onsider the following partially completed one-way analysis of variance table:
Source of variation
SS
df
MS
F
Treatments
583.2
(b)
(c)
(e)
Errors
(a)
15
(d)
Total
714.0
20
The missing value indicated by (d) is: _______
130.8
116.64
5
8.72
To find the missing value indicated by (d) in the table, we can use the formula:
MS = SS / df
From the given information, we can calculate MS for "Errors" as:
MS(Error) = SS(Error) / df(Error)
MS(Error) = (Total SS - Treatment SS) / df(Error)
MS(Error) = (714.0 - 583.2) / 15
MS(Error) = 130.8 / 15
MS(Error) = 8.72
Therefore, the missing value indicated by (d) is "8.72".
To find the missing value indicated by (d), we can use the formula:
SS = MS * df
First, we need to find the value of MS (Mean Square) for the Treatments and Errors using the available information.
For Treatments:
MS = SS / df = 583.2 / b
For Errors:
MS = SS / df = a / 15
Next, we can use the given information to write the equation:
SS (Treatments) + SS (Errors) = Total SS
583.2 + a = 714.0
Solving for "a" gives:
a = 714.0 - 583.2 = 130.8
Now, we can substitute this value of "a" in the equation for Errors' MS:
MS (Errors) = a / 15 = 130.8 / 15 = 8.72
Therefore, the missing value indicated by (d) is 8.72.
To find the missing value indicated by (d) in the one-way analysis of variance table, we need to calculate the mean square (MS) value for errors.
The formula to calculate the mean square for errors is MS = SS / df, where SS is the sum of squares and df is the degrees of freedom.
We have the total sum of squares (SS) for errors as 714.0 and the degrees of freedom (df) for errors as 15. Using the formula, we can calculate the mean square for errors:
MS = 714.0 / 15 = 47.6
So, the missing value indicated by (d) is 47.6.
The production manager of Raylite Batteries, a car battery manufacturer, wants to know whether the three machines used for the production process (labelled A,B and C) produce equal amounts of rejects. A random sample of shifts for each machine was selected and the number of rejects produced per shift was recorded. The data are showed in table below:
Machine A
Machine B
Machine C
11
7
14
9
10
13
6
8
11
12
13
16
14
16
11
Give the df of the treatments
2
3
15
12
14
The degrees of freedom (df) for the treatments can be calculated by subtracting 1 from the number of treatments (in this case, the number of machines being compared).
In this case, there are 3 machines, so the df for the treatments is 3-1 = 2.
Therefore, the df of the treatments is 2.
A consultant to a painting contractor is evaluating three brands of exterior paints in terms of drying time (hours). The table below is obtained.
Brand of paint
Drying condition
A
B
C
1. Direct sunlight
11,3
12,4
10,9
2. Shade
14,2
14,3
13,7
3. Humid
14,9
15,2
17,7
4. Dry
12,2
11,3
12,5
Calculate: SST
0.6467
36.9166
6.2734
43.8367
To calculate the total sum of squares (SST), we need to find the sum of the squared deviations from the overall mean.
First, we calculate the overall mean by adding up all the values and dividing by the total number of observations:
Overall Mean = (11.3 + 12.4 + 10.9 + 14.2 + 14.3 + 13.7 + 14.9 + 15.2 + 17.7 + 12.2 + 11.3 + 12.5) / 12 = 144 / 12 = 12
Next, we calculate the squared deviations from the overall mean for each observation and then sum them up:
SST = (11.3 - 12)^2 + (12.4 - 12)^2 + (10.9 - 12)^2 + (14.2 - 12)^2 + (14.3 - 12)^2 + (13.7 - 12)^2 + (14.9 - 12)^2 + (15.2 - 12)^2 + (17.7 - 12)^2 + (12.2 - 12)^2 + (11.3 - 12)^2 + (12.5 - 12)^2
SST = 0.5467 + 0.1767 + 0.6467 + 2.5467 + 2.7967 + 0.0567 + 2.1967 + 3.2367 + 25.5167 + 0.0567 + 0.1767 + 0.4267
SST = 38.9365
Therefore, the total sum of squares (SST) is approximately 38.9365.
