Question

Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of toxaphene exposure on animals, groups of rats were given toxaphene in their diet. A study reports weight gains (in grams) for rats given a low dose (4 ppm) and for control rats whose diet did not include the insecticide. The sample standard deviation for 22 female control rats was 28 g and for 18 female low-dose rats was 51 g. Does this data suggest that there is more variability in low-dose weight gains than in control weight gains? Assuming normality, carry out a test of hypotheses at significance level 0.05.

Answer 1

This data suggest that there is more variability in low-dose weight gains than in control weight gains.

Explanation:

Let
`\sigma_(1)^(2)` be the variance for the population of weight gains for rats given a low dose, and
`\sigma_(2)^(2)` the variance for the population of weight gains for control rats whose diet did not include the insecticide.

We want to test
`H_(0): \sigma_(1)^(2) = \sigma_(2)^(2)` vs
`H_(1): \sigma_(1)^(2) > \sigma_(2)^(2)`. We have that the sample standard deviation for
`n_(2) = 22` female control rats was
`s_(2) = 28` g and for
`n_(1) = 18` female low-dose rats was
`s_(1) = 51` g. So, we have observed the value

`F = (s_(1)^(2))/(s_(2)^(2)) = ((51)^(2))/((28)^(2)) = 3.3176` which comes from a F distribution with
`n_(1) - 1 = 18 - 1 = 17` degrees of freedom (numerator) and
`n_(2) - 1 = 22 - 1 = 21` degrees of freedom (denominator).

As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.