Question

**Meat packaged in ambient air only has a shelf life of
about two days before the quality of the meat begins to degrade due
to microbial growth. A group of scientists performed an experiment
to test alternative methods of meat packing to assess whether any
method reduced microbial growth. The scientists applied four
different packing methods to equally sized batches of meat,
standard air, 100% CO _{2}, vacuum packaging and a mix of
different gases. They then counted the amount of baterial growth
that occured per cm squared and took the natural log of that. They
are interested in knowing if there is a difference in microbial
growth between the four methods. They have decided to conduct their
experiment at significance level α = 0.001. The results of their
experiment are given in the csv file below. Use R to analyze the
results of their experiment. If you can't use the button to
download the csv file for any reason, the data is given in a table
at the bottom of this page.**

**Copy of data in CSV file:**

Air | CO2 | Vacuum | Mix |

9.64 | 7.95 | 7.70 | 8.89 |

8.96 | 9.96 | 8.62 | 6.86 |

10.04 | 8.22 | 10.02 | 9.32 |

9.80 | 8.98 | 7.59 | 7.54 |

9.59 | 7.47 | 9.56 | 8.46 |

10.23 | 9.07 | 9.25 | 10.36 |

7.90 | 9.27 | 9.46 | 8.97 |

9.04 | 10.48 | 7.54 | 9.34 |

9.23 | 9.38 | 9.00 | 8.58 |

7.76 | 8.99 | 9.69 | 9.70 |

8.37 | 8.95 | 8.59 | 8.19 |

8.13 | 9.60 | 9.85 | 8.45 |

9.27 | 8.96 | 9.10 | 9.79 |

7.99 | 7.74 | 8.98 | 8.71 |

9.14 | 9.59 | 9.62 | 9.71 |

9.61 | 9.05 | 9.84 | 8.61 |

9.67 | 9.04 | 7.82 | 8.48 |

10.12 | 9.46 | 8.26 | 8.76 |

**Which of the following is the correct null
hypothesis?**

μ_{1}=μ_{2}=μ_{3}=μ_{4}=0

μ = 0

μ_{1}=μ_{2},μ_{3}=μ_{4}

μ_{1}=μ_{2}=μ_{3}=μ_{4}

**Which of the following is the correct alternative
hypothesis?**

μ_{1} ≠ μ_{2}, μ_{3} ≠ μ_{4}

μ ≠ 0

μ_{i} ≠ μ_{j}, *for*
*some* *j* ≠ *i*

μ_{1} ≠ μ_{2} ≠ μ_{3} ≠
μ_{4}

**You want to calculate the means for each group. Assume
the data from the csv file was just read in and held in the
variable meatdata. Which of the following would be a correct way to
calculate the means for each group?**

means <- colMeans(meatdata)

means <- rowMeans(meatdata)

means <- mean(meatdata)

means <- average(meatdata)

**Compute the value of the grand mean,**

**Calculate the test statistic**

**Which of the following is the correct R function that
can be used to find the p value of an F-distribution?**

pf()

rf()

df()

qt()

**What is the p value?**

**Do we reject the null hypothesis?**

Yes, we reject the null hypothesis

No, we fail to reject the null hypothesis

Answer #1

3) means = colMeans( meatdata) is correct

4)grand mean=8.970278 #obtain in code

5) test statistic value from Anova . # see in code

F value =0.514

6)

Pf is R function to calculate p-value of f statistics.

7) p value from Anova

P value=0.674

8) decision criteria for testing hypothesis is if p value less than alpha

Then reject null hypothesis

Here alpha = 0.001

P-value >0.001

Therefore accept null hypothesis (Ho)

That is all means are equal.

In other words there is no difference in microbial growth between the four treatments or methods.

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