Question

In: Statistics and Probability

An agricultural field trial compares the yield of two varieties of corn. The researchers divide in...

An agricultural field trial compares the yield of two varieties of corn. The researchers divide in half each of 8 fields of land in different locations and plant each corn variety in one half of each plot. After harvest, the yields are compared in bushels per acre at each location. The 8 differences (Variety A - Variety B) give x¯=2.04 and s=6.93. Does this sample provide evidence that Variety A had a higher yield than Variety B?

(a) State the null and alternative hypotheses: (Type "mu" for the symbol μμ , e.g.  mu >> 1 for the mean is greater than 1, mu << 1 for the mean is less than 1, mu not = 1 for the mean is not equal to 1)
H0H0 :  
HaHa :

(b) Find the test statistic, t =  

(c) Answer the question: Does this sample provide evidence that Variety A had a higher yield than Variety B? (Use a 5% level of significance)
(Type: Yes or No)

Solutions

Expert Solution

Solution:-

a)

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

Null hypothesis: u1< u2
Alternative hypothesis: u1 > u2

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

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

SE = sqrt[(s12/n1) + (s22/n2)]
SE = 2.45012
DF = 7

b)

t = [ (x1 - x2) - d ] / SE

t = 0.721

where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is thesize of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

The observed difference in sample means produced a t statistic of 0.721.

Therefore, the P-value in this analysis is 0.247.

c)

Interpret results. Since the P-value (0.247) is greater than the significance level (0.05), we cannot reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that Variety A had a higher yield than Variety B.


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