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Statistics

Autor:   •  May 3, 2018  •  1,316 Words (6 Pages)  •  473 Views

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overall sample mean is = [pic 52][pic 53]

= [pic 54]

= 138

Sum of squares is given by SSTR = [pic 55][pic 56]

= 5(133-1338)2 +5(139-138)2 +5(136-138)2+5(144-138)2[pic 57]

= 330

Mean square is given by MSTR = [pic 58]

= [pic 59]

= 110

The three sample variance are

= 47.5[pic 60]

= 50[pic 61]

= 21[pic 62]

= 54.5[pic 63]

Sum of squares due to error SSE = [pic 64]

= (5-1)47.5 + (5-1)50 + (5-1)21 + (5-1)54.5

= 692

Mean square due to error MSE = [pic 65]

= [pic 66]

= 43.25

ANOVA table for a completely randomized design:

Variation source

Sum of Squares

Degrees

of

Freedom

Mean Square

F

p - value

Treatment

SSTR

k - 1

MSTR= [pic 67]

[pic 68]

Error

SSE

- k[pic 69]

MSE= [pic 70]

Total

SST

- 1[pic 71]

For null hypothesis the mean drying times for the four paints are equal.

: = ==[pic 72][pic 73][pic 74][pic 75][pic 76]

For alternative hypothesis the mean drying times of four paints are not equal.

: ≠ ≠≠[pic 77][pic 78][pic 79][pic 80][pic 81]

ANOVA is calculated to test the above claim by using the information followed by:

Variation source

Sum of Squares

Degree

of

Freedom

Mean Square

F

p - value

Treatment

330

3

110

2.54335

0.09276

Error

692

16

43.25

Total

1022

19

The equality of three means for the test statistic is F = [pic 82]

= [pic 83]

= 2.54335

Degrees of freedom for the numerator is k – 1= 3

Degrees of freedom for the denominator is - k = 20 – 4 [pic 84]

= 16.

Here we reject null hypothesis, because of the greater values of the test statistics.

Here the p-value is the higher tail area of F distribution to the right side of the test statistic.

F= 2.54335

The following shows the area in the upper tail of an F distribution with 3 numerator and 16 denominator degrees of freedom

Area for the Upper Tail

0.1

0.05

0.025

0.01

F value (=3,=16)[pic 85][pic 86]

2.46

3.24

4.08

5.29

F=2.54334 which is in between 2.46 and 3.24

The area in the upper tail, F= 2.54335 is between 0.05 and 0.1.

The p-value is between 0.05 and 0.1.

Therefore P=0.09276

p-value > α = 0.05, hence [pic 87]

Test does not provide enough information so the mean drying times for the four paints are not same.

25) In the Midwest gas price study,

is the gallon mean price for the gasoline for company Shell[pic 88]

is the mean price per gallon for gasoline for company BP[pic 89]

is the mean price per gallon for gasoline for company Marathon[pic 90]

the hypothesis for the mean price of three companies are as follows:

: = =[pic 91][pic 92][pic 93][pic 94]

: Not all population means are equal[pic 95]

F = [pic 96]

F

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