Chapter 9, Problem 9CRE

(a)

To determine

To explain: how to satisfy the random condition in the study.

Explanation of Solution

Given Information:

The following table displays the data on the yield for each pair regular barley seeds were planted in one plot and kiln dried seeds were planted on the other is given.

In each pair of plots the plots are randomly allocated. One plot is allocated for planting the regular barley seeds while the adjacent plot is allocated for the plantation of kiln dried seeds.

(b)

To determine

To explain: which the appropriate test must be performed to help meet the random condition.

Explanation of Solution

Given Information:

The data on yield is given for the given data first find the difference in the yield as follows.

Plot	Regular	Kiln	Difference
$1$	$1903$	$2009$	$106$
$2$	$1935$	$1915$	$-20$
$3$	$1910$	$2011$	$101$
$4$	$2496$	$2463$	$-33$
$5$	$2108$	$2180$	$72$
$6$	$1961$	$1925$	$-36$
$7$	$2060$	$2122$	$62$
$8$	$1444$	$1482$	$38$
$9$	$1612$	$1542$	$-70$
$10$	$1316$	$1443$	$127$
$11$	$1511$	$1535$	$24$

The table shows the difference in the yield

Calculate the mean $\overline{x}$ for the difference in the yields.

Formula used:

  $\overline{x}=\frac{x_1+x_2+\mathrm{...}{x}_n}{n}$

  $s^2=\frac{{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}}{n-1}$

Calculations:

  $\begin{array}{c}\overline{x}=\frac{106-20+101+\mathrm{...}+24}{11}\\ =33.7293\end{array}$

  $\begin{array}{c}{s}^2=\frac{{\left(106-33.723\right)}^2+{\left(-20-33.723\right)}^2+\mathrm{...}{\left(24-33.723\right)}^2}{11-1}\\ =3980.5620\end{array}$

Calculate the standard deviation by taking the square root of variance.

The value of standard deviation $\sigma =\sqrt{3980.5620}$

Simplify to get $\sigma =66.171$

The null hypothesis and the alternate hypothesis are described as follows

The null hypothesis $H_0=\mu$

The alternate hypothesis $H_1:\mu >0$

where $\mu$ denotes the mean yield.

Formula used:

Describe the test statistics $t$ as follows

  $t=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}$

Calculations:

  $\begin{array}{c}t=\frac{33.7273-0}{\frac{66.171}{\sqrt{11}}}\\ =\frac{33.7273}{19.95}\\ =1.6903\end{array}$

The value of test statistics $t=1.6903$

The p value is $1.8$

Interpretations:

Use the test statistics to find the p- value. The p- value gives the evidence whether to accept or reject the null hypothesis.

The degree of freedom is given by $11-1=10$

Using tables, write the value at $\alpha =0.05$ level of significance.

Use the table and check if the probability lies between $0.05<p<0.10$

Result:

Reject null hypothesis if the level of significance is more than p- value.

Here level of significance $\alpha =0.05$ and $0.05<p<0.10$ .

The p- value do not lie in $0.05<p<0.10$

Accept null hypothesis $H_0=\mu$ .

This implies that there is no convincing evidence to prove that the random condition is met