Question

In a scientific study that tries to relate the biodiversity of macro invertebrates to the acidity of the water near mines, samples of macro invertebrates were taken at the rate of a monthly sample for 12 months in two different locations. For the first locality, an average biodiversity index of 3.11 and a sample standard deviation of 0.771 were obtained. For the second locality the average biodiversity index was 2.04 with a standard deviation of 0.448.
to. BUILD and INTERPRET a 90% confidence interval for the difference in the means of the two localities assuming that the populations conform approximately to a normal population and that the variances can be considered equal.

b. Assess whether the evidence supports the presumption of equal variances

c. Would it have been better to use a 95% range? Explain in detail

Answer 1

x1 = 3.11

s1 = 0.771

x2 = 2.04

s2 = 0.448

n1 = 12

n2 = 12

(a) s2p = (12 - 1)*0.771^2 + (12 - 1)*0.448^2/(12 + 12 - 2) = 0.398

se = 0.398*(1/12 + 1/12) = 0.257

The 90% confidence interval for the difference in the means of the two localities is:

= (3.11 - 2.04) 1.72*0.257

= 0.63, 1.51

(b) Ratio of standard deviations = 0.771/0.448 = 1.72 < 2

Since the ratio is less than 2, the evidence supports the presumption of equal variances.

(c) A 95% range would contain 95% of the data giving a wider confidence interval, thus a 95% range would help us strengthen the conclusion by including more data. Thus, it would have been better to use a 95% range.

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