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% CO2, 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?
μ1=μ2=μ3=μ4=0
μ = 0
μ1=μ2,μ3=μ4
μ1=μ2=μ3=μ4
Which of the following is the correct alternative
hypothesis?
μ1 ≠ μ2, μ3 ≠ μ4
μ ≠ 0
μi ≠ μj, for
some j ≠ i
μ1 ≠ μ2 ≠ μ3 ≠
μ4
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
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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