# Analysis Of Variance

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Dmmarathe
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Quizzes Created: 3 | Total Attempts: 16,713
Questions: 15 | Attempts: 6,552  Settings  This test is on the topic Analysis of Variance, subject Quantitative Models in HR.
It is a multiple- choice test.
Questions: 15 questions
Marks: 30
Time: 45 mins
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• 1.

### The number of rows in which total variance in a one way analysis of variance partitioned is:

• A.

2

• B.

3

• C.

4

• D.

None of the above

A. 2
Explanation
In a one-way analysis of variance (ANOVA), the total variance is partitioned into two components: the between-group variance and the within-group variance. The between-group variance represents the differences between the group means, while the within-group variance represents the variability within each group. Therefore, the correct answer is 2, as the total variance is partitioned into two components.

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• 2.

### The number of rows in which total variance in a two way analysis of variance partitioned is:

• A.

2

• B.

3

• C.

4

• D.

None of the above

B. 3
Explanation
In a two-way analysis of variance, the total variance is partitioned into three components: the variance due to the main effect of the first factor, the variance due to the main effect of the second factor, and the variance due to the interaction between the two factors. Therefore, the correct answer is 3.

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• 3.

### Any difference among the population means in the analysis of variance will inflate the expected value of

• A.

SSE

• B.

MS Columns

• C.

MSE

• D.

All of the above

B. MS Columns
Explanation
In the analysis of variance, the expected value of MS Columns will be inflated if there is any difference among the population means. This is because MS Columns is calculated by dividing the sum of squares of the columns by the degrees of freedom. If there are differences among the population means, it means that the sum of squares of the columns will be larger, resulting in a higher expected value of MS Columns. Therefore, any difference among the population means will inflate the expected value of MS Columns.

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• 4.

### If data in a two way classification is displayed in r rows and k columns, then the degrees of freedom for interaction will be

• A.

R-1

• B.

k-1

• C.

R(k-1)

• D.

(r-1)(k-1)

D. (r-1)(k-1)
Explanation
In a two-way classification, the degrees of freedom for interaction can be calculated by multiplying the degrees of freedom for the rows (r-1) with the degrees of freedom for the columns (k-1). Therefore, the correct answer is (r-1)(k-1).

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• 5.

### The degrees of freedom for k columns of size n will be

• A.

k-1

• B.

n-1

• C.

nk-1

• D.

none of these

A. k-1
Explanation
The degrees of freedom for k columns of size n will be k-1. This is because in a contingency table with k columns, the last column can be determined by the values in the first k-1 columns. Once the values in the first k-1 columns are known, the last column is fixed and does not have any freedom to vary. Therefore, the degrees of freedom for the last column is k-1.

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• 6.

### The degrees of freedom associated with the denominator (r row & k column) of  F-test in the analysis of variance one way are

• A.

k(r-1)

• B.

n(k-1)

• C.

nk-1

• D.

None of these

A. k(r-1)
Explanation
The degrees of freedom associated with the denominator in the F-test in the analysis of variance one way are determined by the number of columns (k) and the number of groups or treatments (r) minus 1. This is because the denominator degrees of freedom represent the variability within the groups, which is calculated by subtracting the overall mean from each individual value and summing the squared differences. Since there are r groups, each with k observations, there are r*k total observations. However, one degree of freedom is lost when calculating the overall mean, resulting in k(r-1) degrees of freedom for the denominator.

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• 7.

### The sum of squares due to column factor for one way analysis of variance is given by:

• A.

SST+ SSE

• B.

SSE-SST

• C.

SST-SSE

• D.

None of these

C. SST-SSE
Explanation
In one way analysis of variance, the sum of squares due to the column factor represents the variation between the column means. SST (sum of squares total) represents the total variation in the data, while SSE (sum of squares error) represents the variation within each column. Therefore, the correct answer is SST-SSE, as it calculates the variation between the column means by subtracting the within-column variation from the total variation.

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• 8.

### The error sum of squares in a one way analysis of variance is given by

• A.

∑∑ (xij - x j )2

• B.

∑∑X2 ij – n (x)2

• C.

∑∑ (xij – x) 2

• D.

None of these

D. None of these
• 9.

### The error sum of squares can be obtained from the equation in 2 way ANOVA

• A.

SSE = SST+ SSR + SSC

• B.

SSE= SSR + SSC – SST

• C.

SSE= SST – SSR – SSC

• D.

None of these

C. SSE= SST – SSR – SSC
Explanation
The equation SSE = SST - SSR - SSC is the correct answer because it correctly represents the calculation of the error sum of squares in a two-way ANOVA. In this equation, SSE represents the sum of squares of errors, SST represents the total sum of squares, SSR represents the sum of squares due to regression, and SSC represents the sum of squares due to the interaction between the two factors. By subtracting SSR and SSC from SST, we can obtain the error sum of squares.

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• 10.

### The degrees of freedom for the error sum of squares in 2 way ANOVA

• A.

(k-1)(r-1)

• B.

Rk-r -k

• C.

rk-r -k+1

• D.

None

C. rk-r -k+1
Explanation
The correct answer is rk-r -k+1. In a 2-way ANOVA, the degrees of freedom for the error sum of squares is calculated by subtracting the degrees of freedom for the main effects and the interaction from the total degrees of freedom. The formula for the error degrees of freedom is (r-1)(k-1), where r is the number of levels in one factor and k is the number of levels in the other factor. Therefore, the correct answer is rk-r -k+1.

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• 11.

### Which of these distributions is used for accepting Ho

• A.

Binominal

• B.

Chi- Square

• C.

Poisson

• D.

None of these

D. None of these
Explanation
None of these distributions is used for accepting Ho (null hypothesis). The correct test for accepting or rejecting a null hypothesis depends on the specific situation and the type of data being analyzed. Common tests for accepting Ho include t-tests, F-tests, and z-tests, among others. Therefore, the answer "none of these" is correct as none of the given distributions are specifically used for accepting Ho.

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• 12.

### To test equality of means of more than 2 populations which of the following techniques is used

• A.

Interval estimate

• B.

Analysis of variance

• C.

Chi-square test

• D.

None of these

B. Analysis of variance
Explanation
Analysis of variance (ANOVA) is used to test the equality of means of more than two populations. ANOVA compares the variation between groups with the variation within groups to determine if there is a significant difference in means. It is a statistical technique that allows for the comparison of multiple groups simultaneously. Interval estimate is used to estimate a population parameter, while the chi-square test is used for testing the independence of categorical variables. Therefore, the correct technique to test equality of means of more than 2 populations is analysis of variance.

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• 13.

### Which of the following assumptions of ANOVA can be discarded in case the sample size is large

• A.

Each population has equal variance

• B.

Samples are drawn from a normal population

• C.

Both(a ) and (b)

• D.

none of these

B. Samples are drawn from a normal population
Explanation
In case the sample size is large, the assumption that the samples are drawn from a normal population can be discarded. This is because, according to the Central Limit Theorem, as the sample size increases, the sampling distribution of the mean approaches a normal distribution regardless of the shape of the population distribution. Therefore, even if the population is not normally distributed, the assumption of normality can be relaxed when the sample size is large.

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• 14.

### Test statistic for equality of r population means is

• A.

MSC / MSE

• B.

MSC/ MST

• C.

SSC / SST

• D.

None of these

A. MSC / MSE
Explanation
The test statistic for equality of r population means is calculated by dividing the mean square between groups (MSC) by the mean square error (MSE). This ratio helps determine if there is a significant difference between the means of the different groups. Therefore, the correct answer is MSC / MSE.

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• 15.

### One way ANOVA was conducted for 4 columns with 6 elements in each column. What is decision?Due to column factor  Source of Veriation SS df MS F Calc Critical Due to column factor 33       3.098 Error           Total 145

• A.

Accept + Ho

• B.

Reject Ho

• C.

Test is inappropriate

• D. Back to top