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.

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

Step-by-step explanation:

Let be the variance for the population of weight gains for rats given a low dose, and the variance for the population of weight gains for control rats whose diet did not include the insecticide.

We want to test vs . We have that the sample standard deviation for female control rats was g and for female low-dose rats was g. So, we have observed the value

which comes from a F distribution with degrees of freedom (numerator) and 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.

the f

last graph i believe

step-by-step explanation:

worked it out i believe

answer: procent 54 wynosi 33,33%