Question

The zoology department of a small college did a survey of the number of organisms per square meter at two different locations in the Roanoke River drainage basin. At each location they created a grid and took a random sample of 8 square-meter parcels in which they counted organisms. Here is the data:

	Number of organisms per square meter
Location 1	5030	2200	11400	10730	15040	7330	6890	2800
Location 2	2800	4670	6890	7720	3320	7330	2190	7720

They are interested in testing to see if the average densities of organisms is the same at both locations. They checked that the conditions for doing a two-sample t test are satisfied and computed the P value to be

	A.	0.232
	B.	0.215
	C.	less than 0.05
	D.	0.108

The degrees of freedom for the test above is (to two decimal places)

Answer 1

Claim: The average densities of organisms is the same at both locations.

The null and alternative hypothesis is

Level of significance = 0.05

For doing this test first we have to check the two groups have population variances are equal or not.

The null and alternative hypothesis is

Test statistic is

F = largest sample variance / Smallest sample variances

n1 = 8

= 7677.5

= 19846336

n2 = 8

= 5330

= 5513429

F = 18846336 / 5513429 = 3.60

Degrees of freedom = n1 - 1 , n2 - 1 = > 8 - 1 , 8 - 1 = > 7 , 7

Critical value = 3.783

F < critical value we fail to reject null hypothesis.

Conclusion: Population variances are equal, so we have to use pooled variance.

Test statistic is

Degrees of freedom = n1 + n2 - 2 = 8 + 8 -2 = 14

P-value = 2*P(T > 1.32) = 2*0.116 = 0.232

P-value > 0.05 we fail to reject null hypothesis.

Conclusion:

The average densities of organisms is the same at both locations.