Chapter 27, Problem 27.8AYK

(a)

To determine

To explain is there evidence that the presence of nine weeds per meter reduces corn yields when compared with weed-free corn and use Wilcoxon rank sum test with the data given and part of the data from example $27.1$ .

Answer to Problem 27.8AYK

We have sufficient evidence to conclude that the presence of nine weeds per meter reduces corn yields when compared with weed-free corn.

Explanation of Solution

In the question, it is given that the corn yield study of example $27.1$ also examined yields in four plots having nine plants per meter. We have to compare it with the weed free corn. So, the data will be as:

0-weed	9-weed
166.7	162.8
172.2	162.7
165	162.4
176.9	142.4

Now, let us use the software to conduct the Wilcoxon test. The hypotheses are defined as: Null hypothesis: There is no difference between them and Alternative hypothesis: The zero-weed field increases corn yield than nine-weed yield. Thus, we have the result as:

n	sum of ranks
4	26	0-weed
4	10	9-weed
8	36	total

18.00	expected value
3.46	standard deviation
2.17	z
.0152	p-value (one-tailed, upper)

No.	Label	Data	Rank
1	0-weed	166.7	6
2	0-weed	172.2	7
3	0-weed	165	5
4	0-weed	176.9	8
5	9-weed	162.8	4
6	9-weed	142.4	1
7	9-weed	162.7	3
8	9-weed	162.4	2

Thus, we have test statistics vale and P-value as:

  $\begin{array}{l}Z=2.17\\ P=0.0152\end{array}$

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 presence of nine weeds per meter reduces corn yields when compared with weed-free corn.

(b)

To determine

To compare the results from part (a) with those from the two-sample t test for these data.

Answer to Problem 27.8AYK

We have sufficient evidence to conclude that the presence of nine weeds per meter reduces corn yields when compared with weed-free corn.

Explanation of Solution

In the question, it is given that the corn yield study of example $27.1$ also examined yields in four plots having nine plants per meter. We have to compare it with the weed free corn. So, the data will be as:

0-weed	9-weed
166.7	162.8
172.2	162.7
165	162.4
176.9	142.4

Thus, the hypotheses will be defined as:

  $\begin{array}{l}{H}_0:{\mu }_0={\mu }_9\\ H_a:{\mu }_0>{\mu }_9\end{array}$

Thus, for testing the hypothesis we will use the calculator $\text{TI}-89$ as, first we will go in the STAT TESTS menu, choose $4:2-$ SampTTest. Then you must specify if you are using data stored in two lists or if you will enter the means, standard deviations, and sizes of both samples. You must also indicate whether to pool the variances (when in doubt, say no) and specify whether the test is to be two-tail, lower-tail, or upper-tail.

Thus, by using the calculator

  $\text{TI}-89$ , the test statistics and the P-value is as:

  $\begin{array}{l}t=2.20\\ P=0.0463\\ \text{Mean}=12.63\\ \text{St}\text{. Dev}\text{.}=5.74\end{array}$

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 presence of nine weeds per meter reduces corn yields when compared with weed-free corn. So, we can see that the P-value for both the tests are less than the level of significance and thus, the conclusion for both is same. Also the test statistics value is approximately equal.

(c)

To determine

To repeat the Wilcoxon test and t analyses by removing the outlier $142.4$ from the data with nine weeds per meter and explain by how much did the outlier reduce the mean yield in its group and by how much it increase the standard deviation and did it have a practically important on your conclusion.

Explanation of Solution

In the question, it is given that the corn yield study of example $27.1$ also examined yields in four plots having nine plants per meter. We have to compare it with the weed free corn. So, the data will be as:

0-weed	9-weed
166.7	162.8
172.2	162.7
165	162.4
176.9

Now, we have to conduct both the test by removing the outlier $142.4$ from the data with nine weeds per meter. So, let us conduct the Wilcoxon test first by the help of the software, so, the result will be as:

n	sum of ranks
4	22	0-weed
3	6	9-weed
7	28	total

16.00	expected value
2.83	standard deviation
1.94	z
.0259	p-value (one-tailed, upper)

No.	Label	Data	Rank
1	0-weed	166.7	5
2	0-weed	172.2	6
3	0-weed	165	4
4	0-weed	176.9	7
5	9-weed	162.8	3
6	9-weed	162.7	2
7	9-weed	162.4	1

Now, if we compare it with the above result in part (a), we can see that both the P-values are less than the level of significance so the conclusion will be the same but the mean is decreased by two and the standard deviation is decreased by:

  $\begin{array}{l}=3.46-2.83\\ =0.63\end{array}$

Now, let us conduct the two-sample t test, thus, for testing the hypothesis we will use the calculator $\text{TI}-89$ as, first we will go in the STAT TESTS menu, choose $4:2-$ SampTTest. Then you must specify if you are using data stored in two lists or if you will enter the means, standard deviations, and sizes of both samples. You must also indicate whether to pool the variances (when in doubt, say no) and specify whether the test is to be two-tail, lower-tail, or upper-tail.

Thus, by using the calculator $\text{TI}-89$ , the test statistics and the P-value is as:

  $\begin{array}{l}t=2.79\\ P=0.0342\end{array}$

Now, if we compare it with the above result in part (b), we can see that both the P-values are less than the level of significance so the conclusion will be the same but the standard deviation is decreased by two and the mean is decreased by:

  $\begin{array}{l}=12.63-7.57\\ =5.06\end{array}$