Mean Degrees of Sum of Square Consider an experiment with Source Freedom Squares (Variance) F four groups, with five values in Among C-1 =? SSA=? MSA=95 FSTAT =? each. For the ANOVA summary groups table shown to the right, fill in all Within n - c=? SSW = 304 MSW =? the missing results. groups Total n-1=? SST=? . . . Complete the ANOVA summary table below. Degrees of Sum of Mean Square Source Freedom Squares (Variance) F Among groups C - 1 = SSA = MSA = 95 FSTAT Within groups n - C= SSW = 304 MSW = Total n - 1 = SST = (Simplify your answers.)An experiment has a single factor with six groups and ve values in each group. In determining the amonggroup variation, there are 5 degrees of freedom. In determining the within-group variation, there are 24 degrees of freedom. In determining the total variation, there are 29 degrees of freedom. Also. note that SSA = 125, SSW = 120, SST = 245, MSA = 25, MSW = 5. and FSTAT = 5. Complete parts (a) through (d). Click here to view p_age 1 of the F table. Click here to view page 2 of the F table. Click here to view [age 3 of the F table. Click here to view p_age 4 of the F table. E) a. Construct the ANOVA summary table and ll in all values in the table. Degrees of Sum of Mean Square Source Freedom Squares (Variance) F Among groups B D D D Within groups B D _ Q Total D D (Simplify your answers.) The table available below shows three samples obtained independently from three populations. a. Conduct a one-way analysis of variance on the data. Use a = 0.05. b. If warranted, use the Tukey-Kramer procedure to determine which populations have different means. Use an experiment-wide error rate of 0.05. g Click the icon to view the sample data. 3 Click the icon to view a table of q-values. a. What are the appropriate hypotheses for the test? 0 A- Ho: #1 =l12=l13 H A: All of the population means are different. 0 3- Ho: H1P2P3 H A: At least two of the population means are equal. 0 C. Ho5l11=l12=113 H A: At least two of the population means are different. 0 9- Ho5l11l~12l~13 HA: All of the population means are equal. State the decision rule. Select the correct choice below and fill in any answer boxes to complete your choice. (Type an integer or a decimal.) O A. If p-value reject the null hypothesis. Otherwise, do not reject. O B. If p-value , reject the null hypothesis. Otherwise, do not reject. Determine the value of the test statistic. FSTAT = (Type an integer or decimal rounded to three decimal places as needed) Determine the value of the p-value. p-value = (Type an integer or decimal rounded to three decimal places as needed) Reach a decision. the null hypothesis. There is evidence that of the population means areb. Which populations, if any, have different means? Comparison Significant Difference Group 1 and Group 2 Group 1 and Group 3 Group 2 and Group 3\fAs part of an initial investigation exploring foreign market entry, 10 countries were selected from each of four global regions. The cost associated with importing a standardized cargo of goods by sea transport in these countries (in US$ per container) is stored in the accompanying data table. Complete (a) through (d) below. Click on the icon to view the data table. . . . a. At the 0.05 level of significance, is there evidence of a difference in the mean cost of importing across the four global regions? Determine the hypotheses. Choose the correct answer below. OA. Ho: My = 12= . . . = H10 OB. Ho: My = H2= . . . = HA H1: Not all u; are equal H: Not all u; are equal (where j = 1,2,..., 10) (where j = 1,2,...,4) OC. Ho: My = H2= . . . . . =H4 OD. Ho: My = 12= . . . = H10 Hy: My = Hy # . . . # HA Hy: H F U , # . . . F H10 Find the test statistic. The test statistic is = (Round to two decimal places as needed.) Determine the p-value. p-value = (Round to three decimal places as needed.)b. The critical value is D (Round to three decimal places as needed.) The degrees of freedom of the numerator (Source: Among Groups) is D The degrees of freedom of the denominator (Source: Within Groups) is D Costs of importing a standardized cargo of goods (-35) a Region 1 Region 2 Region 3 Region 4 899 2307 2261 1320 607 1626 1606 1000 564 898 2269 750 655 977 962 3400 422 1376 2835 1325 654 911 1022 1200 435 1494 1550 1313 728 2910 1023 1023 743 1227 883 1044 603 2161 2872 1635 An experiment is designed to compare ve different advertisements for a pen. AdvertisementA greatly undersells the pen. Advertisement B slightly undersells the pen. Advertisement C slightly oversells the pen. Advertisement D greatly oversells the pen. Advertisement E correctly states the pen. A sample of 30 adult respondents is randomly assigned to the ve advertisements. After reading the advertisement, all respondents unknowingly receive the same pen to evaluate and are then asked to rate the pen from 1 to 7 (lowest to highest) on three characteristics. The combined scores of three ratings for the 30 respondents are in the data table. Complete parts (a) through (d) below. a Click on the icon to view the data table. <:> a. At the 0.01 level of signicance, is there evidence of a difference in the mean rating of the pens following exposure to ve advertisements? Determine the hypotheses. Choose the correct answer below. OA- H0:|.l1=l.l2="'=|15 OB. H0:|.l1=|.l.2="'=l.l6 H1: Not all the means are equal. H1: p1p2 . - . gaps OC.H0:|11=IJ.2="'=|.I5 OD_ H0:p1=l_12=ooo=l_l6 H1: p1p2 - - - if p5 H1: Not all the means are equal. Find the test statistic. b. The test statistic is '1 (Round to two decimal places as needed.) The critical value is El (Round to two decimal places as needed.) The degrees of freedom of the numerator (AMONG) = D The degrees of freedom of the denominator (WITHIN) = D Advertising Scores A B E 13 16 6 10 19 17 7 19 18 JOAAWO 20 10 16 17 17 16 14 21 19 12 19 18 19 13 12 16An experiment has a single factor with six groups and two values in each group. In determining the among-group variation, there are 5 degrees of freedom. In determining the within-group variation, there are 6 degrees of freedom. In determining the total variation, there are 11 degrees of freedom. Also, note that SSA = 90, SSW = 36, SST = 126, MSA = 18, MSW = 6, and FSTAT = 3. Complete parts (a) through (d). E) a. Construct the ANOVA summary table and ll in all values in the table. Degrees of Sum of Mean Square Source Freedom Squares (Variance) F Among groups D D D D Within groups B D D Total D B (Your answers above must all be integers .) b. At the 0.025 level of signicance. what is the upper-tail critical value from the F distribution? F0025 = D (Round to two decimal places as needed.) c. State the decision rule for testing the null hypothesis that all six groups have equal population means. cl. What is your statistical decision? Since FSTAT is El than the upper-tail critical value, |:| H0. There is El evidence to conclude there is a difference in the population means for the six groups. The ANOVA summary table for an experiment with six Degrees of Sum of Mean Square groups, with five values in each group, is shown to the Source Freedom Squares (Variance) F right. Complete parts (a) through (d) below. Among C - 1 =5 SSA = 750 MSA = 150 FSTAT = 5.00 groups Within n - c = 24 SSW = 720 MSW = 30 groups Total n - 1= 29 SST = 1470 . . . a. At the 0.05 level of significance, state the decision rule for testing the null hypothesis that all six groups have equal population means. Determine the hypotheses. Choose the correct answer below. O A. Ho: My = H2= . . . = HG OB. Ho: My = 12= . . . = H6 H: My # H 2 7 . . . # HG H,: Not all the means are equal. O C. Ho: My = H2= . . . = H5 OD. Ho: My = H2= . . . = H5 H,: Not all the means are equal. H: My # Hz # . . . # 15 Find the test statistic. FSTAT = (Round to two decimal places as needed.) Determine the critical value at the 0.05 level of significance For = (Round to two decimal places as needed.)b. What is your statistical decision? Ho. There is evidence to conclude that there is a difference in the population means of the groups