91Ó°ÊÓ

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

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

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.

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.

Let ${\mathrm{μ}}_1$and ${\mathrm{μ}}_2$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

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

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

$\text{(or)}{H}_a:{\mathrm{μ}}_1≠{\mathrm{μ}}_2$

Step 2:

Decide on the significance level, a.

The test is to be performed at the 5 % significance level, or $\mathrm{Î\pm }=0.05$

Compute the value of the test statistic:

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

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

$\stackrel{¯}{d}=\frac{∑d_i}{n}$

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

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

$=37.74$

Consequently the value of the test statistic is

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

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

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 $Â\pm t_{\mathrm{Î\pm }/2}=Â\pm t_{0.05/2}=Â\pm t_{0.025}=Â\pm 2.145$(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.

(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 $Â\pm t_{\mathrm{Î\pm }2}$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 $Â\pm t_{\mathrm{Î\pm }/2}=Â\pm t_{0.01/2}=Â\pm t_{0.005}=Â\pm 2.977$(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.