Making Sense of Statistics

8/13/2019 Making Sense of Statistics

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by

Jason SamuelsCUNY-BMCC

AMATYC 39, 2013-11-2


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Students Dont Get StatisticsAfter years of Algebra courses, Statistics requires a very

different way of thinking

Whats the formula? ome steps requ re a ormu a e.g. find the z-score

Some steps dont

e.g. find the z-score

wait, what?

n

xz

2

2

2

1

2

1

21

nn

xxz

n

pp

ppz

)1(

2

22

1

11

21

)1()1(

n

pp

n

pp

ppz


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Which Topics Can Be Unified? Doing calculations with standard data distributions

Find the data value, z-score, probability

Normal distribution, t-distribution, etc. on ence nterva s

Hypothesis tests

Some ideas so these topics make sense to students


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Key idea #1: Describe the distribution

Orients the students toward the values they will use inthe problem and in their calculations

escr e t e str ut on o t e ata:

Center (mean)

Spread (standard deviation)

Shape (which distribution: normal, t, etc.)


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Describe the distribution an example

Ex) A college has an average of 23.7 students in eachclass, with a standard deviation of 5.6. What is theprobability that a sample of 35 classes has an average of

Get the facts: =23.7 =5.6 n=35 want P( > 25)

Describe the distribution of Mean:

Standard deviation:

Shape: n>30 so its normal

x

x7.23x

95.035

6.5x


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Key Idea #2: Draw the Graph

All values can be organized and connected using onegraph:


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Draw the graph example continued

From before

Get the facts: =23.7 =5.6 n=35 want P( > 25)

Describe the distribution of : normalx 7.23

x 95.0

x

x

Now draw the graph:

z

x 23.7 25


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Key Idea #3:

The Flow ChartAlmost every calculation

students will do withstan ar istri utions isguided by this flow chart:


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Key Idea #4: The Formula There is only one formula students need to know:

deviationstandard

(mean)value)(datastatistictest

Or, equivalently:

data value = (mean)+(test statistic)(standard deviation)

For a single data value:

For a sample mean:

For a sample proportion:

zor

xxz

xx

x

x zxx

z

...or...

pp

p

pzp

pz

...or...


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BenefitStudents learn that z has one meaning the number of

standard deviations from the mean so z has one formula

Never again will students use these varied, complex formulas:

Students make fewer order-of-operation calculation errors

n

xz

n

pp

ppz

)1(

2

2

2

1

2

1

21

nn

xxz

2

22

1

11

21

)1()1(

n

pp

n

pp

ppz


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The formula example continuedFrom before

Get the facts:

=23.7 =5.6 n=35 want P( > 25)

x Describe the distribution of

normal

Now:

Find the z-score:

x

7.23x 95.035

6.5x

37.195.0

7.2325

x

xxz


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Flow Chart & Graph - together

Probability

Z-score

Data value


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Flow Chart & Graph - example continued

Get the facts: =23.7 =5.6 n=35 want P( > 25)Describe the distribution of : normal

Now, fill in the graph following the flowchart:

7.23x 95.0xx

x

Probability

Z-score

Data value

1.37

.9147.0853

,


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Putting it together: an exercise

The mean time for all flight delays is 21 minutes with astandard deviation of 12 minutes. What is theprobability that a sample of 36 flights has a delay


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Step 1: get the facts =21

=12

n=36

(1) Get the facts:=21 =12 n=36find P( >26)

(2) Describe the distribution:

(3) Draw the graph:

x

(4) Do the calculations:

(5) Conclusion:


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Step 2: describe the distribution

Center:

mean =21

Spread:


(2) Describe the distribution:= 21 =2 Normal

(3) Draw the graph:

x x

x

x

standard deviation

Shape:

n>30, so the distribution is normal(4) Do the calculations:

(5) Conclusion:

26

12

36

12

nx


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Step 3: Draw the graph(1) Get the facts:

=21 =12 n=36find P( >26)


(3) Draw the graph:

x x

x


(5) Conclusion:


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Step 4: Do the calculations

z-score:



(3) Draw the graph:

x x

x

2126 xx

Areas: using technology

area to the left = .9937area to the right = .0063

(4) Do the calculations:z = 2.5 area=.9937 & .0063

(5) Conclusion:

.2

x

2.5


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Step 5: Write the conclusion

The probability is .0063



(3) Draw the graph:

x x

x

(4) Do the calculations:z = 2.5 area=.9937 & .0063

(5) Conclusion:The probability is .0063


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A harder exercise (thats not harder)

Ex) for United, the mean delay time is 18 minutes,st.dev.=11 minutes. For Delta, the mean delay time is 22minutes, st.dev.=14 minutes. Find the probability that,

for a sample of 32 United flights and 34 Delta flights,Delta has a higher mean delay time by over 2 minutes.


