Statistics and the TI-83
Lesson #12 - Analysis of Variance
I. The F -distribution.
1. Fpdf (value, df of numerator, df of denominator) gives the value of the F-curve at a given point on the scale (for numbers 0), for given numbers of degrees of freedom in the numerator and denominator.
Exercise 1. What is the F distribution functional (ordinate) value for
a) F =3, df=8, 10 b) F =10, df=8,10 c) b) F =1, df=8,10
d) F=1.21, df=8, 10 e) F=.99, df=8,10 f) F=.4, df=8, 10
· 2nd DISTR 8 Fpdf (3, 8, 10) ENTER answer: .0510001543
Use 2nd ENTRY, INS to add digits and DEL to eliminate digits.
· 2nd ENTRY (change 3 to 10) ENTER answer: 2.960297745E-4
· 2nd ENTRY (change 10 to 1) ENTER answer: .5781831533
· 2nd ENTRY (change 1 to 1.21) ENTER answer: .4588262362
· 2nd ENTRY (change 1.21 to .99) ENTER answer: .5839578692
· 2nd ENTRY (change .99 to .4) ENTER answer: .6032989541
Exercise 2. Draw the graph of the F-distribution functions
F(x, 3, 8) F(x, 7, 15) F(x, 21, 18)
· WINDOW Xmin=-.05 Xmax=5 Ymin=-.05 Ymax=.8
· Y= (deselect all plots and functions)
· Y1= 2nd DISTR 8
· Y1=Fpdf(x, 3, 8) GRAPH
· Y= (deselect all plots and functions)
· Y2= 2nd DISTR 8
· Y2=Fpdf(x, 7, 15) GRAPH
· WINDOW Xmin=-.05 Xmax=5 Ymin=-.05 Ymax=1
· Y= (deselect all plots and functions)
· Y3= 2nd DISTR 8
· Y3=Fpdf(x, 21, 18) GRAPH
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II. Fcdf (lower boundary, upper boundary, df of numerator, df of denominator) gives the AREA or Probability below the F curve between the given boundaries.
Exercise 3. Compute P (0< F < 1.943, df = 6, 8) and draw the area.
· 2nd DISTR 9 Fcdf (0 , 1.943, 6, 8) ENTER answer: 0.8110409686
· 2nd DRAW 1 (ClrDRAW) ENTER CLEAR (done)
· WINDOW Xmin=-1 Xmax= 5, Ymin=-.05 Ymax= 1
· 2nd DISTR DRAW 4
· Shade F (0, 1.943, 6, 8) ENTER
Exercise 4. . Compute P ( F > 2.99, df =16, 12) and draw the area.
· 2nd DISTR 9 Fcdf (2.99 , EE5, 16, 12 ) ENTER answer: 0.0304064245
· 2nd DRAW 1 (ClrDRAW) ENTER (done) CLEAR
· WINDOW Xmin=-1 Xmax= 5, Ymin=-.05 Ymax= .8
· 2nd DISTR DRAW 4
· Shade F (2.99, 20, 16, 12) ENTER
Exercise 5.
a) Find the critical value of F, df=(5, 12) with area to the right =.025
· MATH 0 (solver) Ý CLEAR
· eqn:0= 2nd DISTR 9
· eqn:0 = Fcdf(x, EE5, 5, 12) - 0.025 ENTER
· guess x=3 ALPHA SOLVE answer: 3.8911339332
b) Find the critical value of F, df=(10, 15) with area to the right =.025
· MATH 0 (solver) Ý CLEAR (or change 5, 12 to 10, 15
· eqn:0= 2nd DISTR 9
· eqn:0 = Fcdf(x, EE5, 10, 15) - .025 ENTER
· guess x=3 ALPHA SOLVE answer: 3.0601968514
c) Find the critical value of F, df=(15, 11) with area to the right =.725
· MATH 0 (solver) Ý CLEAR
· eqn:0= 2nd DISTR 9
· eqn:0 = Fcdf(x, EE5 , 15, 11) - .725 ENTER
· guess x=.9 ALPHA SOLVE answer: 0 .72360640544
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III. 2-Sample F test.