Consider the following partially completed ANOVA:
ANOVA
Source of variation
SS
df
MS
Fstat.
Ftab.
Treatments
(a)
2
85.3000
(e)
(f)
Errors
199.1334
(c)
(d)
Total
(b)
14
Give the value at (d).
42.65
16.5945
85.3
15.3180
49.7834
To find the value at (d) in the ANOVA table, we can use the formula:
MS = SS / df
From the given information, we can calculate MS for "Errors" as:
MS(Errors) = SS(Errors) / df(Errors)
MS(Errors) = 199.1334 / c
We also know that the degrees of freedom for Errors is equal to the Total degrees of freedom minus the degrees of freedom for Treatments:
df(Errors) = Total df - Treatment df
df(Errors) = b - 2
Now, we can substitute the values in the formula:
MS(Errors) = 199.1334 / (b - 2)
From the given table, we know that MS(Treatments) = 85.3000. Since the number of treatments is 2, we can calculate the SS(Treatments) as:
SS(Treatments) = MS(Treatments) * df(Treatments)
SS(Treatments) = 85.3000 * 2 = 170.6000
Also, we know that the Total SS is equal to the sum of SS(Treatments) and SS(Errors):
Total SS = SS(Treatments) + SS(Errors)
Total SS = 170.6000 + 199.1334
Total SS = 369.7334
Since the Total df is given as 14, we can substitute the values in the formula:
MS(Treatments) = SS(Treatments) / df(Treatments)
85.3000 = 170.6000 / 2
Finally, we can solve the equations to find the missing values:
(b) = Total SS = 369.7334
(c) = df(Errors) = b - 2 = 14 - 2 = 12
(d) = MS(Errors) = 199.1334 / (b - 2)
(d) = 199.1334 / 12 ≈ 16.5945
(e) = df(Treatments) = 2
(f) = Ftab. = Fstat. (we are not given the F-statistic, so we cannot determine the value)
Therefore, the value at (d) is approximately 16.5945.
Prior to submitting a bid for a construction job, cost engineering prepare a detailed analysis of the estimated labour and materials costs required to complete the job. This estimate will depend on the engineer who performs the analysis. An overly large estimate will reduce the chance of acceptance of a company’s bid price, whereas an estimate that it too low, will reduce the profit or even cause the company to lose money on the job. A company that employs three job cost engineers wanted to compare the mean level of the engineers’ estimates. This was done by having each engineer estimate the cost of the same four jobs. The data are shown in the table below.
Jobs
1
2
3
4
Engineer
1
4.6
6.2
5.0
6.6
2
4.9
6.3
5.4
6.8
3
4.4
5.9
5.4
6.3
Give the null hypothesis appropriate to test for the difference between the block means.
Both a and b
The null hypothesis appropriate to test for the difference between the block means is that there is no significant difference in the mean estimates of the cost of the jobs among the three job cost engineers.
So, the null hypothesis can be stated as:
H0: μ1 = μ2 = μ3
where μ1, μ2, and μ3 represent the mean estimates of the cost of the jobs by engineer 1, engineer 2, and engineer 3, respectively.
Therefore, the correct answer is "Both a and b"
The analysis of variance for a randomised complete block design produced the ANOVA table entries shown below.
Source
SS
Vg/Df
MS
F
Treatments
27.1
3
(e)
(g)
Blocks
(a)
5
14.90
(h)
Residuals
33.4
(c)
(f)
Total
(b)
(d)
Testing at 1% significance level, give the critical value of the blocks.
9.72
4.565
5.425
26.87
5.56
To find the critical value of the blocks, we need to consider the F-statistic at the 1% significance level for the Residuals as the F-critical value for the blocks.
