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An ecologist studying differences in populations of a certain species of lizards on two different islands collects lizards in live traps, weighs them, and then releases them again. (She marks them so she won’t weigh the same lizard twice). During one study period, she collects the data below. All weights are in grams. Is there convincing evidence of a difference in the mean weights of the lizards on the two different islands? Use α = 0.05.
Sample Size Mean (gm) Standard Deviation (gm)
Bear Island 34 43.5 5.33
Goat Island 40 45.9 6.21

Respuesta :

Cricetus

Answer:

The correct solution is "[tex]t=\frac{43.5-45.9}{\sqrt{\frac{5.33^2}{34} +\frac{6.21^2}{40} } }[/tex]".

Step-by-step explanation:

The given values are:

Bear island:

n1 = 34

[tex]\bar{x_1}[/tex] = 43.5

S₁ = 5.33

Goat island:

n2 = 40

[tex]\bar{x_2}[/tex] = 45.9

S₂ = 6.21

Now,

⇒  [tex]t=\frac{\bar{x_1}-\bar{x_2}}{\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2} } }[/tex]

On substituting the given values, we get

⇒  [tex]t=\frac{43.5-45.9}{\sqrt{\frac{5.33^2}{34} +\frac{6.21^2}{40} } }[/tex]

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