Chapter 9.2, Problem 44E

To determine

The provided evidences are enough or not to show the chemical is effective for the variety of grass is to be described.

Answer to Problem 44E

Yes. The chemicals are effective for variety of grass of the germination process.

Explanation of Solution

Given:

  $\begin{array}{l}n=400\\ x=339\\ p=0.05\end{array}$

Concept Used:

The test statistic is calculated by using the following expression:

  $z=\frac{\widehat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}}$

Where,

Sample proportion: $\widehat{p}$

Sample size: $n$

Calculation:

The null hypothesis for the given problem condition is written as:

  $H_0:p=0.80$

The alternative hypothesis for the given problem condition is written as:

  $H_a:p\ne 0.80$

Firstly need to calculate the sample proportion which is the ratio of the number of successes and the sample size:

  $\begin{array}{l}\widehat{p}=\frac{x}{n}\\ \widehat{p}=\frac{339}{400}\\ \widehat{p}=0.8475\end{array}$

The standard deviation is calculated as:

  $\begin{array}{l}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ {\sigma }_{\widehat{p}}=\sqrt{\frac{0.8\left(1-0.8\right)}{400}}\\ {\sigma }_{\widehat{p}}=\sqrt{\frac{0.8\left(0.2\right)}{400}}\\ {\sigma }_{\widehat{p}}=0.02\end{array}$

The test statistic or $Z-$ score is calculated by using the following expression:

  $\begin{array}{l}Z=\frac{\widehat{p}-p}{{\sigma }_{\widehat{p}}}\\ Z=\frac{0.8475-0.80}{0.02}\\ Z=2.375\end{array}$

Now the probability value of the $Z-$ score $Z=2.375$ is:

  $\begin{array}{l}P\left(x<Z\right)=0.99123\\ P\left(x>Z\right)=1-0.99123=0.0087745\end{array}$

So, $P$ value is $0.0087745$ which less than the given significance level is of $0.05$ . So, it leads to reject the null hypothesis $H_0:p=0.80$ .

So, it clearly shows that it has enough evidence say that chemical is effective for this variety of grass.

Conclusion:

Hence, $\begin{array}{l}P<p\\ 0.0087745<0.05\end{array}$