Work out the Chi-Squared Goodness of Fit for Mendel's actual data on stem length: "Expt. 7: Length of stem. –– Out of 1,064 plants, in 787 cases the stem was long, and in 277 short. Hence a mutual ratio of 2.84:1. In this experiment the dwarfed plants were carefully lifted and transferred to a special bed. This precaution was necessary, as otherwise they would have perished through being overgrown by their tall relatives. Even in their quite young state they can be easily picked out by their compact growth and thick dark–green foliage." F2: O E O-E (O-E)^2 (O-E)^2/E # short 277 266 11 121 0.455 # tall 787 798 -11 121 0.152 sums: 1,064 1064 0.607 degrees of freedom = 2-1=1 Chi-squared value = .607 p-value: 0.25 < p < 0.50
- Work out the Chi-Squared Goodness of Fit for Mendel's actual data on stem length:
"Expt. 7: Length of stem. –– Out of 1,064 plants, in 787 cases the stem was long, and in 277 short. Hence a mutual ratio of 2.84:1. In this experiment the dwarfed plants were carefully lifted and transferred to a special bed. This precaution was necessary, as otherwise they would have perished through being overgrown by their tall relatives. Even in their quite young state they can be easily picked out by their compact growth and thick dark–green foliage."
|
F2: |
O |
E |
O-E |
(O-E)^2 |
(O-E)^2/E |
|
# short |
277 |
266 |
11 |
121 |
0.455 |
|
# tall |
787 |
798 |
-11 |
121 |
0.152 |
|
sums: |
1,064 |
1064 |
0.607 |
degrees of freedom = 2-1=1
Chi-squared value = .607
p-value: 0.25 < p < 0.50
- Work out the Chi-Squared Goodness of Fit for Mendel's actual data on flower position:
"Expt. 6: Position of flowers. –– Among 858 cases 651 had inflorescences axial and 207 terminal. Ratio, 3.14:1."
|
F2: |
O |
E |
O-E |
(O-E)^2 |
(O-E)^2/E |
|
# terminal |
207 |
215 |
-8 |
64 |
0.298 |
|
# axial |
651 |
644 |
8 |
49 |
0.077 |
|
sums: |
858 |
0.375 |
degrees of freedom = 2-1=1
Chi-squared value = 0.375
p-value: 0.50 < p < 0.75
- Work out the Chi-Squared Goodness of Fit for Mendel's actual data:
Expt. 5: Color of the unripe pods. –– The number of trial plants was 580, of which 428 had green pods and 152 yellow ones. Consequently these stand in the ratio of 2.82:1.
|
F2: |
O |
E |
O-E |
(O-E)^2 |
(O-E)^2/E |
|
# yellow pod |
428 |
435 |
-7 |
49 |
0.113 |
|
# green pod |
152 |
145 |
7 |
49 |
0.338 |
|
sums: |
580 |
0.451 |
degrees of freedom = 2-1=1
Chi-squared value = 0.451
p-value: 0.50 < p < 0.75
please tell me if these are correct or not I believe they are wrong.
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