Answer to Question #336708 in Statistics and Probability for JEan

Question

an agronomist believes that a newly developed fertilizer will increase the mean harvest of eggplants by more than 2.5 kg. twenty six plants are treated with fertilizer and have a mean of 10.5 kg with standard deviation of 1.2. it is known that the population mean was 7.5 kg. test the claim at 0.01 level of significance  

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le10$

$H_a:\mu>10$

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ $df=n-1=25$ degrees of freedom, and the critical value for a right-tailed test is $t_c = 2.485107.$

The rejection region for this right-tailed test is $R = \{t: t > 2.485107\}$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{10.5-10}{1.2/\sqrt{26}}\approx2.1246$$

Since it is observed that $t = 2.1246 \le 2.485107=t_c ,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value for right-tailed, $df=25$ degrees of freedom, $t=2.1246$ is $0.021843,$ and since $p= 0.021843>0.01=\alpha,$

it is concluded that the null hypothesis is not rejected.

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

is greater than 10, at the $\alpha = 0.01$ significance level.

Therefore, there is not enough evidence to claim that a newly developed fertilizer will increase the mean harvest of eggplants by more than 2.5 kg, at the $\alpha = 0.01$ significance level.