Question
In: Statistics and Probability
find the expected count and the contribution to the chi square statistic for the (control, strongly...
find the expected count and the contribution to the chi square
statistic for the (control, strongly agree) cell i. yhe two way
table below
st agr. agr. neu. dis. str dis
cont 36. 48. 4. 15. 12
treat. 55. 41. 8. 4. 1
Solutions
Expert Solution
| str agr. | agr. | neu. | dis. | str dis | Totals | |
| cont. | 36=f11 | 48=f12 | 4=f13 | 15=f14 | 12=f15 | 36+48+4+15+12=115=f10 |
| treat. | 55=f21 | 41=f22 | 8=f23 | 4=f24 | 1=f25 | 55+41+8+4+1=109=f20 |
| Totals | 91=f01 | 89=f02 | 12=f03 | 19=f04 | 13=f05 | 224=N |
| str agr. | agr. | neu. | dis. | str dis. | |
| cont. | 115*91/224=46.7188 | 115*89/224=45.6920 | 115*12/224=6.1607 | 115*19/224=9.7545 | 115*13/224=6.6741 |
| treat. | 109*91/224=44.2813 | 109*89/224=43.3080 | 109*12/224=5.8393 | 109*19/224=9.2455 | 109*13/224=6.3259 |
| str. agr | agr. | neu. | dis. | str. dis. | |
| cont. | (36-46.7188)^2/46.7188=2.4592 | (48-45.6920)^2/45.6920=0.1166 | (4-6.1607)^2/6.1607=0.7578 | (15-9.7545)^2/9.7545=2.8208 | (12-6.6741)^2/6.6741=4.2500 |
| treat. | (55-44.2813)^2/44.2813=2.5946 | (41-43.3080)^2/43.3080=0.1230 | (8-5.8393)^2/5.8393=0.7995 | (4-9.2455)^2/9.2455=2.9761 | (1-6.3259)^2/6.3259=4.4840 |
the expected count for the (control, strongly agree) =46.7188
the contribution to the chi square statistic for the (control, strongly agree) cell=2.4592
orchestra answered 1 year agoRelated Solutions
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31
39
39
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B
74
87
68
53
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C
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35
27
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134
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