Are yields for organic farming different from conventional farming yields? Independent random samples from method A (organic farming) and method B (conventional farming) gave the following information about yield of lima beans (in tons/acre). Method A 1.39 1.87 1.29 1.53 1.47 1.79 1.77 1.83 1.61 2.15 2.33 Method B 1.45 2.11 1.85 2.07 1.89 2.03 1.95 1.41 1.73 1.27 1.93 1.13 Use a 5% level of significance to test the hypothesis that there is no difference between the yield distributions. (a) What is the level of significance? State the null and alternate hypotheses. Ho: Distributions are the same. H1: Distributions are the same.Ho: Distributions are the same. H1: Distributions are different. Ho: Distributions are different. H1: Distributions are different.Ho: Distributions are different. H1: Distributions are the same. (b) Compute the sample test statistic. (Use 2 decimal places.) What sampling distribution will you use? uniformStudent's t normalchi-square What conditions are necessary to use this distributions? Both sample sizes must be less than 10.At least one sample size must be greater than 10. At least one sample size must be less than 10.Both sample sizes must be greater than 10. (c) Find the P-value of the sample test statistic. (Use 4 decimal places.) (d) Conclude the test? At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. (e) Interpret your conclusion in the context of the application. Reject the null hypothesis, there is sufficient evidence that the yield distributions are different.Fail to reject the null hypothesis, there is sufficient evidence that the yield distributions are different. Fail to reject the null hypothesis, there is insufficient evidence that the yield distributions are different.Reject the null hypothesis, there is insufficient evidence that the yield distributions are different.

Are yields for organic farming different from conventional farming yields? Independent random samples from method A (organic farming) and method B (conventional farming) gave the following information about yield of lima beans (in tons/acre). Method A 1.39 1.87 1.29 1.53 1.47 1.79 1.77 1.83 1.61 2.15 2.33 Method B 1.45 2.11 1.85 2.07 1.89 2.03 1.95 1.41 1.73 1.27 1.93 1.13 Use a 5% level of significance to test the hypothesis that there is no difference between the yield distributions. (a) What is the level of significance? State the null and alternate hypotheses. Ho: Distributions are the same. H1: Distributions are the same.Ho: Distributions are the same. H1: Distributions are different. Ho: Distributions are different. H1: Distributions are different.Ho: Distributions are different. H1: Distributions are the same. (b) Compute the sample test statistic. (Use 2 decimal places.) What sampling distribution will you use? uniformStudent's t normalchi-square What conditions are necessary to use this distributions? Both sample sizes must be less than 10.At least one sample size must be greater than 10. At least one sample size must be less than 10.Both sample sizes must be greater than 10. (c) Find the P-value of the sample test statistic. (Use 4 decimal places.) (d) Conclude the test? At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. (e) Interpret your conclusion in the context of the application. Reject the null hypothesis, there is sufficient evidence that the yield distributions are different.Fail to reject the null hypothesis, there is sufficient evidence that the yield distributions are different. Fail to reject the null hypothesis, there is insufficient evidence that the yield distributions are different.Reject the null hypothesis, there is insufficient evidence that the yield distributions are different.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

Method A	1.39	1.87	1.29	1.53	1.47	1.79	1.77	1.83	1.61	2.15	2.33
Method B	1.45	2.11	1.85	2.07	1.89	2.03	1.95	1.41	1.73	1.27	1.93	1.13

Use a 5% level of significance to test the hypothesis that there is no difference between the yield distributions.

(a)

What is the level of significance?

State the null and alternate hypotheses.

H_o: Distributions are the same. H₁: Distributions are the same.H_o: Distributions are the same. H₁: Distributions are different. H_o: Distributions are different. H₁: Distributions are different.H_o: Distributions are different. H₁: Distributions are the same.

(b)

Compute the sample test statistic. (Use 2 decimal places.)

What sampling distribution will you use?

uniformStudent's t normalchi-square

What conditions are necessary to use this distributions?

Both sample sizes must be less than 10.At least one sample size must be greater than 10. At least one sample size must be less than 10.Both sample sizes must be greater than 10.

(c)

Find the P-value of the sample test statistic. (Use 4 decimal places.)

(d)

Conclude the test?

At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.

(e)

Interpret your conclusion in the context of the application.

Reject the null hypothesis, there is sufficient evidence that the yield distributions are different.Fail to reject the null hypothesis, there is sufficient evidence that the yield distributions are different. Fail to reject the null hypothesis, there is insufficient evidence that the yield distributions are different.Reject the null hypothesis, there is insufficient evidence that the yield distributions are different.

Expert Solution

Step 1

Use the given data to form the excel table:

	Method A	$(x - μ)$	${(x - μ)}^{2}$
	1.39	-0.34	0.1156
	1.87	0.14	0.0196
	1.29	-0.44	0.1936
	1.53	-0.2	0.04
	1.47	-0.26	0.0676
	1.79	0.06	0.0036
	1.77	0.04	0.0016
	1.83	0.1	0.01
	1.61	-0.12	0.0144
	2.15	0.42	0.1764
	2.33	0.6	0.36
Sum	19.03	0	1.0024
Mean	1.73	0

	Method B	$(x - μ)$	${(x - μ)}^{2}$
	1.45	-0.285	0.081225
	2.11	0.375	0.140625
	1.85	0.115	0.013225
	2.07	0.335	0.112225
	1.89	0.155	0.024025
	2.03	0.295	0.087025
	1.95	0.215	0.046225
	1.41	-0.325	0.105625
	1.73	-0.005	0.000025
	1.27	-0.465	0.216225
	1.93	0.195	0.038025
	1.13	-0.605	0.366025
Sum	20.82	0	1.2305
Mean	1.735	0	0.102541666666667

Determine the mean and standard deviation of both the methods using table:

$\begin{array}{rcl} x_{1} & = & \frac{\sum x}{n_{1}} \\ = & \frac{19.03}{11} \\ = & 1.73 \end{array}$

$\begin{array}{rcl} x_{2} & = & \frac{\sum x}{n_{2}} \\ = & \frac{20.82}{12} \\ = & 1.735 \end{array}$

$\begin{array}{rcl} s_{1} & = & \sqrt{\frac{{(x - x)}^{2}}{n_{1} - 1}} \\ = & \sqrt{\frac{1.0024}{11 - 1}} \\ = & 0.3166 \end{array}$

$\begin{array}{rcl} s_{2} & = & \sqrt{\frac{{(x - x)}^{2}}{n_{2} - 1}} \\ = & \sqrt{\frac{1.2305}{12 - 1}} \\ = & 0.3345 \end{array}$

Step 2

H₀: Distributions are the same.

H₁: Distributions are different.

Null hypothesis: $H_{0} : μ_{1} = μ_{2}$
Alternative hypothesis: $H_{A} : μ_{1} \neq μ_{2}$

This hypotheses constitute a two-tailed test and significance level for the test is 0.05.

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

Step 3

$\begin{array}{rcl} S . E & = & \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}} \\ = & \sqrt{\frac{(0.3166)^{2}}{11} + \frac{(0.3345)^{2}}{12}} \\ = & 0.1358 \end{array}$

$\begin{array}{rcl} D . F & = & n_{1} + n_{2} - 2 \\ = & 11 + 12 \\ = & 21 \end{array}$

$\begin{array}{rcl} t & = & \frac{(x_{1} - x_{2}) - d}{S . E} \\ = & \frac{(1.73 - 1.735) - 0}{0.1358} \\ = & - 0.036818851 \\ \approx & - 0.04 \end{array}$

We should use Student's t statistics. The condition of both sample sizes must be greater than 10 is satisfied.

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