Question

A biologist wants to estimate the difference between the mean body lengths of green and brown stinkbugs. A random sample of 20 green stinkbugs has a mean body length of 16.22 millimeters (mm) and a standard deviation of 1.34 mm. A random sample of 20 brown stinkbugs has a mean body length of 13.41 mm and a standard deviation of 0.73 mm. What is the standard error of the difference (green - brown) between the sample means?

Answer 1

The standard error of the difference between the mean body lengths of green and brown stinkbugs is calculated using the sample standard deviations and sizes, resulting in approximately 0.3412 mm.

Step-by-step explanation:

The standard error of the difference between the sample means for the green and brown stinkbugs can be calculated using the formula for the standard error of the difference between two independent sample means. The formula is:

SEdifference = √[(s12/n1) + (s22/n2)]

Where:

• s1 is the standard deviation of the first sample (green stinkbugs).
• n1 is the sample size of the first sample.
• s2 is the standard deviation of the second sample (brown stinkbugs).
• n2 is the sample size of the second sample.

Plugging the provided values into the formula, we get:

SEdifference = √[(1.342/20) + (0.732/20)]

SEdifference = √[(1.7956/20) + (0.5329/20)]

SEdifference = √[0.08978 + 0.026645]

SEdifference = √[0.116425]

SEdifference = 0.3412 mm

Therefore, the standard error of the difference between the mean body lengths of green and brown stinkbugs is approximately 0.3412 mm.

Answer 2

Answer: The standard error is 0.074.

Step-by-step explanation:

Since we have given that

n₁= 20

n₂ = 20

μ₁ = 16.22 mm

μ₂ = 13.41 mm

σ₁ = 1.34 mm

σ₂ = 0.73 mm

So, we get :

`s_1=(\sigma_1)/(√(n_1))=(1.34)/(√(20))=0.29\\\\s_2=(\sigma_2)/(√(n_2))=(0.73)/(√(20))=0.16`

So, the standard error of the difference between the sample means would be :

`\sqrt{(s^2_1)/(n_1)+(s^2_1)/(n_2)}\\\\=\sqrt{(0.29^2)/(20)+(0.16^2)/(20)}\\\\=0.074`

Hence, the standard error is 0.074.