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ANOVA test, a table is helpful to organize values in calculation of the test statistic and correspondin urce of Variation Treatments Error hary statistics are below. ge Group Youths imple Size x S 118 2.00 Degrees of Freedom 1.49 k-1 N-k Young Adults Sum of Squares 245 309 3.60 3.18 1.77 1.57 SSTr SSE Adults Seniors 48 2.72 1.99 Mean Square SST k-1 SSE N-K -MSTr - MSE F= F MSTr MSE value. Recall the format a partial ANOVA table, given below. us on the column for Degrees of Freedom. The value of k is the number of populations being compared. Here, we are comparing the means of four age groups, so we have k-4 and k-1-[ s the total number of observations in the data set, so we have N-118 +245 + 309+48-[ wes of freedom for Error is N-k-[ degrees of freedom for Treatments.

ANOVA test, a table is helpful to organize values in calculation of the test statistic and correspondin urce of Variation Treatments Error hary statistics are below. ge Group Youths imple Size x S 118 2.00 Degrees of Freedom 1.49 k-1 N-k Young Adults Sum of Squares 245 309 3.60 3.18 1.77 1.57 SSTr SSE Adults Seniors 48 2.72 1.99 Mean Square SST k-1 SSE N-K -MSTr - MSE F= F MSTr MSE value. Recall the format a partial ANOVA table, given below. us on the column for Degrees of Freedom. The value of k is the number of populations being compared. Here, we are comparing the means of four age groups, so we have k-4 and k-1-[ s the total number of observations in the data set, so we have N-118 +245 + 309+48-[ wes of freedom for Error is N-k-[ degrees of freedom for Treatments.

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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A hypothesis test is to be conducted to determine if the mean social marginality scores are not the same for all four age groups. That is, the following hypotheses will be tested. Ho: Youths Young Adults Adults = μSeniors H₂: at least two of the four μ's are different Since we are told to assume that it is reasonable to regard the four samples as representative of the U.S. population in the corresponding age groups and that the distributions of social marginality scores for these four groups are approximately normal with the same standard deviation, the assumptions for the ANOVA test have been met. When performing an ANOVA test, a table is helpful to organize values in calculation of the test statistic and corresponding P-value. Recall the format of a partial ANOVA table, given below. Source of Variation Treatments Error The given summary statistics are below. Age Group Youths Sample Size x S 118 Degrees of Freedom 2.00 1.49 k-1 N-K Young Adults 245 3.60 1.77 Sum of Squares Mean Square 309 Adults Seniors 3.18 SSTr 1.57 SSE 48 2.72 1.99 SSTr k-1 SSE N-k = MSTr F = = MSE F MSTr MSE We will first focus on the column for Degrees of Freedom. The value of k is the number of populations being compared. Here, we are comparing the means of four age groups, so we have k = 4 and k - 1 = The value of N is the total number of observations in the data set, so we have N = 118 +245 + 309 + 48 = Thus, the degrees of freedom for Error is N-k= degrees of freedom for Treatments.
Transcribed Image Text:A hypothesis test is to be conducted to determine if the mean social marginality scores are not the same for all four age groups. That is, the following hypotheses will be tested. Ho: Youths Young Adults Adults = μSeniors H₂: at least two of the four μ's are different Since we are told to assume that it is reasonable to regard the four samples as representative of the U.S. population in the corresponding age groups and that the distributions of social marginality scores for these four groups are approximately normal with the same standard deviation, the assumptions for the ANOVA test have been met. When performing an ANOVA test, a table is helpful to organize values in calculation of the test statistic and corresponding P-value. Recall the format of a partial ANOVA table, given below. Source of Variation Treatments Error The given summary statistics are below. Age Group Youths Sample Size x S 118 Degrees of Freedom 2.00 1.49 k-1 N-K Young Adults 245 3.60 1.77 Sum of Squares Mean Square 309 Adults Seniors 3.18 SSTr 1.57 SSE 48 2.72 1.99 SSTr k-1 SSE N-k = MSTr F = = MSE F MSTr MSE We will first focus on the column for Degrees of Freedom. The value of k is the number of populations being compared. Here, we are comparing the means of four age groups, so we have k = 4 and k - 1 = The value of N is the total number of observations in the data set, so we have N = 118 +245 + 309 + 48 = Thus, the degrees of freedom for Error is N-k= degrees of freedom for Treatments.
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