I need this in 2 days the is an excel assignment

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Objectives:

Create 3 Pivot Tables with Analysis ChartsCreate Conditional Formatting for easy viewing of key informationApply Sort and Filters to data setPresent a professionally formatted report with custom formatting


Part 1: Pivot Tables and Charts

Using the data file:

  • DAR Assignment 3 Data3.xlsx

    120 KB


Create a total of three (3) pivot tables on separate worksheets with the following data fields:


  1. Sum of Revenue by Store and Region


  2. Sum of COGS by Store and Region


  3. Sum of Profit by Store and Region

Sample of the basic format of the pivot table (5×5 box, numbers will be different)

LINK TO PART 1 OF VIDEO: https://www.youtube.com/watch?v=rx1LeeW76AU


Design a properly formatted chart for

each

pivot table with a directed point of view chart title, axis titles, and accurate chart type.


Part 2: Conditional Formatting


Copy the data sheet provided into a new worksheet within the same workbookOn the copied worksheet, apply two different custom conditional formats in the COGS (Cost of Goods Sold) column.



Criteria to include:


  • Top 10% COGS figures as one format


  • Bottom 10% COGS figures as a second format


  • Format should be unique/customized and NOT one of the defaults set by Excel


Apply another Conditional Format in the Regions Column.



Criteria to include:


  • Region contains the word “East”


  • Format should be unique/customized and NOT one of the defaults set by Excel

LINK TO PART 2 OF THE VIDEO: https://www.youtube.com/watch?v=ydNAQzY3I_A


Apply all worksheet design guidelines to all worksheets*

*

Note:

It is important to do Currency/Accounting formatting prior to Part 3 or all of your figures may not format properly due to the filter


Part 3: Sort and Filter


Using the same worksheet that you have applied conditional formatting to:


  • Sort the data with the criteria of Revenues as

    Largest to Smallest


  • Apply Filter#1 in the Profits Column to show only the

    Top 25 Profit stores

    (show items, not %)


  • Apply Filter#2 in the Region Column to show only data in the

    East Region


Adjust sizing of rows and columns as needed to be able to read all data presented.

Note: your final sheet may show less than 25 stores in the filter due to the top 25 having other locations such as West or Central.  You will not be able to see all of your conditional formats on the screen (bottom % for example) due to the filter; however, they are still applied in the background if created correctly in Step 2.

LINK TO PART 3 VIDEO: https://www.youtube.com/watch?v=mI2vLhWkRxc

THIS IS AN EXCEL ASSIGNMENT BEFORE AGREE TO ASSIGNMENT PLEASE MAKE SURE YOU ARE GOOD IN EXCEL. THANK YOU….

PLEASE MAKE SURE YOU READ THE DIRECTIONS CAREFULLY AND LOOK AT THE VIDEOS AND THE ATTACHMENTS.

ANOVA

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See the posted sample SAS Univ_TTest_ANOVA_REG program, HBAT-tabs dataset, and SAS Univ_TTest_ANOVA_REG output, for the analysis of X3-Firm Size along with X19-Satisfaction and X20-Likelihood of Recommendation.

Now perform a similar statistical data analysis using X5-Distribution System as the Class Independent predictor variable and X19-Satisfaction and X20-Likelihood of Recommendation as the Dependent response variables, including assessment of testing the differences of X19-Satisfaction and X20-Likelihood of Recommendation across levels of X5-Distribution System.

Provide a brief 3/4-1 page summary along with the program and output as an appendix.

