Respuesta :
Answer:
This data suggest that there is more variability in low-dose weight gains than in control weight gains.
Step-by-step explanation:
Let [tex]\sigma_{1}^{2}[/tex] be the variance for the population of weight gains for rats given a low dose, and [tex]\sigma_{2}^{2}[/tex] the variance for the population of weight gains for control rats whose diet did not include the insecticide.
We want to test [tex]H_{0}: \sigma_{1}^{2} = \sigma_{2}^{2}[/tex] vs [tex]H_{1}: \sigma_{1}^{2} > \sigma_{2}^{2}[/tex]. We have that the sample standard deviation for [tex]n_{2} = 22[/tex] female control rats was [tex]s_{2} = 28[/tex] g and for [tex]n_{1} = 18[/tex] female low-dose rats was [tex]s_{1} = 51[/tex] g. So, we have observed the value
[tex]F = \frac{s_{1}^{2}}{s_{2}^{2}} = \frac{(51)^{2}}{(28)^{2}} = 3.3176[/tex] which comes from a F distribution with [tex]n_{1} - 1 = 18 - 1 = 17[/tex] degrees of freedom (numerator) and [tex]n_{2} - 1 = 22 - 1 = 21[/tex] 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.