Problem-1

Answer: according to the question, we can get the analysis of variance by using the excel

schedule 1 | schedule 2 | schedule 3 | 50 | 60 | 70 | 60 | 65 | 75 | 70 | 66 | 55 | 45 | 54 | 40 | 40 | 57 | 55 |

Anova：single factor | | | | | | | | | | | | SUMMARY | | | | | | | Groups | Count | Sum | Average | Variance | | | schedule 1 | 5 | 265 | 53 | 145 | | | schedule 2 | 5 | 302 | 60.4 | 26.3 | | | schedule 3 | 5 | 295 | 59 | 192.5 | | | | | | | | | | | | | | | | | ANOVA | | | | | | | Source of Variance | SS | df | MS | F | P-value | F crit | Between Group | 154.5333 | 2 | 77.26667 | 0.637163 | 0.545774 | 3.885294 | Within Groups | 1455.2 | 12 | 121.2667 | | | | | | | | | | | Total | 1609.733 | 14 | | | | |

From the above Anova, we find that the F < F crit, therefore we can conclude that at 95% confidence determine, there is not a significant difference in the mean weekly units of production for the three types of work schedules.

Pronlem-2

In the question, first we can get the specific data for the three diets, and make a table for them:

Diet A | Diet B | Diet C | 14 | 12 | 25 | 18 | 10 | 32 | 20 | 22 | 18 | 12 | 12 | 14 | 20 | 16 | 17 | 18 | 12 | 14 |

Then we can use the excel to get the analysis of variance:

Anova：single factor | | | | | | | | | | | | SUMMARY | | | | | | | Groups | Count | Sum | Average | Variance | | | Diet A | 6 | 102 | 17 | 10.8 | | | Diet B | 6 | 84 | 14 | 19.2 | | | Diet C | 6 | 120 | 20 | 50.8 | | | | | | | | | | | | | | | | | Anova | | | | | | | Source of Variance | SS | df | MS | F | P-value | F crit | Between Group | 108 | 2 | 54 | 2.00495 | 0.169175 | 3.68232 | Within Groups | 404 | 15 | 26.93333 | | | | | | | | | | | Total | 512 | 17 | | | | |

From the above table, we find that the value of F is smaller than the value of F crit, so we can conclude that there is no difference in the effectiveness of the three diets, and I would not advise the dietician about the effectiveness of the three diets.

Problem-3

After analyzing the data got from the question, we can make a table for all the data: Store 1 | Store 2 | Store 3 | 46 | 34 | 33 | 47 | 36 | 31 | 45 | 35 | 35 | 42 | 39 | | 45 | | |

Then we can use the Excel to analyze the variance, and get the analysis of the variance for all the data above:

Anova：single factor | | | | | | | | | | | | SUMMARY | | | | | | | Groups | Count | Sum | Average | Variance | | | Store 1 | 5 | 225 | 45 | 3.5 | | | Store 2 | 4 | 144 | 36 | 4.666667 | | | Store 3 | 3 | 99 | 33 | 4 | | | | | | | | | | | | | | | | | Anova | | | | | | | Source of Variance | SS | df | MS | F | P-value | F crit | Between Group | 324 | 2 | 162 | 40.5 | 3.16E-05 | 4.256495 | Within Groups | 36 | 9 | 4 | | | | | | | | | | | Total | 360 | 11 | | | | |

So we get the value of F is much higher than the value of F crit, and we can get the conclusion that there is a significant difference in the average sale of the three stores.

The mean sales of the Store 1 is the highest, therefore we can conclude that the place where the store locates has the highest sales.

Problem-4

Red | Yellow | Blue | 43 | 52 | 61 | 52 | 37 | 29 | 59 | 38 | 38 | 76 | 64 | 53 | 61 | 74 | 79 | 81 | | |

Analyze this data with Excel and we can get the analysis of variance for them:

Anova：single factor | | | | | | | | | | | | SUMMARY | | | | | | | Groups | Count | Sum | Average | Variance | | | Red | 6 | 372 | 62 | 205.6 | | | Yellow | 5 | 265 | 53 | 261 | | | Blue | 5 | 260 | 52 | 384 | | | | | | | | | | | | | | | | | Anova | | | | |…