Question

Perhaps the mice with light at night in Exercise 8.28 gain more weight because they are exercising less. The conditions for an ANOVA test are met and computer output is shown for testing the average activity level for each of the three light conditions. Is there a significant difference in mean activity level? State the null and alternative hypotheses, give the F-statistic and the p-value, and clearly state the conclusion of the test.\(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { DM } & 10 & 2503 & 1999 \\ \text { LD } & 8 & 2433 & 2266 \\ \text { LL } & 9 & 2862 & 2418\end{array}\)

Short answer

As we lack the F-statistic and p-value in the provided data, we can't determine whether there is a significant difference in mean activity levels under different light conditions.

Step-by-step solution

State the Hypotheses

The null hypothesis (\(H_0\)) is that there is no significant difference between the means of the three light conditions (DM, LD, LL). The alternative hypothesis (\(H_A\)) is that there is a significant difference between the means in at least one of the light conditions.

Compute the F-Statistic

The F-statistic in an ANOVA test is calculated by dividing the variance between the groups by the variance within the groups. However, the F-statistic is usually provided in computer output in statistical software which is in this case not given. So the exact value of the F-statistic can't be calculated here.

Compute the p-value

The p-value, similar to the F-statistic, is a value that is usually provided in the computer output which is not given in this case. As such, we don't have precise numbers to determine this in our exercise.

State the Conclusion

Normally, had we the F-statistic and the p-value, we'd compare the p-value with our significance level (\(\alpha\)), which is often 0.05. If the p-value was less than \(\alpha\), we would reject the null hypothesis in favor of the alternate one, indicating a significant difference in mean activity levels. If it was higher, we would fail to reject the null thus saying there's insufficient evidence to say the means vary. However, we don't have these values in this case.