Question

State 6. The GSS 2014 he GSS 2014 asked respondents to identify what was important for truly being merican." When asked "How important to have American ancestry?", answers were measured on a 4-point scale: 1 = very important, 2 fairly important, 3 not very important, and 4 = not important at all. We selected a sample of 20 GSS respondents and present their individual responses. Complete the five-step model for these data, using alpha = .01 to assess the significance of the model. portant to what was im in scale: 1 ht, and of important at all. Wortant, 2 = fairly imporsty answers were White Black Other

Answer 1

ANSWER:

Ert = Z1 = IH : H H, : Atleast two 4; differ х X2 X3 9. 16 4 4 4 4 4 4 4 4 3 3 4 16 4 4 16 4 ΣΧ ΣΧ Σ ΣΧ3 Σ ΣΧ) 20 62 11 19 15 41

k = 3 (Number of groups or treatment Conditions) n = 20 (Number of data points.) %3D (ΣΧ+. +Σ x)2 Ση SSTotal = (Exỉ +...+Ex3) - ... %3D (20 + 11+ 15)² (62 + 19+ 41) %3D 20 = 122 – 105.8 = 16.2 SSTotal 16.2 %3D (Σx)? (ΣΧ+. +Σx)2 Ση (Σx)2 ) - Ext+E»}* SSAmong ... %3D n3 ni 20 20 20 – 105.8 = 111.928571 – 105.8 = 6.128571 %3D %3D SSAmong = 6.1286 %3D SSwithin = SSTotal – SSAmong = 16.2 – 6.128571 = 10.071429 %3D %3D SSwithin = 10.0714

Significance level = 0.01 MSAmong MSwithin %3D 6.128571/2 10.071429/17 SSAmong /k – 1 SSwithin/n - 3 Test statistic, F 3.064286 2 5.172 0.592437 Test statistic, F = 5.172 formulas used in ANOVA table (for demonstration) MS SS df Source MSAmong MSwithin SSAmong MSAmong SSAmong k - 1 Among k - 1 SSwithin MSwiahin SSwithin Within SSTotal Total п- 1

ANOVA Source df MS 6.128571 3.064286 5.172 Among 2 Within 10.071429 0.592437 17 Total 19 16.2

Degrees of freedom for Numerator, d fN = k – 1 = 2 Degrees of freedom for Denominator, dfp = n- k = 17 To get F-critical-value, we can use excel fucntion F.INV.RT (0.01,2,17) 2 6.112 Rejection region is to the right of critical value, i.e., we reject null hypothesis if test statistic is greater than 6.112 To get p-value, we can use excel fucntion F.DIST.RT (5.172 340,2,17) = 0.017595 294 x 0.0176 %3D Since test statistic is not greater than critical value, we fail to reject null hypoth- esis. At 1% level, there is NOT sufficient evidence that atleast one of the means differs from the rest.