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 21 female control rats was 30 g and for 19 female low-dose rats was 58 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

Answer:

Step-by-step explanation:

Hello!

The objective of this experiment is to test if the presence of toxaphene in the diet reduces the gain of weight. For these two random samples of female rats were made, one was fed with a diet that contained a low dose of toxaphene and the other, called the control group, was fed with the same type of food but without toxaphene. The weight of the rats was registered in both groups obtaining:

Sample 1 (with toxaphene)

n₁= 19 female rats

S₁= 58g

Sample 2 (control group)

n₂= 21

S₂= 30g

The hypothesis is that the presence of low-dose insecticide in the diet increases the variability in weight gain. Symbolically: σ₁² > σ₂²

The hypothesis is:

H₀: σ₁² ≤ σ₂²

H₁: σ₁² > σ₂²

α: 0.05

This hypothesis test is for the variance ratio, the statistic to use is:

F= (S₁²/S₂²)*( σ₁²/σ₂²)~F$_{n_1-1;n_2-1}$

This test is one-tailed (right) and the critical value is:

$F_{n_1-1; n_2-1; 1-\alpha } = F_{18;20;0.95} = 2.19$

If F ≥ 2.19, you reject the null hypothesis.

If F < 2.19, you support the null hypothesis.

F= (S₁²/S₂²)*( σ₁²/σ₂²) = (3364/900)*1 = 3.737

Since the calculated value is greater than the critical value, the decision is to reject the null hypothesis. The variability of the weight gain of rats fed with a low dose of insecticide is greater than the variability of the control group.

I hope it helps!