Answer to Question #337583 in Statistics and Probability for sheshe

Question

A hog raiser in a certain province uses two methods of pig-farming: intensive pig farming, where pigs are housed indoors in group housing or straw-lined sheds; and extensive pig farming, where pigs are allowed to wander around the farm or fence. Test the hypothesis whether or not the mean weight of pigs in intensive farming is better than the extensive farming based from the mean weight of the pigs in the sample with data shown below. Use a one-tailed test at x = 1%. Find the computed value of the test statistics.

Intensive farming: x₁=85 kg s = 10 kg

Extensive farming: x2= 79 kg s2 = 6 kg

ni 55

n2=45

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1\le\mu_2$

$H_a:\mu_1>\mu_2$

This corresponds to a right-tailed test, and a z-test for two means, with known population standard deviations will be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ degrees of freedom, and the critical value for a right-tailed test is $z_c = 2.3263.$

zc

​=2.33.

The rejection region for this right-tailed test is $R = \{z: z > 2.3263\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\sigma_1^2/n_1+\sigma_2^2/n_2}}$$

$$=\dfrac{79-85}{\sqrt{6^2/45+10^2/55}}=-3.7081$$

Since it is observed that $z = -3.7081 \le2.3263= z_c ,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p = P(Z>-3.7081)=0.999896,$ and since $p = 0.999896 \ge 0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu_1$

is greater than $\mu_2,$ at the $\alpha = 0.01$ significance level.