Exercise 6. Consider the following information obtained from two samples selected at random from two normal populations.
Sample 1 / Sample 235 / 27
28 / 41
24 / 37
33 / 35
31 / 39
25 / 32
24 / 38
a) Store sample 1 and sample 2 as list L1 and L2
· {35, 28, 24, 33, 31, 25, 24} STOÞ L1 ENTER
· {27, 41, 37, 35, 39, 32, 38} STOÞ L2 ENTER
b) Using the 2-SAMPLE F TEST, test at a 5% significance level whether
the variances of the populations from which these samples were
drawn are equal.
· STAT Ü TESTS ALPHA D Þ DATA ENTER
2-SampFTest
Inpt: DATA ENTER
List 1: L1
List 2: L2
Freq1: 1
Freq2: 1
s1:≠ s2
· Calculate ENTER
Conclusions:
F=.8968
p-value=.8982>.05 Do not reject s1:= s2
Sx1=4.50396651 Sx2=4.75594866
X1=28.57 X2=35.57 n1=7 n2=7
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c) Using a t-test, test at a 5% significance level whether the means of the populations from which these samples were drawn are equal?
· STAT Ü TESTS 4 Þ DATA ENTER
2-SampFTest
Inpt: DATA
List 1: L1
List 2: L2
Freq1: 1
Freq2: 1
m1 ≠m2
Pooled: Yes
· Calculate ENTER
t=-2.827445952
p-value= .0152863047<.05 Reject m1 = m2
Sx1=4.50396651 Sx2=4.75594866
X1=28.57 X2=35.57 n1=7 n2=7
Conclusion: reject H0, we statistical evidence at a 5% level, to believe that the
mean of the populations are not equal
d) Using the one-way ANOVA test, test at a 5% significance level whether the
means of the populations from which these samples were drawn are equal?
· STAT Ü TESTS ALPHA F
· ANOVA(L1, L2) ENTER
F=7.99445061 Note: the F -value obtained is the square of the
t-value obtained previously
p-value=.0152478274 <.05 Reject m1 = m2
ANOVA TABLE (summary of calculator results)
Source / df / SS / MS / FFactors or Treatments
(between) / 1 / 171.5 / 171.5 / =7.99
Error
(within) / 12 / 257.4 / 21.45 / p-value = Fcdf(7.99, EE5, 1,12)
=0.015
Total / 13 / 428.9
Reject H0, we have statistical evidence at a 5% level of significance to believe that the means of the population from which these samples were drawn are not equal.
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Exercise 7. At a 1% level of significance, can you conclude that the mean
yield of wheat for each of the following three brands of fertilizer is the same?
Use an ANOVA test.
Fertilizer A / Fertilizer B / Fertilizer C68 / 45 / 60
75 / 47 / 54
38 / 59 / 70
64 / 52 / 63
68 / 63
73
· {68, 75, 38, 64, 68, 73} STOÞ L3 ENTER
· {45, 47, 59, 52} STOÞ L4 ENTER
· {60. 54. 70. 63. 63} STOÞ L5 ENTER
· STAT ÜTESTS F (ANOVA)
· ANOVA(L3, L4, L5) ENTER
F=2.455944257
p-value=0.1276247294>0.01 Do not reject m1 = m2 = m3
Do not eject H0, we have statistical evidence at a 5% level of significance to believe that the means yield of wheat for each of three brands of fertilizer is the same.
ANOVA TABLE
Source / df / SS / MS / FFactors or Treatments
(between) / 2 / 474.85 / 237.425 / = 2.46
Error
(within) / 12 / 1160.08333 / 96.6736111 / p-value = Fcdf(2.46, EE5, 2,12)
=0.127> 0.01
Total / 14 / 1634.93333
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