Chapter 11.10, Problem 1.6A

Solution

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

  $z=1.775$

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.

To test fertilizer B, the researcher chooses 35 plants at random. The bean plants growth averages 20.3 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, $\overline{x}=20.3,\mu =20,\sigma =1,n=35$

Plugging the values:

  $\begin{array}{c}z=\frac{20.3-20}{\frac{1}{\sqrt{35}}}\\ =0.3\left(\sqrt{35}\right)\\ =1.775\end{array}$

Conclusion:

  $z=1.775$