# 9.12: Chapter 12 Solution (Practice + Homework)

- Page ID
- 51861

Figure \(\PageIndex{10}\)

While there are differences in spread, it is not unreasonable to use ANOVA techniques. Here is the completed ANOVA table:

Source of variation | Sum of squares (\(SS\)) | Degrees of freedom (\(df\)) | Mean square (\(MS\)) | \(F\) |
---|---|---|---|---|

Factor (Between) | \(37.748\) | \(4 – 1 = 3\) | \(12.5825\) | \(26.272\) |

Error (Within) | \(11.015\) | \(27 – 4 = 23\) | \(0.4789\) | |

Total | \(48.763\) | \(27 – 1 = 26\) |

\(P(F > 1.5521) = 0.2548\)

Since the p-value is so large, there is not good evidence against the null hypothesis of equal means. We cannot reject the null hypothesis. Thus, for 2012, there is not any have any good evidence of a significant difference in mean number of wins between the divisions of the American League.

**64**.

\(SS_{between} = 26\)

\(SS_{within} = 441\)

\(F = 0.2653\)

**67**.

\(df(denom) = 15\)

**69**.

- \(H_{0} : \mu_{L}=\mu_{T}=\mu_{J}\)
- \(H_a\): at least any two of the means are different
- \(df(num) = 2; df(denom) = 12\)
- \(F\) distribution
- 0.67
- 0.5305
- Check student’s solution.
- Decision:Cannot reject null hypothesis; Conclusion: There is insufficient evidence to conclude that the means are different.

**72**.

- \(H_{a} : \mu_{C}=\mu_{n}=\mu_{h}\)
- At least any two of the magazines have different mean lengths.
- \(df(num) = 2, df(denom) = 12\)
- \(F\) distribution
- \(F = 15.28\)
- \(p\)-value = 0.001
- Check student’s solution.

**74**.

- \(H_{0} : \mu_{o}=\mu_{h}=\mu_{f}\)
- At least two of the means are different.
- \(df(n) = 2, df(d) = 13\)
- \(F_{2,13}\)
- 0.64
- 0.5437
- Check student’s solution.

**76**.

- \(H_{0} : \mu_{p}=\mu_{m}=\mu_{h}\)
- At least any two of the means are different.
- \(df(n) = 2, df(d) = 12\)
- \(F_{2,12}\)
- 3.13
- 0.0807
- Check student’s solution.

**78**.

The data appear normally distributed from the chart and of similar spread. There do not appear to be any serious outliers, so we may proceed with our ANOVA calculations, to see if we have good evidence of a difference between the three groups.

\(H_{0} : \mu_{1}=\mu_{2}=\mu_{3}\), \(H_{a} : \mu_{1} \neq \mu_{2} \neq \mu_{3}\);

Define \(\mu_{1}, \mu_{2}, \mu_{3}\), as the population mean number of eggs laid by the three groups of fruit flies.

\(F\) statistic = 8.6657;

\(p\)-value = 0.0004

Figure \(\PageIndex{12}\)

** Decision:** Since the \(p\)-value is less than the level of significance of 0.01, we reject the null hypothesis.

** Conclusion:** We have good evidence that the average number of eggs laid during the first 14 days of life for these three strains of fruitflies are different.

Interestingly, if you perform a two sample \(t\)-test to compare the RS and NS groups they are significantly different (\(p = 0.0013\)). Similarly, SS and NS are significantly different (\(p = 0.0006\)). However, the two selected groups, RS and SS are not significantly different (\(p = 0.5176\)). Thus we appear to have good evidence that selection either for resistance or for susceptibility involves a reduced rate of egg production (for these specific strains) as compared to flies that were not selected for resistance or susceptibility to DDT. Here, genetic selection has apparently involved a loss of fecundity.