ANOVA
SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM] HBAT – Two-Sample T-Test ObsIDX1 X2X3X4X5 X6X7X8X9X10 X11X12X13X14X15X16X17X18X19X20X21 X22X23 1 1201118.5 3.92.55.9 4.84.96.06.84.74.35.05.13.78.28.08.465.1 1 2 2310008.2 2.75.17.2 3.47.93.15.35.54.03.94.34.95.76.57.567.1 0 3 3301119.2 3.45.65.6 5.47.45.84.56.24.65.44.04.58.98.49.072.1 1 4 4111106.4 3.37.03.7 4.74.74.58.87.03.64.34.13.04.86.07.240.1 0 5 5201019.0 3.45.24.6 2.26.04.56.86.14.54.53.53.57.16.69.057.1 0 6 6110106.5 2.83.14.1 4.04.33.78.55.19.53.64.73.34.76.36.150.1 0 7 7111106.9 3.75.02.6 2.12.35.48.94.82.52.14.22.05.77.87.241.1 0 8 8201106.2 3.33.94.8 4.63.65.16.95.44.84.36.33.76.35.87.756.1 0 9 9211105.8 3.65.16.7 3.75.95.89.35.94.44.46.14.67.07.58.256.1 1 10 10 101106.4 4.55.16.1 4.75.75.78.45.45.34.15.84.45.55.96.759.1 0 11 11 301018.7 3.24.64.8 2.76.84.66.85.87.53.83.74.07.47.08.468.1 0 12 12 101106.1 4.96.33.9 4.43.96.48.25.85.93.04.93.26.06.36.653.1 0 13 13 110019.5 5.64.66.9 5.06.96.67.66.55.35.14.54.48.48.47.958.1 1 14 14 310019.2 3.95.75.5 2.48.44.87.16.73.04.52.64.27.66.98.272.1 1 15 15 201116.3 4.54.76.9 4.56.85.98.86.05.44.86.25.28.07.07.662.1 1 16 16 300008.7 3.24.06.8 3.27.83.84.96.15.04.33.94.56.66.47.171.1 0 17 17 210115.7 4.06.76.0 3.35.55.16.26.75.44.26.24.56.47.57.250.1 1 18 18 201105.9 4.15.57.2 3.56.45.58.46.26.35.75.84.87.46.98.258.1 1 19 19 211105.6 3.45.16.4 3.75.75.69.15.46.15.06.04.56.87.57.955.1 0 20 20 301109.1 4.53.66.4 5.35.37.18.45.86.74.56.14.47.68.58.867.1 1 21 21 100105.2 3.87.15.2 3.94.35.08.47.14.63.34.93.35.45.57.050.1 0 22 22 311119.6 5.76.85.9 5.48.37.84.56.46.54.33.04.39.99.69.970.1 1 23 23 200018.6 3.67.45.1 3.57.34.73.76.76.04.83.44.07.07.18.160.1 0 24 24 301119.3 2.42.67.2 2.27.24.56.26.44.26.74.44.58.68.18.065.1 1 25 25 100106.0 4.15.34.7 3.55.35.38.06.53.94.75.34.04.84.95.555.1 0 26 26 201106.4 3.66.66.1 4.03.95.37.16.13.75.66.63.96.66.87.058.1 0 27 27 300008.5 3.07.25.8 4.17.63.74.86.96.75.33.84.46.37.17.070.1 0 28 28 110107.0 3.35.45.5 2.64.84.29.06.55.94.35.23.75.45.55.655.1 0 29 29 300008.5 3.05.76.0 2.37.63.74.85.86.05.73.84.46.36.97.270.1 0 30 30 111107.6 3.63.04.0 5.14.24.67.74.97.24.75.53.55.45.56.252.1 0 31 31 110016.9 3.48.54.3 4.56.44.75.27.73.33.72.73.36.16.87.144.1 0 32 32 101108.1 2.57.24.5 2.35.13.86.66.86.13.03.53.06.45.86.251.1 0 33 33 111106.7 3.76.55.3 5.35.14.99.25.74.23.54.53.45.46.57.644.1 0 34 34 211108.0 3.36.15.7 5.54.64.78.75.93.84.76.64.27.37.59.062.1 1 35 35 101106.7 4.05.23.9 3.05.46.88.46.26.02.54.33.56.36.66.754.1 0 36 36 100008.7 3.26.14.3 3.56.12.95.66.16.53.12.92.55.44.67.151.1 0 37 37 200019.0 3.45.94.6 3.96.04.56.86.44.33.93.53.57.18.07.257.1 0 38 38 301119.6 4.16.27.3 2.97.75.57.76.14.45.24.64.98.79.99.977.1 1 39 39 211108.2 3.63.96.2 5.84.95.09.05.27.14.76.94.57.66.97.665.1 1 40 40 100106.1 4.93.04.8 5.13.96.48.25.16.84.54.93.26.05.55.853.1 0 41 41 211108.3 3.43.35.5 3.14.65.29.14.11.74.65.83.97.07.58.461.1 1 SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM] 4242 210019.4 3.84.75.4 3.86.54.98.54.96.24.14.54.17.68.07.961.1 1 4343 301019.3 5.14.66.8 5.86.66.37.45.14.14.64.64.38.97.87.672.1 1 44 44 211115.1 5.16.66.9 4.45.47.85.97.25.24.96.34.57.67.98.455.1 1 45 45 310008.0 2.54.77.1 3.67.73.05.25.13.94.34.24.75.55.66.565.1 0 46 46 201105.9 4.15.75.9 5.86.45.58.46.45.15.25.84.87.48.67.758.1 1 47 47 3100110.0 4.37.16.3 2.95.44.53.86.73.75.04.03.57.18.88.067.1 1 48 48 211105.7 3.86.87.5 5.75.76.08.26.64.86.57.35.27.67.67.160.1 0 49 49 300119.9 3.73.76.1 4.27.06.76.85.97.24.53.43.98.78.18.567.1 1 50 50 311017.9 3.94.35.8 4.46.95.84.75.23.64.14.24.38.67.87.661.1 1 51 51 101106.7 3.65.94.2 3.44.74.87.25.75.34.03.62.85.47.57.248.1 0 52 52 310008.2 2.73.77.4 