STA 106 Discussion 8
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STA 106 Discussion 8
Problem I
1. A study gave three different beers “head formation” score (the “head” of a beer is the foam at the top after you pour the beer). Larger values correspond to larger “heads” on beer. Some information follows:
MSE = 347.3535, MSA = 5036.275, nt = 59, 95% CI for σµ(2)/σY(2) : (0.1154137, 0.9657261) (a) Interpret the confidence interval given in terms of the problem.
We are 95% confident that the proportion of variance in beer “head” score explained by the random effects of beer is between 11.541374% and 96.5726098%.
(b) Can we tell from this confidence interval if the random effects are statistically significant? Explain your answer.
No. The range is large, and besides that it is not a test for statistically significant effects. That can only be done through a hypthesis test.
(c) Calculate and interpret and estimate for σY .
n* = 
( i ni ′ 戎(戎)n(n)t戎i ) = 
(59 ′ 
) = 19.6610169
σˆ2 µ = 
= 
= 238.4882. 
e(2) = MSE = 347.3535.
Y = ←
µ(2) + 
µ(2) = ≥347.3535 + 238.4882 = 24.2042
A typical deviation of beer “head” score from the mean score is: 24.2042
(d) The F test-statistic and p-value for Single Factor ANOVA are : 14.163 , 0.00001053. Do you belive we should use Single Factor ANOVA, or the Random Effects model for this data? Explain your answer.
Random effects - this is because the 3 beers selected certainly do not represent all possible beers there are, and they were likely randomly selected from a larger population. Therefore, whatever effect we see from beer was random.
2. Continue with the data from problem 1. A boxplot of the three beers “head” formation scores follow, along with some summary statistics:
|
|
B1 |
B2 |
B3 |
Overall |
|
Group Mean |
168.60 |
137.20 |
157.42 |
154.36 |
|
Group SD |
11.36 |
29.75 |
6.43 |
22.79 |
|
Rank Mean |
45.30 |
20.23 |
24.18 |
30.00 |
|
Sample Size |
20 |
20 |
19 |
59 |
(a) Would it be appropriate to use Random Effects ANOVA for this problem? Explain why or why not.
No. The variances are wildly different, and there appear to be outliers. It is quite likely that the ANOVA assumptions are violated (although we could check them).
(b) Calculate the value of KWOBS , assuming SR(2) = 294.47. Why might it be appropriate to use the ranks of the data in this case?
KWOBS = 
i(a)=1 ni (
i ′ n乞
)2
= 
[20(45.3 ′ 
)2 + 20(20.225 ′ 
)2 + 19(24.1842105 ′ 
)2 )2 ] = 0.0033959(4681.8 + 1911.0125 + 642.6447) = 24.5714925
(c) For 1000 random permutations, the distribution of KW values follows:
Using this, estimate the p-value for the Kruskal-Wallis test.
|
KW* |
2 |
4 |
6 |
8 |
10 |
12 |
14 |
|
(#KWi 一 KW * ) |
737 |
284 |
94 |
32 |
11 |
2 |
2 |
We estimate the probability is less than 2/2000, or 0.001 (since 24.5711) is larger than any given value on the table).
(d) State the appropriate null and alternative, and interpret the p-value from (c) in terms of the problem.
H0 : The distribution of ranks for all three types of beer are equal. HA : At least one distribution of ranks for the types of beer is different.
If in reality the distribution of ranks for the three beers were equal, we would observe our data or more extreme with probability < 0.001.
2022-05-25