Question

Researchers studying starfish collected two independent random samples of 40 starfish. One sample came from an ocean area in the north, and the other sample came from an ocean area in the south. Of the 40 starfish from the north, 6 were found to be over 8 inches in length. Of the 40 starfish from the south, 11 were found to be over 8 inches in length. Which of the following is the test statistic for the appropriate test to investigate whether there is a difference in proportion of starfish over 8 inches in length in the two ocean areas (north minus south)?A. 6â11640+1140âB. 6â110.1540+0.27540âC. 0.15â0.275(0.15)(0.275)(140+140)âD. 0.15â0.275(0.2125)(0.7875)(140+140)âE. 0.15â0.275(0.2125)(0.7875)140+140â

Answer 1

Answer:

The test statistic for the appropriate test is $\frac{0.15-0.275}{\sqrt{0.2125(1-0.2125)[\frac{1}{40}+\frac{1}{40}]}}$.

Step-by-step explanation:

The experiment conducted here is to determine whether there is a difference in proportion of starfish over 8 inches in length in the two ocean areas, i.e. north and south.

The hypothesis to test this can be defined as follows:

H₀: There is no difference in proportion of starfish over 8 inches in length in the two ocean areas, i.e. p₁ = p₂.

Hₐ: There is a difference in proportion of starfish over 8 inches in length in the two ocean areas, i.e. p₁ ≠ p₂.

The two-proportion z-test would be used to perform the test.

A sample of n = 40 starfishes are selected from both the ocean areas.

It provided that of the 40 starfish from the north, 6 were found to be over 8 inches in length and of  the 40 starfish from the south, 11 were found to be over 8 inches in length.

Compute the sample proportion of starfish from north that were over 8 inches in length as follows:

$\hat p_{n}=\frac{6}{40}=0.15$

Compute the sample proportion of starfish from south that were over 8 inches in length as follows:

$\hat p_{s}=\frac{11}{40}=0.275$

The test statistic is:

$z=\frac{\hat p_{n}-\hat p_{s}}{\sqrt{P(1-P)[\frac{1}{n_{n}}+\frac{1}{n_{s}}]}}$

Compute the combined proportion P as follows:

$P=\frac{X_{n}+X_{s}}{n_{n}+n_{s}}=\frac{6+11}{40+40}=0.2125$

Compute the test statistic value as follows:

$z=\frac{\hat p_{n}-\hat p_{s}}{\sqrt{P(1-P)[\frac{1}{n_{n}}+\frac{1}{n_{s}}]}}$

  $=\frac{0.15-0.275}{\sqrt{0.2125(1-0.2125)[\frac{1}{40}+\frac{1}{40}]}}$

Thus, the test statistic for the appropriate test is $\frac{0.15-0.275}{\sqrt{0.2125(1-0.2125)[\frac{1}{40}+\frac{1}{40}]}}$.