Question

1. In a plant nursery, the owner thinks that the length of seedlings in a box sprayed with a new kind of fertilizer has an average height of 26cm after. One researcher randomly selected 18 such seedlings and calculated the mean  height to be 20cm and the standard deviation was 10cm.  Proceed with the test using ∞ = 0.05

Answer 1

Given $\mu_0=26, \bar{x}=20, s=10, n=18.$

The following null and alternative hypotheses need to be tested:

$H_0:\mu=26$

$H_1:\mu\not=26$

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

Based on the information provided, the significance level is $\alpha=0.05,$ and the critical value for a two-tailed test with degrees of freedom $df=n-1=18-1=17$ is $t_c=2.109816.$

The rejection region for this two-tailed test is $R=\{t:|t|>2.109816\}.$  

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu_0}{s/\sqrt{n}}=\dfrac{20-26}{10/\sqrt{18}}=-2.545584$$

Since it is observed that $|t|=2.545584>2.109816=t_c,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than $26\ cm,$ at the $0.05$ significance level.

Using the P-value approach:

The p-value for two-tailed, $df=17, \alpha=0.05, t=-2.545584$ is $p=0.020881,$ and since $p=0.020881<0.05=\alpha,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than $26\ cm,$ at the $0.05$ significance level.