A farmer is concerned that a change in fertilizer to an organic variant might change his crop yield. He subdivides 6 lots and uses the old fertilizer on one half of each lot and the new fertilizer on the other half. The following table shows the results.

| Lot | Crop Yield Using Old Fertilizer | Crop Yield Using New Fertilizer |
|-----|----------------------------------|---------------------------------|
| 1 | 11 | 12 |
| 2 | 11 | 9 |
| 3 | 9 | 13 |
| 4 | 8 | 9 |
| 5 | 13 | 12 |
| 6 | 11 | 12 |

Let the difference be defined as Old – New.

a. Specify the competing hypotheses that determine whether there is any difference between the average crop yields from the use of the different fertilizers.

A. \( H_0: \mu_D \leq 0; \, H_A: \mu_D > 0 \)
B. \( H_0: \mu_D = 0; \, H_A: \mu_D \neq 0 \)
C. \( H_0: \mu_D \geq 0; \, H_A: \mu_D < 0 \)

b. Assuming that differences in crop yields are normally distributed, calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and your final answer to 2 decimal places.)

c. Find the p-value.

A. \( p\text{-value} < 0.01 \)
B. \( 0.01 \leq p\text{-value} < 0.02 \)
C. \( 0.02 \leq p\text{-value} < 0.05 \)
D. \( 0.05 \leq p\text{-value} < 0.10 \)
E. \( p\text{-value} \geq 0.10 \)

The calculated test statistics (t) is around -0.83, and the comparing p-value is around 0.8972, demonstrating no significant difference. The right reply for the multiple-choice question is p-value > 0.10.

How to find the p-value

a. The competing speculations that decide whether there's any distinction between the normal crop yields from the utilization of diverse fertilizers are:

H0: μD = 0; HA: μD ≠

b. To calculate the esteem of the test statistic, we have to discover the cruel distinction (¯D) and the standard deviation of the contrasts (SD).

Cruel contrast (¯D):

¯D = Σ(D) / n

where D speaks to the contrasts (Ancient - Unused) and n is the number of perceptions.

Calculating the mean difference:

11 - 12 + 11 - 9 + 9 - 13 + 8 - 9 + 13 - 12 + 11 - 12 = -3

(¯D) = (-3 / 6)= -0.5

The standard deviation of the contrasts (SD):

sD = √[Σ(D - ¯D)² / (n - 1)]

where Σ(D - ¯D)² speaks to the whole of squared contrasts.

Calculating the standard deviation of the contrasts:

-0.5 - (-0.5)² + 1.5 - (-0.5)² + -4.5 - (-0.5)² + -1.5 - (-0.5)² + 0.5 - (-0.5)² + -0.5 - (-0.5)² = 22.5

sD = √(22.5 / 5) ≈ 1.89

The test measurement (t) is calculated as:

t = ¯D / (SD / √n)

Substituting the values:

t = (-0.5) / (1.89 / √6) ≈ -0.83

c. With 5 degrees of opportunity (n - 1 = 6 - 1 = 5) and a two-tailed test, the p-value is roughly 0.4486. Since it's a two-tailed test, we twofold this esteem to urge the ultimate p-value.