Testing multiple means with ANOVA

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Question description

BUS 308 wk 3 assignment

Week 3 Testing multiple means with ANOVA <Note: use right click on row numbers to insert rows to perform analysis below any question>
For questions 3 and 4 below, be sure to list the null and alternate hypothesis statements.  Use .05 for your significance level in making your decisions.
For full credit, you need to also show the statistical outcomes - either the Excel test result or the calculations you performed.
1.  Based on the sample data, can the average(mean) salary in the population be the same for each of the grade levels? (Assume equal variance, and use the analysis toolpak function ANOVA.)
Set up the input table/range to use as follows:  Put all of the salary values for each grade under the appropriate grade label.
Be sure to incllude the null and alternate hypothesis along with the statistical test and result.
A B C D E F Note: Assume equal variances for all grades.
2.  The table and analysis below demonstrate a 2-way ANOVA with replication.  Please interpret the results.
Gender A B C D E F
M 24 27 40 47 56 76 The salary values were randomly picked for each cell.
25 28 47 49 66 77
F 22 34 41 50 65 75
24 36 42 57 69 77
Ho: Average salaries are equal for all grades
Ha: Average salaries are not equal for all grades
Ho: Average salaries by gender are equal
Ha: Average salaries by gender are not equal
Ho: Interaction is not significant
Ha: Interaction is significant
Perform analysis:
Anova: Two-Factor With Replication
Count 2 2 2 2 2 2 12
Sum 49 55 87 96 122 153 562
Average 24.5 27.5 43.5 48 61 76.5 46.8333
Variance 0.5 0.5 24.5 2 50 0.5 364.515
Count 2 2 2 2 2 2 12
Sum 46 70 83 107 134 152 592
Average 23 35 41.5 53.5 67 76 49.3333
Variance 2 2 0.5 24.5 8 2 367.333
Count 4 4 4 4 4 4
Sum 95 125 170 203 256 305
Average 23.75 31.25 42.5 50.75 64 76.25
Variance 1.58333 19.5833 9.66667 18.9167 31.3333 0.91667
Source of Variation SS df MS F P-value F crit
Sample 37.5 1 37.5 3.84615 0.07348 4.74723
Columns 7841.83 5 1568.37 160.858 1.5E-10 3.10588 Note: a number with an E after it (E9 or E-6, for example)
Interaction 91.5 5 18.3 1.87692 0.17231 3.10588 means we move the decimal point that number of places.
Within 117 12 9.75 For example, 1.2E4 becomes 12000; while 4.56E-5 becomes 0.0000456
Total 8087.83 23
Do we reject or not reject each of the null hypotheses?  What do your conclusions mean about the population values being tested?
3.  Using our sample results, can we say that the compa values in the population are equal by grade and/or gender, and are independent of each factor?
Grade Be sure to include the null and alternate hypothesis along with the statistical test and result.
Gender A B C D E F <Randomly pick compas to fill each cell - for exampe, a compa
M for the intersection of M and A might be 1.043.>
<If desired, you can use the compa values that relate to the
F salary values used in question 2 for a more direct comparison of the two
Conduct and show the results of a 2-way ANOVA with replication using the completed table above.  The results should look something like those in question 2.
Interpret the results. Are the average compas for each gender (listed as sample) equal?  For each grade?  Do grade and gender interaction impact compa values?
4.  Pick any other variable you are interested in and do a simple 2-way ANOVA without replication.  Why did you pick this variable and what do the results show?
Variable name: Be sure to include the null and alternate hypothesis along with the statistical test and result.
Gender A B C D E F
M Hint: use mean values in the boxes.
5.  Using the results for this week, What are your conclusions about gender equal pay for equal work at this point?

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