Respuesta :
Answer:
standard error of the difference (green - brown) between the sample means
[tex]se(x^{-} _{1} -x^{-} _{2}) = 0.3411[/tex]
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
[tex]se(x^{-} _{1} -x^{-} _{2}) = \sqrt{\frac{s^2_{1} }{n_{1} } +\frac{s^2_{2} }{n_{2} } }[/tex]
[tex]se(x^{-} _{1} -x^{-} _{2}) = \sqrt{\frac{(1.34)^2}{20}+\frac{(0.73)^2}{20} }[/tex]
[tex]se(x^{-} _{1} -x^{-} _{2}) = 0.3411[/tex]
conclusion:-
standard error of the difference (green - brown) between the sample means
[tex]se(x^{-} _{1} -x^{-} _{2}) = 0.3411[/tex]
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 :
[tex]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[/tex]
So, the standard error of the difference between the sample means would be :
[tex]\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[/tex]
Hence, the standard error is 0.074.
