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. The article “Reproduction Study of Toxaphene in the Rat” (J. of Environ. Sci. Health, 1988: 101-126) 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 23 female control rats was 32 g and for 20 female low-dose rats was 54 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

Solution

State the hypotheses.

That is, there is no evidence to conclude that there is more variability in low-dose weight gains than in control weight gains.

That is, there is evidence to conclude that there is more variability in low-dose weight gains than in control weight gains.

Obtain the critical value.

From the information, given that

Use EXCEL Procedure for finding the critical value of F at the 0.05 significance level.

Follow the instruction to obtain the critical value of F:

1. Open EXCEL
2. Go to Formula bar.
3. In formula bar enter the function as“=FINV”
4. Enter the probability as 0.05
5. Enter the numerator degrees of freedom as 19.
6. . Enter the numerator degrees of freedom as 22.
7. Click enter

EXCEL output:

From the EXCEL output, the critical value of F at the 0.05 significance level is 2.0837.

Thus, the critical value of F at the 0.05 significance level is 2.0837.