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The presence of harmful insects in farm fields is detected by erecting boards covered with a sticky material and then examining the insects trapped on the board. To investigate which colors are most attractive to cereal leaf beetles, researchers placed six boards of each of four colors in a field of oats in July. The table below gives data on the number of cereal leaf beetles trapped. Color Insects trapped

Lemon yellow 45 59 48 46 38 47

White 21 12 14 17 13 17

Green 37 32 15 25 39 41

Blue 16 11 20 21 14 7

a. Is there a significant difference in the average numbers of leaf beetles trapped among the 4 colors? The level of significance is set at 5%. Please carry out the hypothesis testing by setting up null and alternate hypothesis. 𝐻_0: 𝜇_1=𝜇_2=𝜇3= . . . =𝜇𝑘 𝐻_𝑎: At least one mean is different from the others

F^stat > F^critical= Reject H_0 30.55 > 3.10 = Reject H_0

b. If the null hypothesis is rejected in part (a), what can you further conclude as to which group(s) has different averages than others? However, it the null hypothesis is not rejected in part (a), what does it mean in the context of the problem.

Above is the question that I'm trying to solve. I came up with the hypothesis and concluded rejecting H_0, however now I'm stuck here and not sure what to do now. I'm thinking that I need to do a pooled T-test but don't know the formula for a One-way ANOVA pooled T-test. Also, whenever I get my number from the pooled T-test how will this help me to tell if the means are different?

If anyone can help I'd really appreciate it. I'm not just looking for the answer I want to be able to better understand how to solve these types of questions and I've looked into tutoring but my college doesn't have anyone to help tutor for stats so now I'm looking for any help, thanks again.

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Here is a one-factor ANOVA based on your data, according to Minitab.

One-way ANOVA: Y, W, G, B 

Method

Null hypothesis         All means are equal
Alternative hypothesis  At least one mean is different
Significance level      α = 0.05

Equal variances were assumed for the analysis.


Factor Information

Factor  Levels  Values
Factor       4  Y, W, G, B


Analysis of Variance

Source  DF  Adj SS   Adj MS  F-Value  P-Value
Factor   3  4218.5  1406.15    30.55    0.000
Error   20   920.5    46.02
Total   23  5139.0


Model Summary

      S    R-sq  R-sq(adj)  R-sq(pred)
6.78417  82.09%     79.40%      74.21%

The P-value given as 0.000 (meaning < 0.0005) shows that there are highly significant differences among the means. So your null hypothesis in (a) is rejected.

There are various methods of 'ad hoc' analysis to see which colors differ from which other ones. One method is Tukey's HSD method. Because there are ${4 \choose 2} = 6$ pairs of colors to consider, we need to be careful about the 'family error rate'. If we tested each of the six pairs at the 5% level, it is possible that error probabilities would accumulate as we looked at all six. Tukey's method keeps the family error rate below a fixed value, here 5%.

Minitab's first display shows confidence intervals (CIs) for each color separately. Because the CI for Y does not overlap the CI for W, we might speculate that Y and W differ significantly. Making a few such comparisons might be a good way to get an intuitive idea what is going on. But this comparison of individual CIs does not have the 'family rate' protection offered by Tukey's HSD procedure.

Means

Factor  N   Mean  StDev      95% CI
Y       6  47.17   6.79  (41.39, 52.94)
W       6  15.67   3.33  ( 9.89, 21.44)
G       6  31.50   9.91  (25.72, 37.28)
B       6  14.83   5.34  ( 9.06, 20.61)

Pooled StDev = 6.78417

One method of showing Tukey comparisons is shown below. Letters A, B, C are shared by colors that are not significantly different. So we can conclude that colors Y, G, and W are all different from each other. The same is true of colors Y, G, and B. However, colors W and B did not show statistically significant results according to Tukey's HSD method.

Tukey Pairwise Comparisons 

Grouping Information Using the Tukey Method and 95% Confidence

Factor  N   Mean  Grouping
Y       6  47.17  A
G       6  31.50    B
W       6  15.67      C
B       6  14.83      C

Means that do not share a letter are significantly different.

There are dozens of different multiple comparison methods. I don't know which one(s) are covered in your course. Tukey's HSD method is one of the most popular, so I included it, hoping you could find it in your text.

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a) The one-way ANOVA in MS Excel:

$\hspace{1cm}$enter image description here

Since $F=30.55>3.10=F$-critical (or $p=1.15\cdot 10^{-7}<0.05=\alpha$), we reject $H_0$, i.e. not all means are equal.

b) Fisher's least significant difference (LSD) test is used for making pairwise comparisons of population means. The hypotheses:

$$H_0: \ \mu_1=\mu_2; \ \ H_a: \ \mu_1\ne \mu_2.$$ $$LSD=t_{\alpha/2}\sqrt{MSW\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}$$ $$\text{Reject} \ H_0 \ \text{if} \ |\bar{x_1}-\bar{x_2}|>LSD.$$ So: $$t_{\alpha/2,df=n-k}=t_{0.025,20}=2.086;\\ MSW=\frac{\sum_{i=1}^k (n_i-1)s_i^2}{n-k}=\frac{920.5}{20}=46.025;\\ LSD=2.086\sqrt{46.025\left(\frac{1}{6}+\frac{1}{6}\right)}=8.17;\\ |x_1-x_2|=|47.17-15.67|=31.5>8.17 \Rightarrow \text{Reject} \ H_0.$$ Note that when the sample sizes are equal (like in this problem), only one value for LSD is computed. In such cases, we can simply compare the magnitude of the difference between any two means with the value of LSD: $$\begin{align}|x_1-x_3|&=|47.17-31.5|=15.67>8.17 \Rightarrow \text{Reject} \ H_0;\\ |x_1-x_4|&=|47.17-14.83|=32.34>8.17 \Rightarrow \text{Reject} \ H_0;\\ |x_2-x_3|&=|15.67-31.5|=15.83>8.17 \Rightarrow \text{Reject} \ H_0;\\ |x_2-x_4|&=|15.67-14.83|=0.84<8.17 \Rightarrow \text{fail to Reject} \ H_0;\\ |x_3-x_4|&=|31.5-14.83|=16.67>8.17 \Rightarrow \text{Reject} \ H_0.\end{align}$$

Hence, the means of White and Blue are equal, other means are different.

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