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Question
Two fertilizers are applied to randomly selected potato plants. After several weeks, the potatoes are weighed. The weights in kilograms are as follows. Which fertilizer is generally more effective?

Fertilizer A: 27, 20, 16, 18, 22, 19, 23, 21, 17, 19
Fertilizer B: 28, 19, 18, 21, 24, 20, 25, 27, 29, 21

(1 point)
Responses

Fertilizer A is generally more effective, because the mean and median weights of the potatoes treated with it are higher.
Fertilizer A is generally more effective, because the mean and median weights of the potatoes treated with it are higher.

Fertilizers A and B are both ineffective, because there is no difference in the ranges of the weights of the potatoes.
Fertilizers A and B are both ineffective, because there is no difference in the ranges of the weights of the potatoes.

Fertilizer B is generally more effective, because the mean and median weights of the potatoes that were treated with it are higher.
Fertilizer B is generally more effective, because the mean and median weights of the potatoes that were treated with it are higher.

Fertilizers A and B are equally effective, because the ranges of the weights of the potatoes are the same.
Fertilizers A and B are equally effective, because the ranges of the weights of the potatoes are the same.
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Answers

GPT-4o mini
To compare the effectiveness of the two fertilizers, we need to calculate the mean and median weights of the potatoes for each fertilizer.

**Fertilizer A weights:**
- 27, 20, 16, 18, 22, 19, 23, 21, 17, 19

**Calculating Mean for Fertilizer A:**
Mean \( = \frac{(27 + 20 + 16 + 18 + 22 + 19 + 23 + 21 + 17 + 19)}{10} \)
\( = \frac{ 27 + 20 + 16 + 18 + 22 + 19 + 23 + 21 + 17 + 19 }{10} = \frac{ 222 }{10} = 22.2 \)

**Calculating Median for Fertilizer A:**
To find the median, we first order the weights: 16, 17, 18, 19, 19, 20, 21, 22, 23, 27.
The median (middle value) is the average of the 5th and 6th values:
\( = \frac{19 + 20}{2} = 19.5 \)

**Fertilizer B weights:**
- 28, 19, 18, 21, 24, 20, 25, 27, 29, 21

**Calculating Mean for Fertilizer B:**
Mean \( = \frac{(28 + 19 + 18 + 21 + 24 + 20 + 25 + 27 + 29 + 21)}{10} \)
\( = \frac{ 28 + 19 + 18 + 21 + 24 + 20 + 25 + 27 + 29 + 21 }{10} = \frac{ 212 }{10} = 21.2 \)

**Calculating Median for Fertilizer B:**
To find the median, we first order the weights: 18, 19, 20, 21, 21, 24, 25, 27, 28, 29.
The median (middle value) is the average of the 5th and 6th values:
\( = \frac{21 + 24}{2} = 22.5 \)

**Summary of Calculated Values:**

- **Fertilizer A:**
- Mean = 22.2
- Median = 19.5

- **Fertilizer B:**
- Mean = 21.2
- Median = 22.5

### Conclusion:
Fertilizer A has a higher mean weight (22.2 kg vs 21.2 kg), but the median of Fertilizer B (22.5 kg) is higher than that of Fertilizer A (19.5 kg).

Given that the question asks which fertilizer is generally more effective, the response that aligns best is:

**Fertilizer B is generally more effective, because the mean and median weights of the potatoes that were treated with it are higher.**

Thus, the correct choice would be:
**Fertilizer B is generally more effective, because the mean and median weights of the potatoes that were treated with it are higher.**

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