stats quiz 6
Consider the following ANOVA table. Sourceof Variation Sumof Squares Degreesof Freedom MeanSquare F Between Treatments 2073.6 4 Between Blocks 6000 5 1200 Error 20 288 Total 29 The mean square due to treatments equals
518.4
Consider the following ANOVA table. Sourceof Variation Sumof Squares Degreesof Freedom MeanSquare F Between Treatments 2073.6 4 Between Blocks 6000 5 1200 Error 20 288 Total 29 The sum of squares due to error equals
5760
The ANOVA procedure is a statistical approach for determining whether or not the means of
three or more populations are equal.
In the ANOVA, treatments refer to
different levels of a factor
Part of an ANOVA table is shown below. Source ofVariation Sum ofSquares Degreesof Freedom MeanSquare F Between Treatments 64 8 Within Treatments (Error) 2 Total 100 The mean square due to treatments (MSTR) is
16
The critical F value with 6 numerator and 60 denominator degrees of freedom at α = .05 is
2.25
Part of an ANOVA table is shown below. Source ofVariation Sum ofSquares Degrees ofFreedom MeanSquare F Between Treatments 180 3 Within Treatments (Error)TOTAL 480 18 The mean square due to error (MSE) is
20
Part of an ANOVA table is shown below. Source ofVariation Sum ofSquares Degreesof Freedom MeanSquare F Between Treatments 64 8 Within Treatments (Error) 2 Total 100 The number of degrees of freedom corresponding to between-treatments is
4
In an analysis of variance where the total sample size for the experiment is nT and the number of populations is k, the mean square due to error is
SSE/(nT - k)
The mean square is the sum of squares divided by
its corresponding degrees of freedom
Consider the following information. SSTR = 6750H0: μ1 = μ2 = μ3 = μ4SSE = 8000Ha: At least one mean is different If n = 5, the mean square due to error (MSE) equals
500
The F ratio in a completely randomized ANOVA is given by
MSTR/MSE
The critical F value with 8 numerator and 29 denominator degrees of freedom at α = .01 is
3.20
Part of an ANOVA table is shown below. Source ofVariation Sum ofSquares Degrees ofFreedom MeanSquare F Between Treatments 180 3 Within Treatments (Error)TOTAL 480 18 The test statistic is
3.00
To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been randomly assigned to the 3 treatments. You are given the results below. Treatment Observations A 20 30 25 33 B 22 26 20 28 C 40 30 28 22 The mean square due to error (MSE) equals
34
To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been randomly assigned to the 3 treatments. You are given the results below. Treatment Observations A 20 30 25 33 B 22 26 20 28 C 40 30 28 22 The mean square due to treatments (MSTR) equals
36
In an analysis of variance problem if SST = 120 and SSTR = 80, then SSE is
40
An ANOVA procedure is applied to data obtained from 6 samples where each sample contains 20 observations. The critical value of F occurs with
5 numerator and 114 denominator degrees of freedom
When an analysis of variance is performed on samples drawn from k populations, the mean square due to treatments (MSTR) is
SSTR/(k - 1)
Consider the following ANOVA table. Sourceof Variation Sumof Squares Degreesof Freedom MeanSquare F Between Treatments 2073.6 4 Between Blocks 6000 5 1200 Error 20 288 Total 29 The test statistic to test the null hypothesis equals
1.8
Part of an ANOVA table is shown below. Source ofVariation Sum ofSquares Degreesof Freedom MeanSquare F Between Treatments 64 8 Within Treatments (Error) 2 Total 100 If we want to determine whether or not the means of the populations are equal, the p-value is
less than .01
To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been randomly assigned to the 3 treatments. You are given the results below. Treatment Observations A 20 30 25 33 B 22 26 20 28 C 40 30 28 22 The test statistic to test the null hypothesis equals
1.06
An ANOVA procedure is used for data obtained from five populations. Five samples, each comprised of 20 observations, were taken from the five populations. The numerator and denominator (respectively) degrees of freedom for the critical value of F are
