Question

The mean yield of corn in the United States is about 135 bushels per acre. A survey of 50 farmers this year gives a sample mean yield of x = 138.4 bushels per acre. We want to know whether this is good evidence that the national mean this year is not 135 bushels per acre. Assume that the farmers surveyed are an SRS from the population of all commercial corn growers and that the standard deviation of the yield in this population is on 10 bushels per acre.

Required:
a. Conduct a hypothesis test to determine if the corn yield has changed using 0.05 as significant level.
b. Write out the hypotheses and report the P-value. Are you convinced that the population mean is not 135 bushels per acre?

Answer 1

The null and alternative hypothesis are:

`H_0: \mu=135\\\\H_a:\mu\\eq 135`

P-value = 0.0162.

Yes, there is enough evidence to support the claim that the corn yield is significantly different from 135 bushels per acre.

Explanation:

This is a hypothesis test for the population mean.

The claim is that the corn yield is significantly different from 135 bushels per acre.

Then, as it is a two-tailed test (we are looking if the yield has changed) the null and alternative hypothesis are:

`H_0: \mu=135\\\\H_a:\mu\\eq 135`

The significance level is 0.05.

The sample has a size n=50.

The sample mean is M=138.4.

The standard deviation of the population is known and has a value of σ=10.

We can calculate the standard error as:

`\sigma_M=(\sigma)/(√(n))=(10)/(√(50))=1.414`

Then, we can calculate the z-statistic as:

`z=(M-\mu)/(\sigma_M)=(138.4-135)/(1.414)=(3.4)/(1.414)=2.4`

This test is a two-tailed test, so the P-value for this test is calculated as:

`\text{P-value}=2\cdot P(z>2.4)=0.0162`

As the P-value (0.0162) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the corn yield is significantly different from 135 bushels per acre.