Question

Zea Mays. Charles Darwin, author of Origin of Speries, investigated the effect of cross-fertilization on the heights of plants. In one study he planted 15 pairs of Zeu maryx plants. Each pair consisted of one cross-fertilized plant and one self-fertilized plant grown in the same pot. The following table gives the height differences, in eighths of an inch, for the 15 pairs. Each difference is obtained by subtracting the height of the self-fertilized plant from that of the cross-fertilized plant.

a. Identify the variable under consideration.

b. Identify the two populations.

c. Identify the paired-difference variable.

d. Are the numbers in the table paired differences? Why or why not?

e. At the $5\%$5significance level, do the data provide sufficient evidence to conclude that the mean heights of cross-fertilized and self-fertilized Zea mays differ?

f. Repeat part (e) at the 15 significance level.

Short answer

Part a) the variable is "Height of Zea mays plants"

Part b)Cross-fertilized Zea mays plants and Self-fertilized Zea mays plants.

Part c) "the difference between the heights of a crossfertilized Zea mays and a self-fertilized Zea mays grown in the same pot.

Part d)yes

Part e)the mean heights of Cross-fertilized Zea mays plants and Self-fertilized Zea mays plants are different.

Part f)the mean heights of Cross-fertilized Zea mays plants and Self-fertilized Zea mays plants are different.

Step-by-step solution

Step 1:Given information

Step 2:Explanation Part a)

We identify the variable is "Height of Zea mays plants" of two different methods, eyepiece method and TV-screen method.

Step 2:Explanation Part b)

We identify the two populations, which are Cross-fertilized Zea mays plants and Self-fertilized Zea mays plants.

Step 2:Explanation Part c)

We identify the paired difference variable is "the difference between the heights of a crossfertilized Zea mays and a self-fertilized Zea mays grown in the same pot.

Step 2:Explanation Part d)

Yes. Because each number is the difference between the heights of a cross-fertilized Zea mays and a self-fertlized Zea mays grow in the same pot.

Step 2:Explanation Part e)

Let ${\mu }_1$and ${\mu }_2$denote the mean height of Cross-fertilized Zea mays plants and Self-fertilized Zea mays plants respectively. The null and alternative hypotheses are

$H_0:$mean height of Cross-fertilized Zea mays plants is equal to that of

Self-fertilized Zea mays plants

$\text{(or)}{H}_0:{\mu }_1={\mu }_2$

$H_{\mathrm{a}}$: mean height of Cross-fertilized Zea mays plants is not equal to that of Self-fertilized Zea mays plants

$\text{(or)}{H}_a:{\mu }_1\ne {\mu }_2$

Step 2:

Decide on the significance level, a.

The test is to be performed at the 5 % significance level, or $\alpha =0.05$

Compute the value of the test statistic:

$t=\frac{\overline{d}}{s_d/\sqrt{n}}$

We first need to determine the sample mean and standard deviation of the paired differences. We do so in the usual manner.

$\overline{d}=\frac{\sum d_i}{n}$

$=\frac{314}{15}=20.93$

$s_d=\sqrt{\frac{\sum d_i^2-{\left(\sum d_i\right)}^2/n}{n-1}}$

$=37.74$

Consequently the value of the test statistic is

$t=\frac{\overline{d}}{s_d/\sqrt{n}}$

$=\frac{20.93}{37.74/\sqrt{15}}=2.15$

CRITICAL VALUE APPROACH

We have n=15 and a=0.05. From t-tables with $\mathrm{df}=n-1=15-1=14$, we find that the critical values are $\pm t_{\alpha /2}=\pm t_{0.05/2}=\pm t_{0.025}=\pm 2.145$(SEE FIGURE)

Interpret the results of the hypothesis test.

At the $5\%$significance level, the data provide sufficient evidence to conclude that, the mean heights of Cross-fertilized Zea mays plants and Self-fertilized Zea mays plants are different.

Step 2:Explaination f)

(f) From part (e), we evaluate up to step

(4) and find the test statistics as

$t=2.15$

CRITICAL VALUE APPROACH

The critical value for a two -tailed test are $\pm t_{\alpha 2}$with $\mathrm{df}=n-1$. Use t-tables to find the critical values.

We have $n=15$and a=0.01. From t-tables with $\mathrm{df}=n-1=15-1=14$, we find that the critical values are $\pm t_{\alpha /2}=\pm t_{0.01/2}=\pm t_{0.005}=\pm 2.977$(SEE FIGURE)

If the value of the test statistic is t=2.148, this falls in the rejection region. Hence, we reject $H_0$. The test results are statistically significant at the $5\%$level.

At the $1\%$significance level, the data provide sufficient evidence to conclude that, the mean heights of Cross-fertilized Zea mays plants and Self-fertilized Zea mays plants are different.