Question

1.

Consider the partially completed​ one-way ANOVA summary table below.

​a) Complete the remaining entries in the table. ​b) How many population means are being​ tested? ​c) Using alphaαequals=0.05​, what conclusions can be made concerning the population​ means?		Source	Sum of Squares	Degrees of Freedom	Mean Sum of Squares	F
		Between	​?	33	​?	​?
		Within	96	​?	​?
		Total	165	19

Click the icon to view a table of critical​ F-scores for

alphaα equals=0.05

​a) Complete the ANOVA table below.

Source	Sum of Squares	Degrees of Freedom	Mean Sum of Squares	F
Between	?	33	?	?
Within	9696	?	?
Total	165	19

​(Type integers or decimals. Round to three decimal places as​ needed.)

2.

Consider the data in the table collected from four independent populations.		Sample 1	Sample 2	Sample 3	Sample 4
​a) Calculate the total sum of squares​ (SST). ​b) Partition the SST into its two​ components, the sum of squares between​ (SSB) and the sum of squares within​ (SSW).		2020	1515	1414	1717
		1818	1010	2323	1212
		1515	1111	1515	1010
		1616		77

​c) Using alphaαequals=0.10,

what conclusions can be made concerning the population​ means?

Click the icon to view a table of critical​ F-scores for

alphaαequals=0.10.

​a) Determine the value of SST.

SSTequals=nothing

​(Type an integer or a​ decimal.)

Answer 1

1a)

from abvove: SS(between) =165-96 =69

df(within) =19-3 =16

MS(between )=69/3 =23

MS(within) =96/16 =6

F =MS(between )/MS(within)=23/6=3.83

Source	SS	df	MS	F
between	69	3	23.000	3.83
within	96	16	6.000
total	165	19

for (3,16) df and 0.05level, critical value =3.24

since test statistic F =3.83 is greater than critical value , we reject null hypothesis.

And conclude that population mean are different,

2)

Applying one way ANOVA: (use excel: data: data analysis: one way ANOVA: select Array):

Source of Variation	SS	df	MS	F		F crit
Between Groups	56	3	18.67	1.02		2.73
Within Groups	183.5	10	18.35
Total	239.5	13

a)SST =239.5

b)SSB =56

SSW =183.5

c)

for (3,10) df and 0.10 level, critical value =2.73

since test statistic F =1.02 is less than critical value , we fail to reject null hypothesis.

And can not conclude that population mean are different,