4 and 95
An experimental design where the experimental units are randomly assigned to the treatments is known as
completely randomized design
To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been randomly assigned to the 3 treatments. You are given the results below. Treatment Observations A 20 30 25 33 B 22 26 20 28 C 40 30 28 22 The null hypothesis for this ANOVA problem is
μ1 = μ2 = μ3
Consider the following ANOVA table. Sourceof Variation Sumof Squares Degreesof Freedom MeanSquare F Between Treatments 2073.6 4 Between Blocks 6000 5 1200 Error 20 288 Total 29 The null hypothesis for this ANOVA problem is
μ1 = μ2 = μ3 = μ4 = μ5
In an analysis of variance problem involving 3 treatments and 10 observations per treatment, SSE = 399.6. The MSE for this situation is
14.8
Consider the following information. SSTR = 6750H0: μ1 = μ2 = μ3 = μ4SSE = 8000Ha: At least one mean is different The mean square due to treatments (MSTR) equals
2250
An ANOVA procedure is used for data obtained from four populations. Four samples, each comprised of 30 observations, were taken from the four populations. The numerator and denominator (respectively) degrees of freedom for the critical value of F are
3 and 116
An ANOVA procedure is used for data that was obtained from four sample groups each comprised of five observations. The degrees of freedom for the critical value of F are
3 and 16
Consider the following information. SSTR = 6750H0: μ1 = μ2 = μ3 = μ4SSE = 8000Ha: At least one mean is different If n = 5, the test statistic to test the null hypothesis equals
4.50
Part of an ANOVA table is shown below. Source ofVariation Sum ofSquares Degrees ofFreedom MeanSquare F Between Treatments 180 3 Within Treatments (Error)TOTAL 480 18 The mean square due to treatments (MSTR) is
60
Consider the following information. SSTR = 6750H0: μ1 = μ2 = μ3 = μ4SSE = 8000Ha: At least one mean is different The null hypothesis is to be tested at the 5% level of significance ( n = 5). The p-value is
between .01 and .025
To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been randomly assigned to the 3 treatments. You are given the results below. Treatment Observations A 20 30 25 33 B 22 26 20 28 C 40 30 28 22 The null hypothesis is to be tested at the 1% level of significance. The p-value is
greater than .1
Consider the following ANOVA table. Sourceof Variation Sumof Squares Degreesof Freedom MeanSquare F Between Treatments 2073.6 4 Between Blocks 6000 5 1200 Error 20 288 Total 29 The null hypothesis is to be tested at the 5% level of significance. The p-value is
greater than .10
Part of an ANOVA table is shown below. Source ofVariation Sum ofSquares Degreesof Freedom MeanSquare F Between Treatments 64 8 Within Treatments (Error) 2 Total 100 At a 5% level of significance, if we want to determine whether or not the means of the populations are equal, the conclusion of the test is that
not all means are equal
Consider the following information. SSTR = 6750H0: μ1 = μ2 = μ3 = μ4SSE = 8000Ha: At least one mean is different The null hypothesis is to be tested at the 5% level of significance (n=5). The null hypothesis
should be rejected
Consider the following ANOVA table. Sourceof Variation Sumof Squares Degreesof Freedom MeanSquare F Between Treatments 2073.6 4 Between Blocks 6000 5 1200 Error 20 288 Total 29 The null hypothesis is to be tested at the 5% level of significance. The null hypothesis
should not be rejected
To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been randomly assigned to the 3 treatments. You are given the results below. Treatment Observations A 20 30 25 33 B 22 26 20 28 C 40 30 28 22 The null hypothesis is to be tested at the 1% level of significance. The null hypothesis
should not be rejected
In a completely randomized experimental design involving five treatments, 13 observations were recorded for each of the five treatments (a total of 65 observations). Also, the design provided the following information. SSTR = 200 (Sum of Squares Due to Treatments)SST = 800 (Total Sum of Squares) The sum of squares due to error (SSE) is
600