2.77.93.15.35.35.04.54.34.95.77.18.267.1 1 53 53 301119.4 2.54.86.1 3.27.34.66.36.39.24.74.64.68.79.09.066.1 1 54 54 110016.9 3.45.74.4 3.36.44.75.26.44.43.22.73.36.17.07.244.1 0 55 55 211108.0 3.33.85.8 3.24.64.78.75.34.24.96.64.27.38.18.162.1 1 56 56 310009.3 3.87.35.7 3.76.45.57.46.65.94.13.23.47.77.68.959.1 1 57 57 201117.4 5.14.87.7 4.57.26.99.66.47.45.76.55.59.07.98.874.1 1 58 58 310007.6 3.65.25.8 5.66.65.44.46.76.44.63.94.08.27.57.558.1 1 5959 3100010.0 4.35.33.7 4.25.44.53.86.74.53.74.03.57.16.57.067.1 0 60 60 311109.9 2.87.26.9 2.65.83.55.46.27.05.64.94.07.98.58.561.1 1 61 61 300008.7 3.28.46.1 2.87.83.84.97.24.55.43.94.56.66.97.271.1 1 62 62 201118.4 3.86.75.0 4.54.75.96.75.14.22.75.03.68.07.68.863.1 1 63 63 100018.8 3.93.85.1 4.34.74.85.85.07.24.43.72.96.35.58.044.1 0 64 64 101107.7 2.26.34.5 2.44.73.46.26.04.73.33.12.66.06.08.147.1 0 65 65 101106.6 3.65.84.1 4.94.74.87.26.53.93.53.62.85.46.97.148.1 0 66 66 211105.7 3.83.56.7 5.45.76.08.25.45.04.77.35.27.66.99.060.1 1 67 67 210105.7 4.07.96.4 2.75.55.16.27.56.45.06.24.56.45.66.250.1 0 68 68 210115.5 3.74.75.4 4.35.34.96.05.62.54.55.94.36.16.38.248.1 0 69 69 111107.5 3.53.83.5 2.94.14.57.65.15.24.05.43.45.25.85.851.1 0 70 70 201106.4 3.62.75.3 3.93.95.37.15.25.54.76.63.96.66.68.058.1 1 71 71 300109.1 4.56.15.9 6.35.37.18.47.15.75.46.14.47.67.57.767.1 0 72 72 110016.7 3.23.03.7 4.86.34.55.05.22.52.92.63.15.86.07.043.1 0 73 73 201106.5 4.32.76.6 6.56.36.08.74.76.34.65.64.67.96.67.966.1 0 74 74 301119.9 3.77.54.7 5.67.06.76.87.24.64.13.43.98.68.89.866.1 1 7575 201118.5 3.95.35.5 5.04.96.06.85.73.64.45.13.78.27.08.465.1 1 76 76 300009.9 3.06.85.0 5.45.94.84.97.37.63.14.33.87.16.68.963.1 1 77 77 100117.6 3.67.64.6 4.74.65.07.48.16.64.55.83.96.46.97.549.1 0 78 78 210019.4 3.87.06.2 4.76.54.98.57.32.44.34.54.17.67.38.061.1 1 79 79 300019.3 3.56.37.6 5.57.55.94.66.63.15.24.14.68.97.38.172.1 1 80 80 111107.1 3.44.94.1 4.05.05.97.86.13.52.63.12.75.75.87.644.1 0 81 81 301009.9 3.07.44.8 4.05.94.84.95.96.93.24.33.87.17.98.863.1 0 82 82 300008.7 3.26.44.9 2.46.84.66.86.35.14.33.74.07.47.38.068.1 1 83 83 200018.6 2.95.83.9 2.95.64.06.36.14.02.73.03.06.66.18.553.1 0 84 84 110106.4 3.26.73.6 2.22.95.08.47.36.52.03.71.65.05.16.537.1 0 85 85 200017.7 2.66.76.6 1.97.24.35.96.54.14.73.94.38.27.57.752.1 1 SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM] 8686 111107.5 3.54.14.5 3.54.14.57.64.92.83.45.43.45.26.07.251.1 0 8787 100105.0 3.61.33.0 3.54.24.98.24.37.62.44.83.15.25.56.048.1 0 88 88 200017.7 2.68.06.7 3.57.24.35.96.97.75.13.94.38.27.68.252.1 0 89 89 210019.1 3.65.55.4 4.26.24.68.36.54.14.64.33.97.36.57.459.1 0 90 90 210115.5 5.57.77.0 5.65.78.26.37.44.95.56.74.98.27.69.359.1 1 91 91 310009.1 3.77.04.1 4.46.35.47.37.54.64.43.03.37.47.97.958.1 1 92 92 110107.1 4.24.12.6 2.13.34.59.95.53.52.04.02.44.85.06.551.1 0 93 93 311019.2 3.94.65.3 4.28.44.87.16.26.64.42.64.27.67.58.672.1 0 94 94 301119.3 3.55.47.8 4.67.55.94.66.44.94.84.14.68.97.68.972.1 1 95 95 311009.3 3.84.04.6 4.76.45.57.45.34.83.63.23.47.77.38.459.1 1 96 96 110018.6 4.85.65.3 2.36.05.76.75.83.64.93.63.67.38.18.150.1 1 97 97 100117.4 3.42.65.0 4.14.44.87.24.56.44.25.63.76.35.57.248.1 0 98 98 100018.7 3.23.33.2 3.16.12.95.65.04.33.12.92.55.47.07.751.1 0 99 99 210117.8 4.95.85.3 5.25.37.17.96.05.74.34.93.96.47.17.461.1 0 100 100 211107.9 3.04.45.1 5.94.24.89.75.75.83.45.43.56.47.37.057.1 0 SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM] HBAT – Two-Sample T-Test The UNIVARIATE Procedure Variable: X19 (X19 – Satisfaction) Moments N 100Sum Weights 100 Mean 6.918Sum Observations 691.8 Std Deviation 1.19183925Variance 1.42048081 Skewness 0.0781812Kurtosis -0.7913045 Uncorrected SS 4926.5Corrected SS 140.6276 Coeff Variation 17.2280898Std Error Mean 0.11918393 Basic Statistical Measures Location Variability Mean 6.918000 Std Deviation 1.19184 Median 7.050000 Variance 1.42048 Mode 7.600000 Range 5.20000 SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM] 5.084 9.922 SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM] HBAT – Two-Sample T-Test The UNIVARIATE Procedure Variable: X20 (X20 – Likelihood of Recommendation) Moments N 100Sum Weights 100 Mean 7.02Sum Observations 702 Std Deviation 1.04330477Variance 1.08848485 Skewness 0.04392529Kurtosis -0.0883467 Uncorrected SS 5035.8Corrected SS 107.76 Coeff Variation 14.8618913Std Error Mean 0.10433048 Basic Statistical Measures Location Variability Mean 7.020000 Std Deviation 1.04330 Median 7.000000 Variance 1.08848 Mode 7.500000 Range 5.30000 SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM] 5.597 9.938 SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM] HBAT – Two-Sample T-Test The TTEST Procedure Variable: X19 (X19 – Satisfaction) X3 NMean Std Dev Std ErrMinimum Maximum 0 496.6408 1.09160.1559 4.7000 8.9000 1 517.1843 1.23330.1727 4.8000 9.9000 Diff (1-2) -0.5435 1.16600.2333 X3 Method Mean95% CL Mean Std Dev95% CL Std Dev 0 6.64086.32736.9544 1.09160.91031.3638 1 7.18436.83757.5312 1.23331.03191.5330 Diff (1-2) Pooled -0.5435-1.0064-0.0806 1.16601.02321.3557 Diff (1-2) Satterthwaite -0.5435-1.0053-0.0817 Method Variances DFt Value Pr > |t| Pooled Equal 98-2.33 0.0219 Satterthwaite Unequal97.357-2.340.0216 Equality of Variances Method Num DFDen DFF Value Pr > F Folded F 50481.280.3977 SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM] Variable: X20 (X20 – Likelihood of Recommendation) X3 NMean Std Dev Std ErrMinimum Maximum 0 496.7306 1.00630.1438 4.6000 8.8000 1 517.2980 1.01100.1416 5.5000 9.9000 Diff (1-2) -0.5674 1.00870.2018 X3 Method Mean95% CL Mean Std Dev95% CL Std Dev 0 6.73066.44167.0196 1.00630.83921.2571 1 7.29807.01377.5824 1.01100.84591.2568 Diff (1-2) Pooled -0.5674-0.9679-0.1670 1.00870.88511.1728 Diff (1-2) Satterthwaite -0.5674-0.9678-0.1670 Method Variances DFt Value Pr > |t| Pooled Equal 98-2.81 0.0059 Satterthwaite Unequal97.875-2.810.0059 Equality of Variances Method Num DFDen DFF Value Pr > F Folded F 50481.010.9753 SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM] SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM] HBAT – Two-Sample T-Test The ANOVA Procedure Class Level Information Class Levels Values X3 20 1 Number of Observations Read 100 Number of Observations Used 100 SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM] HBAT – Two-Sample T-Test The ANOVA Procedure Dependent Variable: X19 X19 – Satisfaction SourceDFSum of Squares Mean SquareF ValuePr > F Model 17.3817817 7.3817817 5.430.0219 Error 98133.2458183 1.3596512 Corrected Total 99140.6276000 R-Square Coeff Var Root MSE X19 Mean 0.052492 16.85517 1.1660416.918000 Source DFAnova SS Mean Square F ValuePr > F X3 17.38178167 7.38178167 5.430.0219 SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM] HBAT – Two-Sample T-Test The ANOVA Procedure Class Level Information Class Levels Values X3 20 1 Number of Observations Read 100 Number of Observations Used 100 SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM] HBAT – Two-Sample T-Test The ANOVA Procedure Dependent Variable: X20 X20 – Likelihood of Recommendation Source DFSum of Squares Mean SquareF ValuePr > F Model 18.0461144 8.0461144 7.910.0059 Error 9899.7138856 1.0174886 Corrected Total 99107.7600000 R-Square Coeff Var Root MSE X20 Mean 0.074667 14.36904 1.0087067.020000 Source DFAnova SS Mean Square F ValuePr > F X3 18.04611445 8.04611445 7.910.0059 SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM] HBAT – Two-Sample T-Test The REG Procedure Model: MODEL1 Dependent Variable: X19 X19 – Satisfaction Number of Observations Read 100 Number of Observations Used 100 Analysis of Variance Source DFSum of Squares Mean Square F Value Pr SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM] HBAT – Two-Sample T-Test The REG Procedure Model: MODEL1 Dependent Variable: X19 X19 – Satisfaction SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM] SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM] SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM] HBAT – Two-Sample T-Test The REG Procedure Model: MODEL1 Dependent Variable: X20 X20 – Likelihood of Recommendation Number of Observations Read 100 Number of Observations Used 100 Analysis of Variance Source DFSum of Squares Mean Square F Value Pr SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM] HBAT – Two-Sample T-Test The REG Procedure Model: MODEL1 Dependent Variable: X20 X20 – Likelihood of Recommendation SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM] SAS Output HBAT_Univ_TTest_ANOVA_REG_Prog_SAS_Output_Fa16.htm[9/26/2016 9:00:05 PM]
ANOVA
*; *; * HBAT – Two -Sample T -Test and Univariate Examples; *; *; ods graphics on ; *; options ls =80 ps =50 nodate pageno= 1; *; Title ‘HBAT – Two -Sample T -Test’ ; *; * Input HBAT ; *; Data HBAT; Infile ‘N: BIA652D_Multivariate Data Analytics_2016_Fall Class_04_Chap 2HBAT_tabs.txt’ DLM = ’09’X TRUNCOVER ; Input ID X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 X15 X16 X17 X18 X19 X20 X21 X22 X23; *; Data HBAT; Set HBAT (Keep = ID X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 X15 X16 X17 X18 X19 X20 X21 X22 X23); Label ID = ‘ID – Identification Number’ X1 = ‘X1 – Customer Type’ X2 = ‘X2 – Industry Type’ X3 = ‘X3 – Firm Size’ X4 = ‘X4 – Region’ X5 = ‘X5 – Distribution System’ X6 = ‘X6 – Product Quality’ X7 = ‘X7 – E-Commerce’ X8 = ‘X8 – Technical Support’ X9 = ‘X9 – Complaint Resolution’ X10 = ‘X10 – Advertizing’ X11 = ‘X11 – Product Line’ X12 = ‘X12 – Salesforce Image’ X13 = ‘X13 – Competitive Pricing’ X14 = ‘X14 – Warranty & Claims’ X15 = ‘X15 – New Products’ X16 = ‘X16 – Order & Billing’ X17 = ‘X17 – Price Flexibility’ X18 = ‘X18 – Delivery Speed’ X19 = ‘X19 – Satis faction’ X20 = ‘X20 – Likelihood of Recommendation’ X21 = ‘X21 – Likelihood of Future Purchase’ X22 = ‘X22 – Current Purchase/Usage Level’ X23 = ‘X23 – Consider Strategic Alliance/Partnership in Future’ ; *; *; Data HBAT; Set HBAT; *; Proc Print Data = HBAT; *; * HBAT – Univariate; *; Proc Univariate Data = HBAT; Var X19 X20; *; * HBAT – Two -Sample T -Test; *; Proc TTest Data = HBAT; Class X3; Var X19 X20; *; *; * HBAT – ANOVA; *; Proc ANOVA Data = HBAT; Class X3; Model X19 = X3; *; Proc ANOVA Data = HBAT; Class X3; Model X20 = X3; *; *; * HBAT – REG; *; Proc REG Data = HBAT; Model X19 = X3; *; Proc REG Data = HBAT; Model X20 = X3; *; * ods graphics off; *; *; Run ; Quit ;
ANOVA
SAS Univariate_TTest_ANOVA_REG – HW Due June 13th See the posted sample SAS Univ_TTest_ANOVA_REG program, HBAT-tabs dataset, and SAS Univ_TTest_ANOVA_REG output, for the analysis of X3-Firm Size along with X19-Satisfaction and X20-Likelihood of Recommendation. Now perform a similar statistical data analysis using X5-Distribution System as the Class Independent predictor variable and X19-Satisfaction and X20-Likelihood of Recommendation as the Dependent response variables, including assessment of testing the differences of X19-Satisfaction and X20-Likelihood of Recommendation across levels of X5-Distribution System. Provide a brief 3/4-1 page summary along with the program and output as an appendix.

