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Suppose the variance in trunk diameter of the giant sequoia tree species is 15.7m^2, while the variance in trunk diameter of the California redwood tree species is 10.6m^2. Let x1 represent the average trunk diameter of four randomly sampled giant sequoia trees, and let x2 represent the average trunk diameter of three randomly sampled California redwood trees. If the random sampling is done with replacement, what is the standard deviation σ(x1−x2) of the sampling distribution of the difference in sample means x1−x2 ?

Sagot :

Ajeigbeibraheemidn

Answer:

[tex]\mathbf{ \sqrt{\dfrac{15.7}{4} + \dfrac{10.6}{3}} }[/tex]

Step-by-step explanation:

Given that:

[tex]\overline x_1 =[/tex] the average diameter of 4 randomly select sequoia trunk from giant trees

[tex]\overline{x_2}[/tex] = the average diameter of 3 randomly selected trunk in California trees.

Thus;

sample size [tex]\mathbf{n_1 = 4 \ \ \& \ \ n_2 = 3}[/tex]

Variance:

[tex]\mathbf{\sigma_1^2 = 15.7 }[/tex]

[tex]\mathbf{\sigma_2^2 = 10.6 }[/tex]

Recall that:

[tex]\mathbf {( \overline x_1 - \overline x_2) \sim Normal \Big ( \mu_1 -\mu_2 \ , \ \dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_12^2}{n_2} \Big )}[/tex]

[tex]\mathbf{Mean ( \overline x_1 - \overline x_2) = \mu_1 - \mu_2}[/tex]

Therefore:

The standard deviation [tex]\mathbf{ \overline x_1 - \overline x_2 = \sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_2^2}{n_2} } }[/tex]

[tex]\mathbf{ \overline x_1 - \overline x_2 = \sqrt{\dfrac{15.7}{4} + \dfrac{10.6}{3}} }[/tex]

MrRoyalidn

The standard deviation of the sampling distribution of the difference in the sample means is 9.95

The given parameters are:

  • Variance: [tex]\sigma_1 ^2 = 15.7[/tex], [tex]\sigma_2 ^2 = 10.6[/tex]
  • Sample size: [tex]n_1 = 4[/tex], [tex]n_2 = 3[/tex].

For a normal distribution we have:

[tex]\bar x_1 - \bar x_2 = \sqrt{\frac{\sigma_1^2}{n_1} +\frac{\sigma_2^2}{n_2}}[/tex]

So, the equation becomes

[tex]\bar x_1 - \bar x_2 = \sqrt{\frac{15.7^2}{4} +\frac{10.6^2}{3}}[/tex]

Evaluate the exponents

[tex]\bar x_1 - \bar x_2 = \sqrt{\frac{246.49}{4} +\frac{112.36}{3}}[/tex]

[tex]\bar x_1 - \bar x_2 = \sqrt{61.62+37.45}[/tex]

Evaluate the sums

[tex]\bar x_1 - \bar x_2 = \sqrt{99.07}[/tex]

Evaluate the square roots

[tex]\bar x_1 - \bar x_2 = 9.95[/tex]

Hence, the standard deviation of the sampling distribution of the difference in the sample means is 9.95

Read more about sampling distribution at:

https://brainly.com/question/17218784

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