Stats Chapter 14 Tools

8/11/2019 Stats Chapter 14 Tools

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CHAPTER 14 TOOLS

14.2

Correlation Coefficient

n 8

r 0.869374114 y x

r=CORREL(array 1, array 2) 0.869374114

Starting

Salary GPA2,700 3.2

2,800 3.4

2,200 2.6

2,700 3.4

3,200 3.7

2,900 3.9

2,100 2.5

2,300 3.3

y x

20,900 26

Significance of the Correlation Coefficient - Hypothesis Testing When the Correlation Coefficient is

(Correlation Coefficient)

0.01

r -0.68377

n (number of ordered pairs) 11

t(Test Statistic for ) = r/ (1-r2

)/(n -2)] -2.81117822

degrees of freedom 9

t/2(critical t score)(see table 5 in appendix) 2.353

14.3

Regression Line Equation and Slope^y = a + bx

n 8

b 1.815730337

a 2.905617978

x

Temperat

ure

Total Sum of Squares 21

SST (equal to SSE + SSR) 730 13

21

SSE (Sum of Squares Error) 179.8337079 20

8

SSR (Sum of Squares Regression) 550.1662921 14

19

Sample Coefficient of Determination (2is the population coefficient of determination)> 21


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R2= SSR / SST 0.753652455

F Test Statistic

F = SSR / (SSE/(n-2)) 18.35583435

x

Critical F Score 137

D1(SSR Degrees of Freedom always equals 1) 1

D2(SSE Degrees of Freedom is n-2) 6

USE THE RIGHT TABLE FOR THE RIGHT

14.4

Confidence Interval and Prediction Interval Construction

Use phStat on the data provided Year GPA

INCLUDE THE DATA COLUMN NAMES 1 3.07

2 2.883 2.96

4 2.85

5 2.98

6 3.05

7 3.07

8 2.99

9 3.19

10 3.14

11 3.16

14.5 CI = b t/2* sb x

a= 17.54340071 Minutes

b= -0.562425684 3

8

n= 10 0

SSE= 30.29726516 6

se= 1.946062215 6

sb= 0.21220652 1

= 4.7 9

0.05 2

df(n-2)= 8 7

t/2= 2.306 5

UCL= -0.07307745

LCL= -1.051773918

SST= 56.9

SSR= 26.60273484 x

47

If actual value isn't specified for the

confidence and prediction interval x value,

insert the last value of x if the data on the x

axis is chronological. You can use the ANOVA

table to find the P value and other fun stuff!


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R2= SSR / SST 0.467534883

F = SSR / (SSE/(n-2)) 7.024458399

t= (b-) / sb (assuming = 0 ) -2.650369483


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y xy x2

y2

272 4961 2513 9978

(see cumulative probabilities for sandard normal distribution table)

In the two tail test, the yielded p is multiplied by two.

y xy x2 y2

Rating 16.98098

14 42 9 196 16.41855 0.562426

13 104 64 169

17 0 0 289

11 66 36 121

15 90 36 225

17 17 1 289

14 126 81 196

20 40 4 400

14 98 49 196

14 70 25 196

y xy x2

y2

149 653 305 2277

variable's total variation that can be explain

independent variable

When 2i s equal to zero, that means that va

outcomes is not explained by x data value

indicates that there is no relationship betwe

The p value is the probability of observing a sample mean at least as extreme as the one

selected for the hypothesis test, assuming the null hypothesis is true.


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7/8

ndent


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ed by an

iation in y

; which

n x and y.

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Stats Chapter 14 Tools