DUE IN 4 HOURS

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discussion ONE PAGE

Watch the following video http://www.ted.com/talks/margaret_wertheim_crochets_the_coral_reef.htmlThe idea of a hyperbolic space is, as the speaker said, about anything that grows so much at the edges that it becomes ‘curly’ or ‘wavy’.  Give an example of an object that you suspect might be a hyperbolic space in this way

Response to Another Student Discussion on tests and measurements

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Directions are attached. This is 1 of 2 responses that will be required. Please follow directions explicitly and cite all references

Response to Another Student Discussion on tests and measurements
DIRECTIONS: I have attached the response of 1 learner who has assumption concerns that differ from my own. Kindly read their discussion and follow the bulleted directions explicitly. Please cite all references used. The directions call for identifying one post the same as my choice and one that is different. Since no one has posted a discussion the same as mine, I am attaching one that is different until someone posts one that is the same as mine. I thank you kindly. Response Guidelines Carefully read the response guidelines regarding the two different submissions required for response posts. Respond to the posts of at least two other learners. Attempt to identify one post that is the same as your choice and one that is different from your choice. Discuss how the content of the post is similar to or different from your post and explore how different experiences and knowledge result in different perspectives. Focus on providing facts, rather than opinion to support the assumption being presented as concerning to them. Keep in mind that the objective of response posts is to stimulate discussion, promote an exchange of ideas, and (most importantly) to generate a sense of community among learners in the courseroom. Post # 1- Hannah Gentry (Assumption 3) Top of Form                To begin, an indication of concern I have for this assumption, I give my concern a rating of 1 = somewhat concerned.
 While the assumption (I believe) remains fairly accurate on the prediction of job-readiness or possibly career options, I question the validity of a written assessment that predicts behavior. While the concept of a test for this purpose seems to hold promise, I have questions of the credibility that a test of any kind, can provide an accurate measure of the behavior that a person has yet to have.                      When predicting an outcome of a career with an aptitude test, the results are measured by certain character strengths that the individual posses. However, when predicting the social behavior of a person involved in crime, raises some concerns. The text mentions a form of psychological testing of behavior predictions for former criminals (Cohen, 2012). I believe that this element of testing for legal matters, lies within the perspective of the administrator controlling the measuring of the test. If the measuring is to prove the criminal has indeed learned from their past behaviors, we can assume that the examination has credibility. For example, the National Council of State Boars of Nursing believe that criminals should have the right to jobs, thus a test should be administered to assess their psychological readiness (National Council of State Boards of Nursing, 2010). With that perspective of testing, the results have a level of integrity and helpfulness attached to the measuring process. However, if the assessment is used to somehow prove that because a person committed a crime in the past, they must have committed a crime they are questioned for. Personally, I feel that the perspective effects the outcome of a test and therefore lowers the rate of credibility.                       It is possible that this assumption could be based off of the success it may have from predicting the future of ones behaviors, which is understandable. However, further into chapter four from our text, the author discusses how a “good test” can be determined; reliability and validity (Cohen, 2013). The use of psychological testing for analyzing ones personality or predicting a future career, has been proven to be a fairly good test based on results of consistency with the tester (Cohen, 2013). My concern for the reliability of the test are with the consistency of the test; does this test have a consistent outcome of success when administered (Cohen, 2013)? Some doubts were raised on the measurement of the test and how the results are gathered. Just as a good test should measure, does this test actually test for it’s purpose (Cohen, 2013)? As our text mentions, “what do these scores really tell us about the construct?” (Cohen, 2013) Does this actually prove anything in relevance with the subject being measured (Cohen, 2013)? 
                 In the future chapters, I believe (hope) that I can gain knowledge on the use of assessments for education (chapter 11). As a former teacher, I always question the credibility of examinations and if there truly is any accuracy in the results. Also, as a student who struggles with test anxiety but thrives in all other assignments, I doubt the results of many tests that I have taken. Finally, as a therapist who will work primarily with children with autism, I question if standard examinations in education could be useful for their growth process. I am thoroughly excited to begin studying this particular chapter and form my own opinion of assessments in education.                                                          References Cohen, R.J., Swerdilk, M.E., & Sturman, E.D. (2013). Chapter four: Of Tests and Testing.Retrieved from: PSYCHOLOGICAL TESTING AND ASSESSMENT:AN INTRODUCTION TO TESTS AND MEASUREMENT, EIGHTH EDITION National Council of State Boards of Nursing, Inc. 2010. Does Past Criminal Behavior Predict Future Criminal Behavior? Retrieved from: http://journalofnursingregulation.com/article/S2155-8256(15)30332-X/fulltext

Response to another student discussion: Odds in logistic regression

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Respond to another student discussion on odds in logistic regression.Respond  by evaluating the learner’s response. Do you agree or disagree? Why? Do you consider this a good answer to the question? Why or why not?

Please cite references. Student discussion is attached.

