Question

In a random sample of 80 ears of corn, Mr. Ntow, a farmer finds that 9 of them have worms. Mr. Ntow claims that fewer than 20% of all his corn has worms. Using the P-value method, test this claim at the 0.05 significance level.

Answer 1

The following information is provided: The sample size is N = 80

N=80, the number of favorable cases is X = 9 and the sample proportion is $\bar p = \frac{X}{N} = \frac{ 9}{ 80} = 0.1125$ at α=0.05.

Step 1: The following null and alternative hypotheses need to be tested:

Ho: p = 0.20

Ha:p<0.20

This corresponds to a left-tailed test, for which a z-test for one population proportion needs to be used.

Step 2:

The z-statistic  is computed as follows:

$z = \frac{\bar p - p_0}{\sqrt{p_0(1-p_0)/n}} = \frac{ 0.1125 - 0.20 }{\sqrt{ 0.20(1- 0.20)/80}} = -1.957$

Step 3: Using the P-value approach: From the Z table , we find the value of -1.957 from the table to find p value. (region left to -1.957)

The p-value is p = 0.0252

and since p=0.0252<0.05, it is concluded that the null hypothesis is rejected.

Step 4: conclusion: It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion p is less than p0, at the $\alpha$ = 0.05 significance level.