Question

Two types of rice varieties are being considered for yield and a comparison is needed. Thirty hectares were planted with the rice varieties exposed to fairly uniform growing conditions.

The results are tabulated below:

variety A variety B

average yield 80 sacks/hectare 35 sacks/hectare

sample variance 5.9 12.1

At .05 significance level, can we conclude that variety A is the better type?

Answer 1

$$\begin{matrix} & Variety\ A & Variety\ B \\ Average\ yield & 80\ sacks/hectare & 35\ sacks/hectare \\ Sample\ variance& 5.9 & 12.1 \\ \end{matrix}$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1\leq \mu_2$

$H_1:\mu_1>\mu_2$

This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Based on the information provided, the significance level is $\alpha=0.05,$ the degrees of freedom are computed as follows, assuming that the population variances are equal:

$$df=n_1+n_2-2=30+30-2=58.$$

The critical value for this right-tailed test $\alpha=0.05, df=58$ degrees of freedom is $t_c=1.671553.$

The rejection region for this right-tailed test is $R=\{t:t>1.671553\}.$

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$$t=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$$

$$=\dfrac{80-35}{\sqrt{\dfrac{(30-1)(5.9)^2+(30-1)(12.1)^2}{30+30-2}(\dfrac{1}{30}+\dfrac{1}{30})}}$$

$$\approx18.309$$

Since it is observed that $t=18.309>1.671553=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for right-tailed test $\alpha=0.05,$ $df=58$ degrees of freedom is $p\approx0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

The degrees of freedom are computed as follows, assuming that the population variances are equal:

$$df=\dfrac{(\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2})^2}{\dfrac{(s_1^2/n_1)^2}{n_1-1}+\dfrac{(s_2^2/n_2)^2}{n_2-1}}$$

$$=\dfrac{(\dfrac{(5.9)^2}{30}+\dfrac{(12.1)^2}{30})^2}{\dfrac{((5.9)^2/30)^2}{30-1}+\dfrac{((12.1)^2/30)^2}{30-1}}\approx42.052091$$

The critical value for this right-tailed test $\alpha=0.05, df=42.052091$ degrees of freedom is $t_c=1.6819.$

The rejection region for this right-tailed test is $R=\{t:t>1.6819\}.$

Since it is assumed that the population variances are unequal, the t-statistic is computed as follows:

$$t=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}}$$

$$=\dfrac{80-35}{\sqrt{\dfrac{(5.9)^2}{30}+\dfrac{(12.1)^2}{30}}}\approx18.309$$

Since it is observed that $t=18.309>1.671553=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for right-tailed test $\alpha=0.05,$ $df=42.052091$ degrees of freedom is $p\approx0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu_1$

is greater than $\mu_2,$ at the $\alpha = 0.05$ significance level.

Therefore, there is enough evidence to conclude that variety A is the better type at the $\alpha = 0.05$ significance level.