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 .05.

Short answer

Reject null hypothesis and there is more variability in the low dose weight gains.

Step-by-step solution

To find the hypothesis

Testing null hypothesis \({H_0}:\sigma _1^2 = \sigma _2^2\)versus alternative hypothesis \({H_a}\)under the assumption that the two populations are normal and independent

Depending on alternative hypothesis \({H_a}\)the \(P\)value is corresponding area under the \({F_{m - 1,n - 1}}\)curve.

The hypotheses of interest are \(H:0:{\sigma _1} = \sigma _2^2\) versus\({H_a}:\sigma _1^2 > \sigma _2^2\), where \({\sigma _1},{\sigma _2}\)denote the variance in weight pain for low-dose treatment and variance in weight gain for control condition, respectively.

The corresponding sample standard deviations are \({s_1} = 54\)and \({s_2} = 32.\)The degrees of freedom are \(m - 1 = 20 - 1 = 19\)and \(n - 1 = 23 - 1 = 22.\)

Finding test statistic

The test statistic is,

\(f = \frac{{s_1^2}}{{s_2^2}}\)

Thus, the value of the statistic is

\(\begin{array}{c}f = \frac{{{{54}^2}}}{{{{32}^2}}}\\ = 2.85\end{array}\)

At significance level \(\alpha = 0.01,\)and mentioned degrees of freedom, value \({F_{0.01,19,22}}\) is

\({F_{0.01,19,22}} = 2.0837\)

Decision rule and conclusion

The test is upper sided, and \({F_{0.05,19,22}} = 2.0837 < 2.85 = f\)

Hence reject null hypothesis at given significance level.

The \(P\)value could have been used as well. For the upper sided test, the \(P\) value is

\(\begin{array}{c}P = P(F > 2.85)\\ = 0.010036\end{array}\)

Where \(F\)has Fisher's distribution with degrees of freedom \({\nu _1} = 19\)and\({\nu _2} = 22\). The value was computed using a software (you can use table estimate the value as mentioned above). Since for the \(P\)the following holds

\(P = 0.010036 < 0.05 = \alpha \)

Reject null hypothesis at any given significance level. Hence there is more variability in the low dose weight gains.