Chapter 11.10, Problem 1.4A

Solution

To find: The z -score for the mean of the sample treated with fertilizer A compared to the population of bean plants without fertilizer.

  $z=2.19$

Given information:

Given,a company that develops fertilizers wants to know whether either of the two new fertilizers they have in development shows a significant difference in the growth of plants based on a 95% confidence level. The company has data on the growth of bean plants without fertilizers. For a growth period of one month, the population of the beans grown without fertilizers have a mean of 20 cm with a standard deviation of 1 cm.

A company researcher chooses 30 plants at random and uses fertilizer A for one month. The researcher finds that after using fertilizer A, the mean of the bean plants growth is 20.4 cm.

Calculation:

To compare the mean of a sample with the mean of a population, the formula is given by $z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}$ .

Here, $\overline{x}=20.4,\mu =20,\sigma =1,n=30$

Plugging the values:

  $\begin{array}{c}z=\frac{20.4-20}{\frac{1}{\sqrt{30}}}\\ =0.4\left(\sqrt{30}\right)\\ =2.19\end{array}$

Here the z -score is more than 1.96 and indicates that at a 95% confidence level the sample mean and the population mean difference is significant.

Conclusion:

  $z=2.19$