Response to another student discussion: Odds in logistic regression
DIRECTIONS: Respond to another student discussion on odds in logistic regression.Respond  by evaluating the learner’s response. Do you agree or disagree? Why? Do you consider this a good answer to the question? Why or why not? PLEASE CITE REFERENCES Robert Laukaitis U05D1 – Interpretation of Odds – R. Laukaitis Top of Form U05D1 – Binary Logistic Regression Odds in logistic regression represent the probability of an event occurring compared to the probability of an event not occurring (George & Mallery, 2013; Warner, 2013). In order to provide an example, an article provided by Szumilas (2010) summarized the following scenario: 186 of the 263 adolescents previously judged as having experienced a suicidal behaviour requiring immediate psychiatric consultation did not exhibit suicidal behaviour (non-suicidal, NS) at six months follow-up. Of this group, 86 young people had been assessed as having depression at baseline. Of the 77 young people with persistent suicidal behaviour at follow-up (suicidal behaviour, SB), 45 had been assessed as having depression at baseline. (para. 6) The hypotheses for this example might state: H0: There is no relationship between depression and suicidal behavior in young people. Ha: There is a relationship between depression and suicidal behavior in young people. Calculations The scenario data was recreated in SPSS (IBM, 2016) and analyzed. Warner (2013) suggested that starting with an odds ratio table would help present data that could help make a decision about H0. Table 1 represents the 2×2 odds ratio table to begin the analysis. Table 1 Odds example   Suicidal Behavior (SB) (Y=0) Non-suicidal Behavior (NS) (Y=1) Total (N) Depression (X=0) 45 (34.4%) 32 (24.2%) 77 No Depression (X=1) 86 (65.6%) 100 (75.8%) 186 Total (N) 131 132 263 Odds for suicidal behavior with depression, suicidal behavior without depression, non-suicidal behavior with depression and non-suicidal behavior without depression. The odds in Table 1 were calculated by calculating the odds that an individual with depression (n=131) had suicidal behavior: 45/131= 34.4%. Next, those with no depression is assumed by calculating the remaining balance of those having suicidal behavior: 1 (100%)-.344 (34.4%) = .656 (65.6%). The same approach was used to calculate those with non-suicidal behavior: 32/132=.242 (24.2%) with the residual representing those with non-suicidal behavior and no diagnosis of depression: 1 (100%)-.242 (24.2%)=.758 (75.8%). Table 2 represents the χ2 test for independence between the frequency of participants having depression and those having suicidal behaviors. In Table 2, the Pearson χ2 results indicated that there was not a significant relationship between those participants diagnosed with depression and exhibiting suicidal behaviors ( χ 2 (1) = 3.245, p > 0.05). Therefore, H0 is accepted. Table 2 Chi-Square Tests   Value df Asymp. Sig. (2-sided) Exact Sig. (2-sided) Exact Sig. (1-sided) Pearson Chi-Square 3.245a 1 .072     Continuity Correctionb 2.775 1 .096     Likelihood Ratio 3.256 1 .071     Fisher’s Exact Test       .079 .048 Linear-by-Linear Association 3.232 1 .072     N of Valid Cases 263         a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 38.35. b. Computed only for a 2×2 table In order to understand the effect size, phi (φ) was calculated to determine the goodness of fit (Warner, 2013). The effect size was calculated as Cramer’s V. According to Warner (2013), the effect size is equal to:   In this case, the df = 1 with φ = .111 represents a relatively small effect size. Table 3 Symmetric Measures   Value Approx. Sig. Nominal by Nominal Phi .111 .072 Cramer’s V .111 .072 N of Valid Cases 263   Odds versus Odds Ratio Warner (2013) suggested that the concept of odds represented the probability of an event occurring and the probability of an event not occurring. Using the example in Table 1, the odds of an individual being diagnosed with depression and displaying suicidal behavior: 34.4%/65.6% ≈ 1:2. The odds of an individual being diagnosed with depression and not displaying suicidal behavior: 24.2/75.8 ≈ 1:3. This indicated that an individual being diagnosed with depression was 1.5 times more likely to display suicidal behavior than an individual not diagnosed with depression. However, Warner (2013) odds ratio for depression is calculated using the depression row (X=0) and the non-depression row (X=1). These two numbers computed as a ratio represent the odds ratio: (X=0;Y=0)/(X=0;Y=1) (X=1;Y=0)/(X=1;Y=1) This would be calculated as 45/32 86/100 = 1.41 .86 = 1.64 his would indicate that a one unit increase in those diagnosed with depression displaying suicidal behavior is equal to 1.64 unit increase in those not diagnosed with depression exhibiting suicidal behavior. References IBM. (2016). Statistical Package for the Social Sciences (SPSS) software. Retrieved Feb 28, 2016, from www.IBM.com: http://www-01.ibm.com/software/analytics/spss/ Szumilas, M. (2010). Explaining odds ratios. Journal of the Canadian Academy of Child and Adolescent Psychiatry, 19(3), 227-229. Retrieved from https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2938757/ Warner, R. M. (2013). Applied statistics: From bivariate through multivariate techniques (2nd ed.). Thousand Oaks, CA: Sage. Retrieved from http://online.vitalsource.com/books/9781452268705

f(x)=2x

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f(x)=2x

Answer Submitted by Coloratus on Fri, 2012-01-06 06:59teacher rated 77 times 4.753245price: $0.00 Linear function

f(x) = 2x

is a form of linear function, generally: f(x) = ax + b

In this case b = 0, meaning that the line (graph of this function) goes through the origin of coordinate system.

a = 2, meaning that the slope of the line is 2, or the angle between the line and +x is bigger than 45°. The function is always rising.

We can draw graph if we know 2 points and draw a line between them:

f(0) = 0

f(1) = 2

So, the points are: (0, 0), (1, 2), and the graph is on the picture

Exercise 10

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Complete Exercise 10 in Statistics for Nursing Research: A Workbook for Evidence-Based Practice, and submit as directed by the instructor.


PLEASE PROVIDE RATIONALES OF EACH ANSWER

Exercise 10
Questions to Be Graded EXERCISE 10 Name: _______________________________________________________ Class: __________ Date: _____________________________________________________________ Follow your instructor’s directions to submit your answers to the following questions for grading. Your instructor may ask you to write your answers below and submit them as a hard copy for grading. Alternatively, your instructor may ask you to use the space below for notes and submit your answers online at http://evolve.elsevier.com/Grove/statistics/ under “Questions to Be Graded.” 1. What demographic variables were measured at the nominal level of measurement in the Oh et al. (2014) study? Provide a rationale for your answer. 2. What statistics were calculated to describe body mass index (BMI) in this study? Were these appropriate? Provide a rationale for your answer. 3. Were the distributions of scores for BMI similar for the intervention and control groups? Provide a rationale for your answer. 4. Was there a significant difference in BMI between the intervention and control groups? Provide a rationale for your answer. 5. Based on the sample size of N = 41, what frequency and percentage of the sample smoked? What frequency and percentage of the sample were non-drinkers (alcohol)? Show your calculations and round to the nearest whole percent. 6. What measurement method was used to measure the bone mineral density (BMD) for the study participants? Discuss the quality of this measurement method and document your response. 7. What statistic was calculated to determine differences between the intervention and control groups for the lumbar and femur neck BMDs? Were the groups significantly different for BMDs? 8. The researchers stated that there were no significant differences in the baseline characteristics of the intervention and control groups (see Table 2). Are these groups heterogeneous or homo-geneous at the beginning of the study? Why is this important in testing the effectiveness of the therapeutic lifestyle modification (TLM) program? 9. Oh et al. (2014, p. 296) stated that “the adherence rate to the TLM program was 99.6%.” Discuss the importance of intervention adherence, and document your response. 10. Was the sample for this study adequately described? Provide a rationale for your answer. Copyright © 2017, Elsevier Inc. All rights reserved.

