grade, 50 bushels of average grade and 50 bushels of poor grade oranges. Is the quality
of oranges from this orchard representative of this area?
(2 = 167.5)
6.
A study was conducted to establish whether there is a difference in susceptibility to
mildew b
Case 1. One plant per pot
The difference between the two plant heights cannot be identified as being due to plant differences
or to pot differences.
Case 2. Two plants per pot and there is no difference between plants in the same pot but there are
2
2
dif
EXERCISES
1.
Suppose 50 years rainfall data are summarized in the table below. Test whether the data
approximate a normal distribution.
Class
1
Midpoint
(inches)
8
Frequency
3
Endpoint
9.5
2
11
9
12.5
3
14
14
15.5
4
17
10
18.5
5
20
7
21.5
6
23
0
24.5
7
26
2
2
2
S = 2cfw_S s / nr + S e / r
d
2
2
Having estimates of S s and S e allows the experimenter to see how changes in n and r will effect
L
2
the magnitude of S d or the confidence interval for a mean difference ( U = d t / 2Sd . For instance S s
2
is rel
SUMMARY
1.
The chisquare distribution can be used to test the goodness of fit of data to a
hypothesized model.
2 =
( 0i E i ) 2
Ei
Where Oi is the observed frequency and Ei is the expected frequency from the
hypothesized model. Assume observations are c
Step 7: Sum of squares and mean square for sampling error.
SSE =
=
MSE =
=
TSS  SSU
62.65  55.71 = 6.94
SSS/kr(n1)
6.94/6(5) (1) = 0.23
Step 8: Calculate F values.
For testing the hypothesis of equal treatment means,
F = MST/MSE = 6.99/0.86 = 8.13
The
basis of the marginal total the expected entry in the upper lefthand corner would be (24)
(10)/40 = 6. After this entry or any other has been calculated, all the remaining entries can be
obtained by subtraction from the marginal totals. Since only one v
degrees of freedom equal to 29, one less than the total number of plots. There are six nitrogen
levels, therefore, there are 5 df for treatments. There are 5 replications per treatment, therefore
there are 4 df per treatment and with 6 treatments, there a
8.
The degrees of freedom for 2 depends on the number of parameters that must be
estimated for computing the expected frequencies. In this case, we have 9 classes,
therefore there are 8 df for classes. This is further reduced by a degree of freedom for
me
2
2
2
3) What are the interpretations of F = MSE/MSS? Since MSE s + n s , and MSS estimates s ,
thus F = MSE/MSS = (MSE/n)/(MSS/n)
2
2
2
( s / n + e ) / s / n
2
2
can be used to test whether e = 0. If F is significantly greater than 1, it implies e > 0 a
2
Table 76 shows how S y changes for the AOV of Table 75 as we change n and r.
2
Table 76. S y for various combinations of n and r, where
2
2
2
S s = 0.31, S e = 0.23 and S y = 0.23 / rn + 0.31 / r
n\r
1
2
3
4
5
6
7
1
.54
.27
.18
.14
.11
.09
.08
2
.43
Has inoculation been effective? (5% level) (No, since adj. 2 = 2.82)
10.
It is suspected that different combinations of temperature and humidity affect the number
of defective articles produced in a certain workroom. Do the following data confirm this
sus
The symbol for a general variate in this twoway table is Yij, where i represents a row from 1 to k and j
represents a column from 1 to r. For example, the variate in row 2 and column 3 is denoted by Y23.
The dot subscript indicates an operation over all
Products
A
C
10
6
5
8
7
3
9
6
4
7
6
2
6
8.
B
5
1
The following are the amounts of corn in bushels obtained from equalsize plots planted to 4
different varieties. Using the 1% level analyze the results for significant differences among
variety mean yields
Table A2. Binomial probabilities [P(r) for values of r]: each value in the table represents the probability
of r occurrences in n trials with single occurrence probability equal to p.
p
n
r
.05
.10
.15
.20
.25
.30
.35
.40
.45
1
0
.950
.900
.850
.800
.750
Y22 + 2aY2 + b
Y32 + 2aY3 + b

Yr2 + 2aYr + b
= Yj2 + 2a Yj + rb
From this example the following rules can be stated:
1.
The summation of a constant is r multiplied by the constant.
2.
The summation of a constant multiplied by a variable is the same as t
Sampling
error
kr(n1)
2
2
ijk ij. / n
SSS/kr(n1)
When subsamples are taken, experimental error consists of two sources of variation, variation among
2
2
2
2
subsamples, s , and variation among units, e , treated alike, i.e., MSE is an estimate of s + n
Chapter 1.
1.1
STATISTICAL NOTATION AND ORGANIZATION
Summation Notation for a OneWay Classification.
In statistical computations it is desirable to have a simplified system of notation to avoid
complicated formulas describing mathematical operations.
A m
SUMMARY
1. The analysis of variance is the most powerful technique to analyze welldesigned experiments. The
completely randomized design is the simplest experimental design. The data obtained from such
designs can be analyzed by the oneway analysis of v
EXERCISES
1.
If the percent germination (probability of germination) of a certain seed is listed as
90%, how many plants would one expect to obtain out of 500 seeds planted? (450)
Answer: Let X be the number
2.
If the probability that it will rain on any
13.
A pastry chef wishes to determine whether the proportion of unsatisfactory bear claws is
affected by oven temperature. He has baked batches of them at 350 , 400 , and 450 , and
has obtained the following results:
350
Number satisfactory
Number unsatis
Table 111.
Observed and expected frequencies to test the goodness of fit of percent sucrose
values to a normal distribution.
Class
Midpoint
Yi
Endpoint
Oi
Obs.
freq.
1
2
3
4
5
6
7
8
9
4.8
63
7.8
9.3
10.8
12.3
13.8
15.3
16.8
5.6
7.1
8.6
10.1
11.6
13.1
14.
$
31 = 40.9  39.6 = 1.3
the sum of the squares of error, i.e.
7.2 Completely Randomized Design With Subsamples
As already discussed, the experimental unit is the unit of research material to which a treatment is
applied. In many experiments, it is commo
13. In determining the relationship of plant growth rate (absolute ratio of dry matter
accumulation per plant) and growth rate of the plant tissue (leaf area growth rate), the
following data were obtained in the time interval from 22 to 29 days after plan
yields in bushels are shown. On the basis of these data, is variety A superior to variety B in yield
at the 5% level of significance?
(t=2.35, df=4)
Block
1
Standard
Variety
50
New
Variety
44
2
45
42
3
60
55
4
55
57
5
60
52
13.
In an experiment testing th