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A (not) harder exercise

Step 1: Get the facts

Delta:

= = =

(1) Get the facts:1=22 1=14 n1=342=18 2=11 n2=32find P( >2)


12 xx

1 1 1United:

2=18 2=11 n2=32

Find P( >2)

(3) Draw the graph:


(5) Conclusion:

21 xx


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Step 2: Describe the distribution

Center

Mean

=22-18=4



Normal

12 xx

421 xx 090.321 xx

2121

xx

Spread

Standard deviation

Shape

n1, n2>30 so its normal

(3) Draw the graph:


(5) Conclusion:

22

2

2

2

1

2

1 )()(or)()(

2121 xxxx

nn

09.332

11

34

14 22


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Step 3: Draw the graph (1) Get the facts:1=22 1=14 n1=342=18 2=11 n2=32find P( >2)


Normal

12 xx

421 xx 090.321 xx(3) Draw the graph:


(5) Conclusion:


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Step 4: Do the calculations

z-score:



Normal

12 xx

421 xx 090.321

xx42

Areas:

area to the left = .2587area to the right = .7413

(3) Draw the graph:

(4) Do the calculations:z = -0.65 areas: .2587 & .7413

(5) Conclusion:

.09.3

-0.65


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Normal

12 xx

421 xx 090.321

xx

(3) Draw the graph:

(4) Do the calculations:z = -0.65 areas: .2587 & .7413

(5) Conclusion:



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A Handy Tool StatDisk

Does all basic statistics calculations with a simplegraphical interface and one or two clicks

.


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The Issue of the Center

First students learn that they know , this defines thecenter of the distribution, and x (the value from thedata) exists relative to that

,

In the case of inference confidence intervals andhypothesis tests (or p) is not known. Rather, weknow (or ) and make an inference about (or p).

What does this mean for the distribution, and thegraph?

x p


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The Issue of the Center Confidence Interval

Formula:

What does this imply for the graph?),( xx zxzx

The center is , not ! We are calculating values for , not

With confidence intervals we just use the formula and ignore it

With hypothesis tests, the issue does not go away

xxzx xzx

xx


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The Issue of the Center

Hypothesis Test

Old way:

Ho: = 0

1 o

and you spend all this time explaining why, even thoughthe hypothesis says > o you shade to the right of

(and I think students still dont understand, they just do it)

o x

x


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Recognizing a Different Center Hypothesis Test

New way:

Ho: = 0

1 o

and now you shade where the claim tells you to shade,and that area is your confidence level

xo


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Why This Makes Sense Shaded area matches the claim

Hypothesis tests and confidence intervals are bothinferences about the population, and they should agree

, , . We are using a distribution of values for

The center is

What does confidence mean? Its a type of probabilistic statement

95% of the time, a conclusion made in this way will becorrect

x


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Different center: an exercise

Ex) We want to find out if the average American familyhas more than 1.8 kids (because that places a strain onmunicipal services). From a survey of 500 families, the

=. . .


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Step 1: Get the facts

=1.92 =0.9 n=500

(1) Get the facts:=1.92 =0.9 n=500

test claim: > 1.8


(3) Draw the graph:

x

x

c a m: .


(5) Conclusion:


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Step 2: describe the distribution of

Center:

(1) Get the facts:=1.92 =0.9 n=500

test claim: > 1.8

(2) Describe the distribution:mean=1.92 stdev=.0402 normal

(3) Draw the graph:

x

mean= .

Spread:

Shape:n>30 so its normal (4) Do the calculations:

(5) Conclusion:

0402.500

9.0st.dev.

n


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Step 3: Draw the graph(1) Get the facts:

=1.92 =0.9 n=500

test claim: > 1.8


(3) Draw the graph:

x


(5) Conclusion:


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Step 4: Do the calculations(1) Get the facts:

=1.92 =0.9 n=500

test claim: > 1.8


(3) Draw the graph:

x

99.2

0402.

92.18.1

z

Areas:

area to the left = .0014

area to the right = .9986(4) Do the calculations:

z = 2.99 areas: .0014 & .9986(5) Conclusion:

-2.99


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We are .9986 confident in the

(1) Get the facts:=1.92 =0.9 n=500

test claim: > 1.8


(3) Draw the graph:

x

.average American family

has more than 1.8 children)

(4) Do the calculations:z=-2.99 areas .0014 & .9986

(5) Conclusion:We have .9986 confidence

that > 1.8


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Big Changes

All the formulas for the test statistic flip For means

the center is

The formula for z is:

For ro ortions

..

0

ds

xz

x

the center is

The formula for z is:

These are equivalent to the confidence interval formulas

(just solve for 0) so we already used them withoutknowing it

The formulas forx & z (given population info) were inverses;

Now the formulas for and z from inference (confidenceintervals & hypothesis tests) are inverses as they should be

..

0

ds

ppz

p


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Jason [email protected]

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Making Sense of Statistics