Response to Another Student Rotation Discussion

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DIRECTIONS: READ THE ATTACHED DISCUSSION BELOW BY ANOTHER STUDENT AND:


Respond by expanding on one of the points made. Cite references.

Response to Another Student Rotation Discussion
DIRECTIONS: READ THE DISCUSSION BELOW BY ANOTHER STUDENT AND: Respond by expanding on one of the points made. Cite references. Rotation _Jessica Coutain Collapse Top of Form Unit 3 Discussion 2 Factor rotation is a technique used to transform factors gained from factor analysis (FA) so that the factor loadings that are small would be minimized, and factor loadings that are large would be maximized, in order to enhance the interpretability of these factors (Field, 2013; Warner, 2013, p. 848). The major difference between orthogonal and oblique rotation is that the orthogonal rotation preserves the orthogonality of the factors (i.e., the correlations between them remain equal to zero), whereas the oblique rotation allows the new factors to be correlated. Some advantages of orthogonal rotations are that they provide results that are simpler due to the preserved orthogonality of factors, and might be easier to interpret. Orthogonal rotations also produce lower sampling errors, and the results from these rotations are more likely to be replicated in further studies (Kieffer, 1998, p. 12). An important limitation is that “underlying factors” are rarely completely uncorrelated if they correspond to something in reality, so orthogonal rotations tend to oversimplify the model (Field, 2013; Kieffer, 1998). On the contrary, oblique rotations allow for the best fit of the model to the gathered data, and they may better correspond to the scholar’s view about the world (Kieffer, 1998; Warner, 2013). However, a limitation is that these results are more difficult to interpret, for oblique rotations produce more data to be assessed (Kieffer, 1998, p. 16). A researcher might want to use oblique rotation if they believe that the orthogonal rotation oversimplifies the data, because “hidden factors,” as was noted, will rarely be uncorrelated if they are to reflect something in reality (Field, 2013, sec. 17.4.6). Also, a substantial reason might be that the positioning of clusters of the original variables is such that an orthogonal rotation will be unsuccessful in maximizing the factor loadings, whereas an oblique rotation will be much more effective (Field, 2013, sec. 17.4.6). References Field, A. (2013). Discovering statistics using IBM SPSS Statistics (4th ed.). Thousand Oaks, CA: SAGE Publications. Kieffer, K. M. (1998). Orthogonal versus oblique factor rotation: A review of the literature regarding the pros and cons. Retrieved from http://files.eric.ed.gov/fulltext/ED427031.pdf Warner, R. M. (2013). Applied statistics: From bivariate through multivariate techniques (2nd ed.). Thousand Oaks, CA: SAGE Publications. Bottom of Form

EX18

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Complete Exercise 18 in Statistics for Nursing Research: A Workbook for Evidence-Based Practice, and submit as directed by the instructor

EX18
Questions to Be Graded EXERCISE 18 Name: ______________________________ Class: __________________________ Date: ________________________________________________________________ Follow your instructor’ s directions to submit your answers to the following questions for grading. Your instructor may ask you to write your answers below and submit them as a hard copy for grading. Alternatively, your instructor may ask you to use the space below for notes and submit your answers online at http://evolve.elsevier.com/Grove/Statistics/ under “Questions to Be Graded.” 1. Mayland et al. (2014) do not provide the degrees of freedom (df ) in their study. Use the degrees of freedom formulas provided at the beginning of this exercise to calculate the group df and the error df. 2. What is the F value and p value for spiritual need—patient? What do these results mean? 3. What is the post hoc result for facilities for the hospital with LCP vs. the hospital without LCP (see Table 2)? Is this result statistically significant? In your opinion, is this an expected finding? 4. What are the assumptions for use of ANOVA? 5. What variable on Table 3 has the result F = 10.6, p < 0.0001? What does the result mean? 6. ANOVA was used for analysis by Mayland et al. (2014). Would t -tests have also been appropriate? Provide a rationale for your answer. 7. What type of post hoc analysis was performed? Is the post hoc analysis performed more or less conservative than the Scheffé test? 8. State the null hypothesis for care for the three study groups (see Table 2). Should the null hypothesis be accepted or rejected? Provide a rationale for your answer. 9. What are the post hoc results for care? Which results are statistically significant? What do the results mean? 10. In your opinion, do the study findings presented in Tables 2 and 3 have implications for end of life care? Provide a rationale for your answer. Copyright © 2017, Elsevier Inc. All rights reserved.

Inferential Statistics and Findings

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Complete

an inferential statistics (hypothesis) test usingthe research question and variables your learning team developed for the Week 2 Business Research Project Part 1 assignment. Include:

  • The research question (Most up to date version using team all team discussion and instructor feedback)
  • Mock data for the independent and dependent variables (from Week 4)


Determine

the appropriate statistical tool to test the hypothesis based on the research question. (t-test, ANOVA, Chi-Squared test, test of correlation coefficient, etc)


Conduct

a hypothesis test with a 0.05 level of significance.


Write

an interpretation of no more than 350-words about the results of the hypothesis test.  Clearly connect the results of the test to your Research Question (RQ).


Include

the Excel output or test calculations in an Appendix.