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