Question

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 26 cm after three days and a standard deviation of 10 cm. One researcher randomly selected 80 such seedlings and calculated the mean height to be 20 cm. Use a=0.05 to test out whether what the plant nursery owner thinks us correct or not.

Answer 1

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 z-test for one mean, with known 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 is $z_c = 1.96.$

The rejection region for this two-tailed test is $R = \{z:|z|>1.96\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{20-26}{10/\sqrt{80}}=-5.3666$$

Since it is observed that $z=5.3666>1.96=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(z<-5.3666)=0,$ 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 $\mu$

is different than 26, at the $\alpha = 0.05$ significance level.