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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

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.

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

1 answer

To determine which fertilizer is generally more effective, we need to calculate the mean and median weights for both fertilizers.

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

  1. Mean for Fertilizer A: \[ \text{Mean} = \frac{(27 + 20 + 16 + 18 + 22 + 19 + 23 + 21 + 17 + 19)}{10} = \frac{ 27 + 20 + 16 + 18 + 22 + 19 + 23 + 21 + 17 + 19 }{10} = \frac{ 202 }{10} = 20.2 \]

  2. Median for Fertilizer A:

    • Sort the weights: 16, 17, 18, 19, 19, 20, 21, 22, 23, 27
    • Since there are 10 numbers, the median is the average of the 5th and 6th numbers: \[ \text{Median} = \frac{19 + 20}{2} = 19.5 \]

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

  1. Mean for Fertilizer B: \[ \text{Mean} = \frac{(28 + 19 + 18 + 21 + 24 + 20 + 25 + 27 + 29 + 21)}{10} = \frac{ 28 + 19 + 18 + 21 + 24 + 20 + 25 + 27 + 29 + 21 }{10} = \frac{ 212 }{10} = 21.2 \]

  2. Median for Fertilizer B:

    • Sort the weights: 18, 19, 20, 21, 21, 24, 25, 27, 28, 29
    • With 10 numbers, the median is the average of the 5th and 6th numbers: \[ \text{Median} = \frac{21 + 24}{2} = 22.5 \]

Now comparing the results:

  • Fertilizer A:

    • Mean: 20.2
    • Median: 19.5

  • Fertilizer B:

    • Mean: 21.2
    • Median: 22.5

From the calculated means and medians, we can see that both the mean and median weights for Fertilizer B are higher than those for Fertilizer A.

Thus, the correct response is:

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