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A researcher is investigating whether a difference exists in the mean weight of green-striped watermelons grown on two different farms: one that uses organic methods and one that uses nonorganic methods. The mean and standard deviation of the weights in a random sample of 43 watermelons from the organic farm were 18 pounds and 2 pounds, respectively. The mean and standard deviation of the weights in a random sample of 40 watermelons from the nonorganic farm were 20 pounds and 1.7 pounds, respectively.

Required:
What represents the standard error of the difference in the mean weights of watermelons from the two farms?

Respuesta :

Answer:

D. 2^2+1.7^243+40−−−−−√

Step-by-step explanation:

The computation is shown below:

Given that

Sample 1

n_1 = 43

mean  = 18

s_1 = 2

Sample 2

n_2 = 40

mean = 20

s_2 = 1.7

Now the standard error is

[tex]= \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} } \\\\= \sqrt{\frac{2^2}{43} + \frac{1.7^2}{40} }[/tex]

= 0.4065

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