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

Answer:

standard error of the difference (green - brown) between the sample means

$se(x^{-} _{1} -x^{-} _{2}) = 0.3411$

Step-by-step explanation:

Explanation:-

Given data 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.

First sample size 'n₁' = 20

mean of the first sample (x₁⁻) =16.22mm

Standard deviation of first sample (S₁) = 1.34mm

First sample size 'n₂' = 20

mean of the first sample (x₂⁻) =13.41mm

Standard deviation of first sample (S₂) = 0.73mm

Standard error of the difference between the sample means is defined by

$se(x^{-} _{1} -x^{-} _{2}) = \sqrt{\frac{s^2_{1} }{n_{1} } +\frac{s^2_{2} }{n_{2} } }$

$se(x^{-} _{1} -x^{-} _{2}) = \sqrt{\frac{(1.34)^2}{20}+\frac{(0.73)^2}{20} }$

$se(x^{-} _{1} -x^{-} _{2}) = 0.3411$

conclusion:-

standard error of the difference (green - brown) between the sample means

$se(x^{-} _{1} -x^{-} _{2}) = 0.3411$

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 :

$s_1=\dfrac{\sigma_1}{\sqrt{n_1}}=\dfrac{1.34}{\sqrt{20}}=0.29\\\\s_2=\dfrac{\sigma_2}{\sqrt{n_2}}=\dfrac{0.73}{\sqrt{20}}=0.16$

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

$\sqrt{\dfrac{s^2_1}{n_1}+\dfrac{s^2_1}{n_2}}\\\\=\sqrt{\dfrac{0.29^2}{20}+\dfrac{0.16^2}{20}}\\\\=0.074$

Hence, the standard error is 0.074.