Question

Exercises 8.41 to 8.45 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { A } & 5 & 10.200 & 2.864 \\ \text { B } & 5 & 16.800 & 2.168 \\ \text { C } & 5 & 10.800 & 2.387\end{array}\) \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 2 & 133.20 & 66.60 & 10.74 & 0.002 \\ \text { Error } & 12 & 74.40 & 6.20 & & \\ \text { Total } & 14 & 207.60 & & & \end{array}\) Is there sufficient evidence of a difference in the population means of the three groups? Justify your answer using specific value(s) from the output.

Short answer

Yes, there is sufficient evidence of a difference in the population means of the three groups. This conclusion is based on the p-value, which is 0.002, far below the typical 0.05 cutoff for significance, and the F value, indicating that the variability among group means significantly exceeds that expected by chance.

Step-by-step solution

Identify the relevant values

From the second table, the F statistic is given as 10.74 and the p-value is given as 0.002.

Analyze the p-value

The p-value is 0.002, which is less than 0.05. Therefore, this means that if the null hypothesis were true (that is, all group means are equal), we would only get an F-statistic this large 0.2% of the time.

Conclusion

Since the p-value is less than the significance level (0.05), this provides sufficient evidence to reject the null hypothesis. Therefore, there is a significant difference in the means of the three groups.