Researchers conducted an experiment to test the effects of alcohol. Errors were recorded in a test of visual and motor skills for a treatment group of 24 people who drank ethanol and another group of 24 people given a placebo. The errors for the treatment group have a standard deviation of 2.40 , and the errors for the placebo group have a standard deviation of 0.75. Use a 0.05 significance level to test the claim that the treatment group has errors that vary significantly more than the errors of the placebo group. Assume that the two populations are normally distributed.

Answer: We reject the null hypothesis and conclude that the treatment group has errors that vary significantly more than the errors of the placebo group.

Step-by-step explanation:

Null hypothesis : Var1 = Var 2

Alternative hypothesis : Var1 > Var 2

The F test statistic is the ratio of squared sample standard deviation:

F = (2.4) ^2 / (0.75) ^2 = 5,76 / 0,5625 = 10.24

Now we have to get the critical from the F table with degrees of freedom (n1-1, n2-1) = (23,23):

From the F table, F (0.05) = between 2.0050 and 2.0476.

General rule : if the value of the test statistic is in a rejection region. Reject We reject the Null hypothesis.

F = 10.24 > 2.0476. Therefore we reject the null hypothesis.

Conclusion: The errors of the treatment group vary more than the error of the placebo group.