Question

The historical mean yield of soybeans from farms in Iowa is 32.9 bushels per acre. Following a dry summer, a random sample of 40 Iowa farms is taken and the mean yield was found to be 30.78 bushels per acre with a standard deviation of 4.3. A hypothesis test is run to test the claim that the true mean yield of the dry summer was less than the historical mean. Which is the correct alternative hypothesis to test this claim?

	1.	Ha: µ > 32.9
	2.	Ha: µ > 30.78
	3.	Ha: µ < 30.78
	4.	Ha: µ < 32.9

The historical mean yield of soybeans from farms in Iowa is 32.9 bushels per acre. Following a dry summer, a random sample of 40 Iowa farms is taken and the mean yield was found to be 31.78 bushels per acre with a standard deviation of 4.3. A hypothesis test is run to test the claim that the true mean yield of the dry summer was less than the historical mean. A hypothesis test at the 5% significance level to determine if the true mean yield of the dry summer was less than the historical mean was run. Find the correct p-value.

The historical mean yield of soybeans from farms in Iowa is 32.9 bushels per acre. A hypothesis test is run to test the claim that the true mean yield of a dry summer was less than the historical mean. If the farmers believe they have evidence of a dry summer (smaller true mean yield), then they will spend a lot of money on irrigation system.

Which of the following describes what making a Type I error for this problem is?

	1.	We support the claim that the true mean yield is less than 32.9 bushels per acre when in fact the true mean is less than 32.9 bushels per acre.
	2.	We support the claim that the true mean yield is less than 32.9 bushels per acre when in fact the true mean is at least 32.9 bushels per acre.
	3.	We do not support the claim that the true mean yield is less than 32.9 bushels per acre when in fact the true mean less than 32.9 bushels per acre.
	4.	We do not support the claim that the true mean yield is less than 32.9 bushels per acre when in fact the true mean is at least 32.9 bushels per acre.

The historical mean yield of soybeans from farms in Iowa is 32.9 bushels per acre. A hypothesis test is run to test the claim that the true mean yield of a dry summer was less than the historical mean. If the farmers believe they have evidence of a dry summer (smaller true mean yield), then they will spend a lot of money on irrigation system.

Which of the following describes what making a Type II error for this problem is?

	1.	We do not support the claim that the true mean yield is less than 32.9 bushels per acre when in fact the true mean is less than 32.9 bushels per acre.
	2.	We support the claim that the true mean yield is less than 32.9 bushels per acre when in fact the true mean is at least 32.9 bushels per acre.
	3.	We support the claim that the true mean yield is less than 32.9 bushels per acre when in fact the true mean is less than 32.9 bushels per acre.
	4.	We do not support the claim that the true mean yield is less than 32.9 bushels per acre when in fact the true mean is at least 32.9 bushels per acre.

Answer 1

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 32.9
4) Alternative hypothesis: u < 32.9

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.6799
DF = n - 1

D.F = 39
t = (x - u) / SE

t = - 3.23

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of - 3.23.

P-value = P(t < - 3.23)

Use the t-value calculator for finding p-values.

P-value = 0.001

Interpret results. Since the P-value (0.001) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that the true mean yield of the dry summer was less than the historical mean was run.

b)

(2) We support the claim that the true mean yield is less than 32.9 bushels per acre when in fact the true mean is at least 32.9 bushels per acre.

c)

(1) We do not support the claim that the true mean yield is less than 32.9 bushels per acre when in fact the true mean is less than 32.9 bushels per acre.