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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: Myouths young Adults Adults Seniors H: at least two of the four ,'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 Degrees of Freedom The given summary statistics are below. Age Group Youths k-1 N-k Young Sum of Squares Mean Square 55Tr SSE Adults Seniors SSTr k-1 = MST SSE NE MSE N-K = F MSTr MSE F ==

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: Myouths young Adults Adults Seniors H: at least two of the four ,'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 Degrees of Freedom The given summary statistics are below. Age Group Youths k-1 N-k Young Sum of Squares Mean Square 55Tr SSE Adults Seniors SSTr k-1 = MST SSE NE MSE N-K = F MSTr MSE F ==

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section: Chapter Questions
Problem 25SGR
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← ✔ All Black - Full Dark Them X WLE3 - MATH142-2_E01_20 X webassign.net/web/Student/Assignment-Responses/tutorial?pos=7&dep=31365371&tags=autosave#question4526942_7 with the same standard deviation. Is there evidence that the mean social marginality scores are not the same for all four age groups? Test the relevant hypotheses using a = 0.01. Step 1 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 Myouths young Adults Adults Seniors Source of Variation Treatments 83 F Cloudy H₂: at least two of the four ,'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. Error The given summary statistics are below. Age Group Youths Sample Size x 5 LTI Launch 328935761_48874....jpg 117 2.00 1.46 Degrees of Freedom k-1 N-k Young Adults 242 3.50 1.76 Sum of Squares 302 Adults Seniors 3.18 55Tr 1.59 55E 42 2.73 1.97 Mean Square SST=MSTR k-1 SSE N-K = MSC ▬▬▬▬ F= Facebook F 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 = 117 +242 + 302 +42 = Thus, the degrees of freedom for Error is N-k= Search MSTr MSE [Solved] Engineering Data X e = =JO 61 ES degrees of freedom for Treatments. Show all 5:38 PM 2/2/2023 x X
Transcribed Image Text:← ✔ All Black - Full Dark Them X WLE3 - MATH142-2_E01_20 X webassign.net/web/Student/Assignment-Responses/tutorial?pos=7&dep=31365371&tags=autosave#question4526942_7 with the same standard deviation. Is there evidence that the mean social marginality scores are not the same for all four age groups? Test the relevant hypotheses using a = 0.01. Step 1 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 Myouths young Adults Adults Seniors Source of Variation Treatments 83 F Cloudy H₂: at least two of the four ,'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. Error The given summary statistics are below. Age Group Youths Sample Size x 5 LTI Launch 328935761_48874....jpg 117 2.00 1.46 Degrees of Freedom k-1 N-k Young Adults 242 3.50 1.76 Sum of Squares 302 Adults Seniors 3.18 55Tr 1.59 55E 42 2.73 1.97 Mean Square SST=MSTR k-1 SSE N-K = MSC ▬▬▬▬ F= Facebook F 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 = 117 +242 + 302 +42 = Thus, the degrees of freedom for Error is N-k= Search MSTr MSE [Solved] Engineering Data X e = =JO 61 ES degrees of freedom for Treatments. Show all 5:38 PM 2/2/2023 x X
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