Question

Consider the data in the table collected from three independent populations. Sample 1 Sample 2 Sample 3 III a) Calculate the total sum of squares (SST) and partition the SST into its two components, the sum of squares between (SSB) and the sum of squares within (SSW) b) Use these values to construct a one- way ANOVA table c) Using a 0.05, what conclusions can be made concerning the population means? Click the icon to view a table of critical F-scores for a = 0.05. a) Determine the values SST = (Type an integer or a decimal.) SSB = (Type an integer or a decimal) SSW (Type an integer or a decimal) b) Complete the one-way ANOVA table below Sum of Degrees of Mean Sum of Source Squares Freedom Squares F Between Within Total (Type integers or decimals. Round to three decimal places as needed.) c) Let 1.1 and 3 be the population means of samples 1, 2, and 3, respectively. What are the correct hypotheses for a one-way ANOVA test? O A H * 2 * 3 OB. Ho H = H2 H3 Hy Not all the means are equal. H + H2 H3 0c. Họ * 15 * 13 0 D. He i = H2 = 13
Falls/ does not fall, Reject/do not reject, Provide/do not provide

Answer 1

Solution:

In excel install analysis tool pak.

Go to data >Data analysis>ANOVA single factor

Home Insert Page Layout Formulas Data Review View Tell me what you want to do... X V for sample-1 Public ment Recovery Internal Confidential D G H Restricted JK ? X Sensitivity: Not set A B C sample-1 sample-2 sample-3 1 5 2 1 66 recovered the following files. Save ou wish to keep Anova: Single Factor e Files Input Input Range: ок | $A$1:$C$5 est_proportions (version 1).xls [.. rsion created from the last Auto... 16/2019 9:57 AM Cancel Grouped By: O Columns Rows Help est_proportions.xls (Original] rsion created last time the user .. 15/2019 12:56 PM Labels in First Row Alpha: 0.05 ūūū o ovou Aw N- Output options ade example(1) (version 1).xlsb- rsion created from the last Auto... 16/2019 9:58 AM Qutput Range: O New Worksheet Ply: New Workbook ade example(1).xlsx (Original] rsion created last time the user ..

You will get

SUMMARY
Groups	Count	Sum	Average	Variance
sample-1	3	18	6	9
sample-2	3	9	3	7
sample-3	4	16	4	6.666667
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	14.1	2	7.05	0.949038	0.431824	4.737414
Within Groups	52	7	7.428571
Total	66.1	9

COMING TO ANSWERS:

SST=66.1

SSB=14.1

SSW=52

Solution-b:

ANOVA
Source of Variation	Sum of squares	Degrees of freedom	Mean sum of squars	F
Between Groups	14.1	2	7.05	0.949038
Within Groups	52	7	7.428571
Total	66.1	9

Solution-c:

Ho:$\mu 1=\mu 2=\mu 3$

Ha:Not all means are equal

F critical=4.737

Since F statistic does not fall in the rejection region,accept Ho.The data so not provide sufficient evidence that the population

means are not all the same.