In the ANOVA table, the F-statistic for the Residuals is denoted by F.
The degrees of freedom for Treatments is given as 3 (Vg/Df) and for Residuals is given as c (df).
Since we are testing at the 1% significance level, the alpha level (α) is 0.01.
Now, we can use a statistical table or calculator to find the F-critical value for the blocks.
Therefore, the critical value of the blocks is denoted as Ftab for the Residuals, which corresponds to the F-statistic at the 1% significance level and with (c) degrees of freedom in the numerator and (f) degrees of freedom in the denominator.
Since the exact values for (c) and (f) are not given, we cannot determine the specific value of the critical value of the blocks without additional information or calculations.
None of the given options match the critical value of the blocks at the 1% significance level.
Prior to submitting a bid for a construction job, cost engineering prepare a detailed analysis of the estimated labour and materials costs required to complete the job. This estimate will depend on the engineer who performs the analysis. An overly large estimate will reduce the chance of acceptance of a company’s bid price, whereas an estimate that it too low, will reduce the profit or even cause the company to lose money on the job. A company that employs three job cost engineers wanted to compare the mean level of the engineers’ estimates. This was done by having each engineer estimate the cost of the same four jobs. The data are shown in the table below.
Jobs
1
2
3
4
Engineer
1
4.6
6.2
5.0
6.6
2
4.9
6.3
5.4
6.8
3
4.4
5.9
5.4
6.3
How many treatments are involved in this experiment?
2
4
3
12
None of the above
In this experiment, the cost estimates for the jobs are compared among the three job cost engineers. Each engineer represents a treatment in the experiment.
Therefore, there are 3 treatments involved in this experiment.
Consider the following partially completed one-way analysis of variance table:
ANOVA
Source of variation
SS
vg / df
MS
F
Treatments
583.2
(b)
(c)
(e)
Errors
(a)
15
(d)
Total
714.0
20
The missing value indicated by (b) is: _______
130.8
116.64
5
8.72
To find the missing value indicated by (b) in the ANOVA table, we can use the formula:
MS = SS / df
From the given information, we can calculate MS for "Treatments" as:
MS(Treatments) = SS(Treatments) / df(Treatments)
MS(Treatments) = 583.2 / b
We also know that the degrees of freedom for Errors is given as 15.
Now, we can substitute the values in the formula:
MS(Treatments) = 583.2 / b
Given that the Total SS is 714.0 and the Total degrees of freedom is 20, we can calculate the SS for Errors as:
SS(Errors) = Total SS - SS(Treatments)
SS(Errors) = 714.0 - 583.2
SS(Errors) = 130.8
Since SS(Errors) = MS(Errors) * df(Errors), we can calculate MS(Errors) as:
MS(Errors) = SS(Errors) / df(Errors)
MS(Errors) = 130.8 / 15
MS(Errors) = 8.72
Therefore, the missing value indicated by (b) is 8.72.
In a study on the effect of 4 fertilizers on varieties of wheat over a period of 4 years the following table is obtained which shows the yields for the 4 varieties of wheat measured in kilograms per area.
Fertilizer
Year
1998
1999
2000
2001
F1
70
75
68
81
F2
66
59
55
63
F3
59
66
39
42
F4
41
57
39
55
Determine:
2 499.9375
55 056.75
57 139
56 196.25
54 639.0625
To determine the grand mean yield, we need to sum all the yields for the 4 varieties of wheat over the 4 years, and then divide by the total number of observations.
Sum of yields = 70 + 75 + 68 + 81 + 66 + 59 + 55 + 63 + 59 + 66 + 39 + 42 + 41 + 57 + 39 + 55
Sum of yields = 895
Total number of observations = 4 varieties x 4 years = 16
Grand mean yield = Sum of yields / Total number of observations
Grand mean yield = 895 / 16
Grand mean yield = 55.9375
Therefore, the grand mean yield is 55.9375. However, none of the given options match this value.