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.

a). 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

a)

90% confidence interval for the difference between the population means μ1​−μ2​, for the case that the population standard deviations are not known.

we assume that the population variances are equal, so then the number of degrees of freedom are df=n1​+n2​−2=12+12−2=22.

The critical value for α=0.1 and df=22 degrees of freedom is tc​=t1−α/2;n−1​=1.717.

Since the population variances are assumed to be equal

90% confidence interval for the difference between the population means 0.628<μ1​−μ2​<1.512

b)

95% confidence interval for the difference between the population means 0.536<μ1​−μ2​<1.604