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Agriculture: Lima Beans Are yields for organic farming different from conventional farming yields? Independent random samples from method \(\mathrm{A}\) (organic farming) and method B (conventional farming) gave the following information about yield of lima beans (in tons/acre). (Reference: Agricultural Statistics, United States Department of Agriculture.) \begin{tabular}{l|lllllllllll} \hline Method A & \(1.83\) & \(2.34\) & \(1.61\) & \(1.99\) & \(1.78\) & \(2.01\) & \(2.12\) & \(1.15\) & \(1.41\) & \(1.95\) & \(1.25\) & \\ \hline Method B & \(2.15\) & \(2.17\) & \(2.11\) & \(1.89\) & \(1.34\) & \(1.88\) & \(1.96\) & \(1.10\) & \(1.75\) & \(1.80\) & \(1.53\) & \(2.21\) \\ \hline \end{tabular} Use a \(5 \%\) level of significance to test the hypothesis that there is no difference between the yield distributions.

Step by step solution

01

State Hypotheses

The first step is to state the null and alternative hypotheses. The null hypothesis (H_{0}) states that the two population means are equal, i.e., there is no difference in yields between the two farming methods. On the other hand, the alternative hypothesis (H_{a}) states that there is a difference in the means.\(H_{0}: \mu_{A} = \mu_{B} \)\(H_{a}: \mu_{A} eq \mu_{B} \)

02

Calculate Means and Standard Deviations

Calculate the sample means and standard deviations for both methods. For Method A, calculate the mean: \\(\bar{x}_A = \frac{1.83 + 2.34 + 1.61 + 1.99 + 1.78 + 2.01 + 2.12 + 1.15 + 1.41 + 1.95 + 1.25}{11}\).Similarly, calculate the standard deviation \(s_A\). Repeat the calculations for Method B with its data points.

03

Determine Degrees of Freedom and Critical Value

Determine the degrees of freedom using the formula for two independent samples: \\( df = n_A + n_B - 2 \).Find the critical value of \\( t \) using the t-distribution table with the obtained degrees of freedom and a 5% significance level (two-tailed).

04

Calculate Test Statistic

Use the calculated means and standard deviations to compute the t-test statistic:\[ t = \frac{\bar{x}_A - \bar{x}_B}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}} \].Plug in the values computed in Step 2 into this formula.

05

Compare Test Statistic and Critical Value

Compare the calculated test statistic from Step 4 to the critical value found in Step 3. If the test statistic is greater than the critical value, reject the null hypothesis. Otherwise, do not reject the null hypothesis.

06

Conclusion

Based on the comparison in Step 5, state the conclusion. If the null hypothesis is rejected, it suggests that there is a statistically significant difference in yield between the two farming methods. If not rejected, there is not enough evidence to conclude that the yields differ.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null Hypothesis

In hypothesis testing, the null hypothesis acts as the starting point. It claims that there is no effect or difference in the population. Think of it as the default assumption. In this exercise, the null hypothesis (\(H_0\)) states that the mean yield of organic farming is equal to that of conventional farming methods. Mathematically, it's expressed as: \(H_0: \mu_A = \mu_B\), where \(\mu_A\) and \(\mu_B\) represent the mean yields of methods A and B respectively. By assuming that these are equal, any difference observed is due to random chance rather than a real effect.
Testing this requires analysis to see if there is enough evidence to refute the null hypothesis, leading us to consider whether the alternative hypothesis might be true.

T-test

The T-test is a statistical method used to determine if there are significant differences between the means of two groups. It's particularly useful when you have two independent samples and a relatively small data size. In this situation, the T-test helps us assess whether any observed differences in the mean yields of organic and conventional farming are statistically significant, rather than due to random variation.
We start by calculating the mean and standard deviation for each method. Then, the t-test statistic is computed using these values with the formula:\[ t = \frac{\bar{x}_A - \bar{x}_B}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}} \]

• \(\bar{x}_A\) and \(\bar{x}_B\) are sample means.
• \(s_A\) and \(s_B\) are standard deviations of the samples.
• \(n_A\) and \(n_B\) stand for sample sizes.

This computation results in a test statistic, which we compare against a critical value to decide if the null hypothesis can be rejected.

Significance Level

The significance level, often denoted by \(\alpha\), is a threshold probability used in hypothesis testing. It helps determine whether the null hypothesis can be rejected. A common choice for the significance level is 5%, or 0.05. This means there is a 5% risk of concluding that there is an effect when in fact there isn't (Type I error).
In the exercise provided, the significance level is set at 5%. It helps in deciding the critical value from the t-distribution. When the calculated t-statistic exceeds this critical value, the null hypothesis is rejected. In simpler terms, it quantifies our "tolerance" for the risk that our conclusion might be wrong because of sampling variability.

Degree of Freedom

Degrees of freedom (df) are vital in statistical testing, especially in determining the critical values from distribution tables. They provide the flexibility in estimation that's crucial for understanding variability in the data. In the context of a two-sample t-test, the degrees of freedom are calculated with:\[ df = n_A + n_B - 2 \]

• \(n_A\) is the number of observations in sample A.
• \(n_B\) is the number of observations in sample B.

For the current problem, calculating the degrees of freedom allows us to find the critical value from a t-distribution table at our chosen significance level. These critical values are then used to compare with the computed t-statistic to make informed decisions on the null hypothesis.