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An agricultural field trial compares the yield of two varieties of corn. The researchers divide in half each of 14 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 14 differences (Variety A - Variety B) give x¯=2.82 and s=6.96. 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 μ , 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) H0 : Ha : (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)

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Answer:

Step-by-step explanation:

1) H0 : Mean of variety A - Variety B = 0

  Ha : Mean of variety A - Variety B>0

(one tailed test)

(b) Find the test statistic, t = Mean difference / std error

Std error = std dev/sqrt n = [tex]\frac{6.96}{\sqrt{14} } \\=1.86[/tex]

(c) Answer the question:

Does this sample provide evidence that Variety A had a higher yield than Variety B?

df = 14-2 =12

p value = 0.043

Since p<0.05, (our significance level) reject H0.

Yes. this sample provides evidence that Variety A had a higher yield than Variety B

Hypothesis are statements which are asserted but not yet verified. We use hypothesis testing to verify them.

The answers are:

a) The hypotheses are

  • Null hypothesis: [tex]H_0 : \mu_1 = \mu_2[/tex]
  • Alternate hypothesis: [tex]H_1 : \mu_A > \mu_B \text{\:(single tailed)}[/tex]

b) The test statistic is t = 1.516 approximately

c) No, the sample doesn't provide evidence that variety A had a higher yield than variety B

How to form the hypotheses?

There are two hypotheses. First one is called null hypothesis and it is chosen such that it predicts nullity or no change in a thing. It is usually the hypothesis against which we do the test. The hypothesis which we put against null hypothesis is alternate hypothesis.

Null hypothesis is the one which researchers try to disprove.

For the given case, we want to test if the variety A had higher yield than variety B.

For null hypothesis, we'll assume that both were same, thus, mean difference is 0.

Let the mean of yield of variety A be [tex]\mu_A[/tex] and of variety B be [tex]\mu_B[/tex]

Then,

Null hypothesis: [tex]H_0 : \mu_1 = \mu_2[/tex]

Alternate hypothesis: [tex]H_1 : \mu_A > \mu_B \text{\:(single tailed)}[/tex]

The test statistic t is calculated as

[tex]t = \dfrac{\text{Mean difference}}{\text{standard error}} = \dfrac{\overline{x}_A - \overline{x}_B}{ s/\sqrt{n}}[/tex]

where s is standard deviation of difference of paired data = 6.96 (as given) and n is number of pairs in sample which is 14

Thus,

[tex]t = \dfrac{\text{Mean difference}}{\text{standard error}} = \dfrac{\overline{x}_A - \overline{x}_B}{ s/\sqrt{n}}\\\\t = \dfrac{2.82}{6.96/\sqrt{14}} \approx 1.516[/tex]

The degree of freedom = [tex]2(n) - 2 = 26[/tex]

Level of significance = 5% = 0.05

The p value for t = 1.516 at 26 degree of freedom for 0.05 level of significance is calculated as p value: 0.1416

Since p value: 0.1416> 0.05, thus, we cannot reject null hypothesis.

Thus,

The answers are:

a) The hypotheses are

  • Null hypothesis: [tex]H_0 : \mu_1 = \mu_2[/tex]
  • Alternate hypothesis: [tex]H_1 : \mu_A > \mu_B \text{\:(single tailed)}[/tex]

b) The test statistic is t = 1.516 approximately

c) No, the sample doesn't provide evidence that variety A had a higher yield than variety B

Learn more about testing of hypothesis here:

https://brainly.com/question/18831983

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