Question

In Exercises 5–16, test the given claim.
Testing Effects of Alcohol 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 22people who drank ethanol and another group of 22 people given a placebo. The errors for the treatment group have a standard deviation of 2.20, and the errors for the placebo group have a standard deviation of 0.72 (based on data from “Effects of Alcohol Intoxication on RiskTaking, Strategy, and Error Rate in Visuomotor Performance,” by Streufert et al., Journal of Applied Psychology, Vol. 77, No. 4). 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.

Short answer

There is enough evidence to support the claim that the treatment group has errors that vary significantly more than the errors of the placebo group.

Step-by-step solution

Given information

The errors that were recordedtest of visual and motor skills have a standard deviation of 2.20 for the treatment group of 22 people who were given ethanol.

The errors that were recordedtest of visual and motor skills have a standard deviation of 0.72 for the group of 22 people who were given a placebo.

It is claimed that the variation in the errors for the treatment group is greater than the variation in the errors for the placebo group.

Hypotheses

Let\({\sigma _1}\)and\({\sigma _2}\)be the population standard deviations of errorsfor the treatment group and the placebo group, respectively.

Null Hypothesis: The populationstandard deviation of errors for the treatment group is equal to the populationstandard deviation of errors for the placebo group.

\({H_0}:{\sigma _1} = {\sigma _2}\)

Alternative Hypothesis: The populationstandard deviation of errors for the treatment group is greater than the populationstandard deviation of errors for the placebo group.

\({H_1}:{\sigma _1} > {\sigma _2}\)

Compute the test statistic

Since two independent samples involve a claim about the population standard deviation, apply an F-test.

Consider the larger sample variance to be\(s_1^2\)and the corresponding sample size to be\({n_1}\).

The following values are obtained:

\({\left( {2.20} \right)^2} = 4.84\)

\({\left( {0.72} \right)^2} = 0.5184\)

Here,\(s_1^2\)is the sample variance corresponding to the treatment group and has a value equal to 4.84.

\(s_2^2\)is the sample variance corresponding to the placebo group and has a value equal to 0.5184.

Substitute the respective values to calculate the F statistic:

\(\begin{array}{c}F = \frac{{s_1^2}}{{s_2^2}}\\ = \frac{{{{\left( {2.20} \right)}^2}}}{{{{\left( {0.72} \right)}^2}}}\\ = 9.336\end{array}\)

Thus, F is equal to 9.336.

Critical value and p-value

The value of the numerator degrees of freedom is equal to:

\(\begin{array}{c}{n_1} - 1 = 22 - 1\\ = 21\end{array}\)

The value of the denominator degrees of freedom is equal to:

\(\begin{array}{c}{n_2} - 1 = 22 - 1\\ = 21\end{array}\)

For the F test, the critical value corresponding to the right-tail is considered.

The critical value can be obtained using the F-distribution table with numerator degrees of freedom equal to 21 and denominator degrees of freedom equal to 21 for a right-tailed test.

The level of significance is equal to 0.05.

Thus, the critical value is equal to 2.0842.

The right-tailed p-value for F equal to 9.336 is equal to 0.0000.

Conclusion

Since the test statistic value is greater than the critical value and the p-value is less than 0.05, the null hypothesis is rejected.

Thus, there is enough evidence to supportthe claimthat the treatment group has errors that vary significantly more than the errors of the placebo group.