Chapter 10, Problem 10.88SE

(a)

To determine

To find: the assumptions made to compare full-sun versus shade plants and explain whether the assumptions are violated or not.

Answer to Problem 10.88SE

There is no reason to believe that the normality assumptions have been violated.

Explanation of Solution

Given:

	Full Sun		Shade
	$\overline{x}$	$s$	$\overline{x}$	$s$
Leaf Area($c{m}^2$)	$128.00$	$43.00$	$78.70$	$41.70$
Overlap Area($c{m}^2$)	$46.80$	$2.21$	$8.10$	$1.26$
Leaf Number	$9.75$	$2.27$	$6.93$	$1.49$
Thickness($mm$)	$0.90$	$0.03$	$0.50$	$0.02$
Length($cm$)	$8.70$	$1.64$	$8.91$	$1.23$
Width($cm$)	$5.24$	$0.98$	$3.41$	$0.61$

Calculation:

Assumptions behind student’s t distribution:

• The samples are randomly and independently selected from normally distributed populations.
• The variances of the populations ${\sigma }_1{}^2$ and ${\sigma }_2{}^2$ are equal.

No. There is no reason to believe that the normality assumptions have been violated. The standard deviations are not different enough to doubt that the distributions may have the same shape.

Conclusion: Thus, there is no reason to believe that the normality assumptions have been violated.

(b)

To determine

To find: whether the given data is an evidence for the difference in mean leaf area for full-sun versus shade plants.

Answer to Problem 10.88SE

There is sufficient evidence to indicate a difference in mean leaf area for full sun versus shade plants.

Explanation of Solution

Given:

	Full Sun		Shade
	$\overline{x}$	$s$	$\overline{x}$	$s$
Leaf Area($c{m}^2$)	$128.00$	$43.00$	$78.70$	$41.70$
Overlap Area($c{m}^2$)	$46.80$	$2.21$	$8.10$	$1.26$
Leaf Number	$9.75$	$2.27$	$6.93$	$1.49$
Thickness($mm$)	$0.90$	$0.03$	$0.50$	$0.02$
Length($cm$)	$8.70$	$1.64$	$8.91$	$1.23$
Width($cm$)	$5.24$	$0.98$	$3.41$	$0.61$

Calculation:

The null and alternative hypotheses are,

  $H_0:({\mu }_1-{\mu }_2)=0$ versus $H_a:({\mu }_1-{\mu }_2)\ne 0$

For Full Sun: $\overline{x_1}=128.00$ , $s_1=43.00$ , and $n_1=16$

For Shade: $\overline{x_2}=78.70$ , $s_2=41.70$ , and $n_2=15$

The ratio of the two sample variances is,

  $\begin{array}{l}\frac{\text{Larger }{s}^2}{\text{smaller }{s}^2}=\frac{s_1{}^2}{s_2{}^2}\\ =\frac{1849}{1738.89}\\ =1.06\end{array}$

The ratio is less than $3$ , which makes the pooled method appropriate as the assumption of a common population variance is reasonable.

The pooled estimator for ${\sigma }^2$ is,

. $\begin{array}{l}{s}^2=\frac{(n_1-1)s_1{}^2+(n_2-1)s_2{}^2}{n_1+n_2-2}\\ =\frac{(16-1){(43.00)}^2+(15-1){(41.70)}^2}{16+15-2}\\ =\frac{52079.46}{29}\\ =1795.8434\end{array}$

The test static is,

  $\begin{array}{l}t=\frac{\overline{x_1}-\overline{x_2}}{\sqrt{s^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}\\ =\frac{128-78.7}{\sqrt{1795.8434\left(\frac{1}{16}+\frac{1}{15}\right)}}\\ =\frac{49.3}{15.2303}\\ =3.237\\ \end{array}$

Since the observed value, $t=3.237$ , lies above $t_{0.005}=2.756$ , the tail area to the right of $2.81$ is below $0.005$ . Since this area represents only half of the p-value, you can write

  $\frac{1}{2}(p-value)<0.005$ or $p-value<0.010$

Since, the p-value is low, we can reject the null hypothesis.

There is sufficient evidence to indicate a difference in mean leaf area for full sun versus shade plants.

Conclusion: Thus, there is sufficient evidence to indicate a difference in mean leaf area for full sun versus shade plants.

(c)

To determine

To find: whether the given data is an evidence to indicate a difference in mean overlap area for full-sun versus shade plants.

Answer to Problem 10.88SE

There is sufficient evidence to indicate a difference in mean overlap area for full sun versus shade plants.

Explanation of Solution

Given:

	Full Sun		Shade
	$\overline{x}$	$s$	$\overline{x}$	$s$
Leaf Area($c{m}^2$)	$128.00$	$43.00$	$78.70$	$41.70$
Overlap Area($c{m}^2$)	$46.80$	$2.21$	$8.10$	$1.26$
Leaf Number	$9.75$	$2.27$	$6.93$	$1.49$
Thickness($mm$)	$0.90$	$0.03$	$0.50$	$0.02$
Length($cm$)	$8.70$	$1.64$	$8.91$	$1.23$
Width($cm$)	$5.24$	$0.98$	$3.41$	$0.61$

Calculation:

The null and alternative hypotheses are:

  $H_0:({\mu }_1-{\mu }_2)=0$ versus $H_a:({\mu }_1-{\mu }_2)\ne 0$

For Full Sun: $\overline{x_1}=46.80$ , $s_1=2.21$ , and $n_1=16$

For Shade: $\overline{x_2}=8.10$ , $s_2=1.26$ , and $n_2=15$

The ratio of the two sample variances is

  $\begin{array}{l}\frac{\text{Larger }{s}^2}{\text{smaller }{s}^2}=\frac{s_1{}^2}{s_2{}^2}\\ =\frac{4.8841}{1.5876}\\ =3.07\end{array}$

The ratio is more than 3, which makes the pooled method inappropriate as the assumption of a common population variance is not reasonable.

The un-pooled test static is,

  $\begin{array}{l}t=\frac{\overline{x_1}-\overline{x_2}}{\sqrt{\left(\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}\right)}}\\ =\frac{46.8-8.1}{\sqrt{\left(\frac{4.8841}{16}+\frac{1.5876}{15}\right)}}\\ =\frac{38.7}{0.6412}\\ =60.36\end{array}$

Since the observed value, $t=60.36$ , lies above $t_{0.005}=2.756$ , the tail area to the right of $60.36$ is below $0.005$ . Since this area represents only half of the p-value, you can write

  $\frac{1}{2}(p-value)<0.005$ or $p-value<0.010$ .

Since, the p-value is low, we can reject the null hypothesis.

There is sufficient evidence to indicate a difference in mean overlap area for full sun versus shade plants.

Conclusion: Thus, there is sufficient evidence to indicate a difference in mean overlap area for full sun versus shade plants.