Chapter 27, Problem 27.22AYK

(a)

To determine

To find the median number of beetles trapped by boards of each color and find out which colors appear more effective.

Answer to Problem 27.22AYK

The color which attracted more insects is Yellow.

Explanation of Solution

In the question, it is given that the researchers want to examine the insects trapped on the boards with different colors. Thus, the data is given on the insects trapped by the different colored boards. Now, to find the median we will use the excel formula which is,

  $=\text{Median}(Number1,Number2,\mathrm{...})$

Thus, the formula used will be as:

	Blue	Green	White	Yellow
	16	37	21	45
	11	32	12	59
	20	20	14	48
	21	29	17	46
	14	37	13	38
	7	32	20	47
Median=	=MEDIAN(AZ3:AZ8)	=MEDIAN(BA3:BA8)	=MEDIAN(BB3:BB8)	=MEDIAN(BC3:BC8)

And the result will be as:

	Blue	Green	White	Yellow
	16	37	21	45
	11	32	12	59
	20	20	14	48
	21	29	17	46
	14	37	13	38
	7	32	20	47
Median=	15	32	15.5	46.5

Now, by looking at the median, the color which attracted more insects is Yellow and then after that most attracted insects by color is White, as their median is largest and second largest.

(b)

To determine

To find out what hypotheses does ANOVA test and Kruskal-Wallis test.

Explanation of Solution

In the question, it is given that the researchers want to examine the insects trapped on the boards with different colors. Thus, the data is given on the insects trapped by the different colored boards. Now, the hypotheses defined by the ANOVA test will be as:

  $\begin{array}{l}{H}_0:{\mu }_1={\mu }_2={\mu }_3={\mu }_4\\ H_1:\text{Not all four means are equal}\text{.}\end{array}$

And the hypotheses defined by the Kruskal-Wallis test will be as:

Null hypothesis: Insect trapped have same distribution in all groups.

Alternative hypothesis: They are all different from each other.

(c)

To determine

To find out what are $k,n_i,N$ and arrange the counts in order and assign ranks.

Answer to Problem 27.22AYK

  $k=4,n_i=6,N=24$ .

Explanation of Solution

In the question, it is given that the researchers want to examine the insects trapped on the boards with different colors. Thus, the data is given on the insects trapped by the different colored boards. Since there are four colors therefore there are four population, thus, $k=4$ . And also the observations under each observations is six, thus, $n_i=6$ . The total number of observations will be therefore:

  $\begin{array}{l}N=4\times 6\\ =24\end{array}$

Now, we will arrange the insects trapped by different colors in order and assign ranks with the help of Excel using the sort and filter command in the home tab. Thus, the result will be as:

Label	Data	Rank
Blue	16	7
Blue	11	2
Blue	20	10
Blue	21	12.5
Blue	14	5.5
Blue	7	1
Green	37	17.5
Green	32	15.5
Green	20	10
Green	29	14
Green	37	17.5
Green	32	15.5
White	21	12.5
White	12	3
White	14	5.5
White	17	8
White	13	4
White	20	10
Yellow	45	20
Yellow	59	24
Yellow	48	23
Yellow	46	21
Yellow	38	19
Yellow	47	22

(d)

To determine

To calculate the Kruskal-Wallis statistics H and find how many degrees of freedom should you use for the chi-square approximation to its null distribution and use chi-square table to give approximate P-value and what does the test lead you to conclude.

Answer to Problem 27.22AYK

  $H=18.51$ and degrees of freedom is three and $P<0.0005$ .

Explanation of Solution

In the question, it is given that the researchers want to examine the insects trapped on the boards with different colors. Thus, the data is given on the insects trapped by the different colored boards. Since there are four colors therefore there are four population, thus, $k=4$ . And also the observations under each observations is six, thus, $n_i=6$ . The total number of observations will be therefore:

  $\begin{array}{l}N=4\times 6\\ =24\end{array}$

And the arrange order is given above in part (c). Thus, we will sum the rank for each color and the result will be as:

Colours	Sum of ranks
Blue	38
Green	90
White	43
Yellow	129

Now, the Kruskal-Wallis statistics H can be calculated as:

  $\begin{array}{l}H=\frac{12}{N(N+1)}{\displaystyle \sum \frac{R_i{}^2}{n_i}-3(N+1)}\\ =\frac{12}{24(24+1)}{\displaystyle \sum (\frac{{38}^2}{6}+\frac{{90}^2}{6}+\frac{{43}^2}{6}}+\frac{{129}^2}{6})-3(24+1)\\ =18.51\end{array}$

The degrees of freedom will be as:

  $\begin{array}{l}k-1=4-1\\ =3\end{array}$

Now, we will look at the chi-square table for the P-value as:

  $P<0.0005$

As we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, so we have,

  $P<0.05\Rightarrow \text{Reject }{H}_0$

Thus, we have sufficient evidence to conclude that the insects trapped by different colors are statistically different from each other.