Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

For Dr. Young\'s experiment, 150 seedlings survived long enough to reach an age

ID: 265772 • Letter: F

Question

For Dr. Young's experiment, 150 seedlings survived long enough to reach an age at which nodulation could occur. The table below presents her observed results.



What value would you use for v (degrees of freedom)?
v = (r-1) * (c-1) =

Fill in the blanks for expected frequencies in the table above, using the following equation. Round to two decimal places.
Expected frequency = (R * C) / N

What is the calculated chi-square value? Calculate chi-square value using the equation below. Do not use the rounded values reported in the table above. Perform all calculations before rounding; then round final answer to two decimal places.

Table 1: Observed and expected values for M. cerifera seedlings inoculated with experimental and reference treatments PR NR BI ML Total Nodulated Obs. 15 13 19 30 77 Exp. Non-Nodulated Obs. 12 11 16 34 73 Exp. Total 27 24 35 64

Explanation / Answer

What value would you use for v (degrees of freedom)?

v = (r-1) * (c-1) = 4-1 * 2-1 = 3.

Expected frequency = (R * C) / N

PR

NR

BI

ML

OBS

15.00

13.00

19.00

30.00

77

EXP

13.86

12.32

17.97

32.85

OBS

12.00

11.00

16.00

34.00

73

EXP

19.89

17.68

25.78

47.14

40.86

36.32

52.97

96.85

150

Null hypothesis:

The observed values are not deviating from the expected values.

Categories

Nodulated

Non-Nodulated

Total

PR

NR

BI

ML

PR

NR

BI

ML

OBS

15

13

19

30

12

11

16

34

150

EXP

13.86

12.32

17.96667

32.85333

13.14

11.68

17.03333

31.14667

150

DEVIATION

1.14

0.68

1.033333

-2.85333

-1.14

-0.68

-1.03333

2.853333

D^2

1.2996

0.4624

1.067778

8.141511

1.2996

0.4624

1.067778

8.141511

D^2/E

0.093766

0.037532

0.059431

0.247814

0.098904

0.039589

0.062688

0.261393

0.901117

SUM

0.901117

Inference: The calculate chisquare value i.e. 0.90 is less than the table value at 0.95 and 3 DF i.e. 7.82, so the null hypothesis is accepted.

PR

NR

BI

ML

OBS

15.00

13.00

19.00

30.00

77

EXP

13.86

12.32

17.97

32.85

OBS

12.00

11.00

16.00

34.00

73

EXP

19.89

17.68

25.78

47.14

40.86

36.32

52.97

96.85

150

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote