable increases, the other increases.

• If the value is negative, as one variable increases, the other decreases.

• The variable, r, will always be a value between –1 and 1 inclusive.

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Scatter Diagrams• A visual aid used with correlation is the scatter

diagram, a plot of points (bivariate data).– The independent variable, x, generally is a

quantity that can be controlled.

– The dependent variable, y, is the other variable.

• The value of r is a measure of how far a set of points varies from a straight line. – The greater the spread, the weaker the

correlation and the closer the r value is to 0.

– The smaller the spread, the stronger the correlation and the closer the r value is to 1.

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Correlation

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Linear Correlation Coefficient

• The formula to calculate the correlation coefficient (r) is as follows:

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Example: Words Per Minute versus Mistakes

There are five applicants applying for a job as a medical transcriptionist. The following shows the results of the applicants when asked to type a chart. Determine the correlation coefficient between the words per minute typed and the number of mistakes.

934Nancy

1041Kendra1253Phillip1167George824Ellen

MistakesWords per MinuteApplicant

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Solution• We will call the words typed per

minute, x, and the mistakes, y.• List the values of x and y and

calculate the necessary sums.

306811156934xy = 2,281

y2 = 510

x2 =10,711

y = 50x = 219

1012118y

Mistakes xyy2 x2x

41536724

WPM

41010016816361442809737121448919264576

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Solution continued

• The n in the formula represents the number of pieces of data. Here n = 5.

r n xy x y

n x2 x 2 n y 2 y 2

r 5 2281 219 50

5 10,711 219 2 5 510 50 2

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Solution continued

11,405 10,950

5 10,711 47,961 5 510 2500

455

53,555 47,961 2550 2500

455

5594 500.86

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Solution continued

• Since 0.86 is fairly close to 1, there is a fairly strong positive correlation.

• This result implies that the more words typed per minute, the more mistakes made.

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Linear Regression

• Linear regression is the process of determining the linear relationship between two variables.

• The line of best fit (regression line or the least squares line) is the line such that the sum of the squares of the vertical distances from the line to the data points (on a scatter diagram) is a minimum.

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The Line of Best Fit

• Equation:

y mx b, where

m n xy x y

n x2 x 2, and b

y m x n

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Example• Use the data in the previous

example to find the equation of the line that relates the number of words per minute and the number of mistakes made while typing a chart.

• Graph the equation of the line of best fit on a scatter diagram that illustrates the set of bivariate points.

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Solution

• From the previous results, we know that

m n xy x y

n x2 x 2

m 5(2,281) (219)(50)

5(10,711) 2192

m 455

5594m 0.081

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Solution

• Now we find the y-intercept, b.

Therefore the line of best fit is y = 0.081x + 6.452

b y m x

n

b 50 0.081 219

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b 32.261

56.452

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Solution continued

• To graph y = 0.081x + 6.452, plot at least two points and draw the graph.

8.88230

8.07220

7.26210

yx

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Solution continued

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Example: page 407 #24a) Draw a scatter

diagram

b) Determine the value of r, rounded to the nearest thousandth

c) Determine whether a correlation exists at … = 0.05

d) Determine whether a correlation exists at …. = 0.01

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Example: continued

a) The first thing

to do is plot the

points.

Here we have

(6,13), (8,11),

(11,9),

(14,10) and (17,7).

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Example: continued

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Example: page 407 #32Determine the equation of the line of best

fit from the data in the exercise indicated. Round both the slope and y

intercept to the nearest hundredth.