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Calculated value <strong>of</strong> χ² = 3.5<br />

Table value <strong>of</strong> χ² = χ²(r-1) (c-1), 5%<br />

Table value <strong>of</strong> χ² = χ² (1) (4), 5%<br />

Table value <strong>of</strong> χ² = 9.48773<br />

Calculated value <strong>of</strong> chi-square is lesser than table value <strong>of</strong> Chi-square. Hence Null hypothesis (H 0 ) is<br />

accepted.<br />

Chi – Square for independence <strong>of</strong> attributes<br />

Null Hypothesis: H 0 : <strong>The</strong>re is no significance relationship between pr<strong>of</strong>ession and convenient outlet.<br />

Pr<strong>of</strong>ession Govt employee Pvt employee Business man House wives Others total<br />

Convenient outlet<br />

Super market 6 8 6 8 2 30<br />

Provision store 2 2 8 20 12 44<br />

Medical store 4 6 2 6 8 26<br />

Total 12 16 16 34 22 100<br />

O E O-E [O-E] 2 [0-E] 2 /E<br />

14 8 6 36 4.5<br />

6 5 1 1 0.2<br />

8 10 -2 4 0.4<br />

6 19 -13 169 8.9<br />

8 7 1 1 0.14<br />

20 15 5 25 1.7<br />

12 10 2 4 0.4<br />

12 8 4 16 2<br />

6 9 3 9 1<br />

8 6 2 4 0.7<br />

Ʃ = 19.94<br />

Calculated value <strong>of</strong> χ² = 19.94<br />

Table value <strong>of</strong> χ² = χ²(r-1) (c-1), 5%<br />

Table value <strong>of</strong> χ² = χ² (2) (4), 5%<br />

Table value <strong>of</strong> χ² = 15.5073<br />

Calculated value <strong>of</strong> chi-square is lesser than table value <strong>of</strong> Chi-square. Hence Null hypothesis (H 0 ) is<br />

accepted.<br />

One way ANOVA<br />

Null Hypothesis: H 0 : <strong>The</strong>re is no significance difference between quantity <strong>of</strong> rice consumed per day<br />

and quantity preferred.<br />

Source <strong>of</strong> variation Sum <strong>of</strong> Squares Degree <strong>of</strong> freedom Mean Square Variance ratio<br />

Between groups 4.409 4 1.102 F = 1.182<br />

Within groups 88.591 95 0.933<br />

Total 93.000 99<br />

<strong>The</strong> test statistic is F =<br />

=<br />

= 1.182<br />

<strong>The</strong>refore calculated F = 1.182<br />

Tabulated F at 5% level for (3, 96) degrees <strong>of</strong> freedom =2.68<br />

www.theinternationaljournal.org > RJS<strong>IT</strong>M: Volume: 02, Number: 02, December-2012 Page 22

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