/Stat._Ch._4_Final_Review_May_8

Page 1

x Sample mean for grouped data x =

Chapter 7

©xf n

Sample standard deviation for grouped data s =

©(x - x)2f ©x2f - (©xf )2!n = n - 1 n - 1 B !"#"$%&'$%(%)*+#,%-./*.0% B

Chapter 4

Pearson product moment correlation coefficient n©xy - (©x)(©y) 2

2

2

Standard deviation of x distrib

Least-squares line yn = a + bx

n©xy - (©x)(©y) n©x 2 - (©x)2

x

s!

Chapter 8 for m x - E 6 m 6 x + E where E = zc

a = y - bx

% Coefficient of determination = r 2 1$ 23%'.#/*.4%5#46%4.#,,7%86.%934.%:#63,*+.;%!8<<36.%"'#"%#%5#4%*6%5'36.+%#"% 4#+=39$%>."%?%@.%"'.%0.*:'"%3A%"'.%5#4%B*+%'8+=4.=6%3A%<38+=6CD%#+=%,."%7%@.%"'.% 9*,.6%<.4%:#,,3+%B9<:C$% % ?% EE% ((% (E% (F% (G% HG% (I% (J% K% EE% EI% 11% E1% 1J% HG% 1F% EL% )*+=%"'.%.M8#"*3+%3A%"'.%,.#6"N6M8#4.6%,*+.$%BO,.#6.%438+=%7384%A*+#,%#+60.4%"3%"03% %

z =

Confidence Interval 2

2n©x - (©x) 2n©y - (©y)

where b =

Mean of x distribution mx = m

Standard score for x;

Regression and Correlation

r =

Raw score x = zs + m

=.5*9#,%<,#5.6$C%

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s when s 1n

s when s i 1n with d.f. = n -

E = tc


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8.

!%B_%"'386#+=6C%

1J%

EL%

1I%

(F%

1G%

IE%

"%B_%"'386#+=6C%

H(G%

EPG%

1JE%

(1H%

1HG%

LJJ%

5. Create a scatter diagram. Using only the scatter diagram, would you estimate the correlation coefficient to be positive, close to zero, or negative? Explain your answer.

9. For the given data compute each of the following. (a) x and y

(b)

! x, ! y, ! x2 , ! y 2 , ! xy

(c) The slope b and y-intercept a of the least squares line; write out the equation for the least squares line.

(d) Graph the least squares line on your scatter plot of problem 5. 10. Compute the sample correlation coefficient r.

11. Suppose that the amount spent on advertising is $37,000. What does the least-squares line predict for the increase in sales?


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