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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) (Reference: Agricultural Statistics, U.S. Department of Agriculture). $$ \begin{array}{l|llllllllllll} \hline \text { Method A } & 1.83 & 2.34 & 1.61 & 1.99 & 1.78 & 2.01 & 2.12 & 1.15 & 1.41 & 1.95 & 1.25 & \\ \hline \text { Method B } & 2.15 & 2.17 & 2.11 & 1.89 & 1.34 & 1.88 & 1.96 & 1.10 & 1.75 & 1.80 & 1.53 & 2.21 \\ \hline \end{array} $$ Use a \(5 \%\) level of significance to test the hypothesis that there is no difference between the yield distributions.

Short Answer

Expert verified
There is no significant difference between the yields of organic and conventional farming at a 5% significance level.

Step by step solution

01

State Hypotheses

We need to state our null and alternative hypotheses. The null hypothesis (\(H_0\)) is that there is no difference in the means of the two methods: \( \mu_A = \mu_B \). The alternative hypothesis (\(H_1\)) is that there is a difference: \( \mu_A eq \mu_B \).
02

Calculate Sample Means

Calculate the mean yield for each method. For Method A: \( \bar{x}_A = \frac{1.83 + 2.34 + 1.61 + 1.99 + 1.78 + 2.01 + 2.12 + 1.15 + 1.41 + 1.95 + 1.25}{11} = 1.76 \) tons/acre. For Method B: \( \bar{x}_B = \frac{2.15 + 2.17 + 2.11 + 1.89 + 1.34 + 1.88 + 1.96 + 1.10 + 1.75 + 1.80 + 1.53 + 2.21}{12} = 1.78 \) tons/acre.
03

Calculate Sample Standard Deviations

Compute the sample standard deviations for each method. For Method A: Find the variance first, \( s_A^2 = \frac{(1.83-1.76)^2 + (2.34-1.76)^2 +... + (1.25-1.76)^2}{10} = 0.152 \), and \( s_A = \sqrt{0.152} = 0.39 \). Similarly, for Method B: \( s_B^2 = 0.125 \), \( s_B = \sqrt{0.125} = 0.35 \).
04

Determine Test Statistic

We use a two-sample t-test with unequal variances (Welch's t-test). The test statistic \( t \) is calculated using \[ t = \frac{\bar{x}_A - \bar{x}_B}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}} \]\( = \frac{1.76 - 1.78}{\sqrt{\frac{0.39^2}{11} + \frac{0.35^2}{12}}} \) which simplifies to \( \approx -0.14 \).
05

Find the Critical Value

With a 5% significance level and degrees of freedom calculated by the Welch-Satterthwaite equation, \( df \approx 20 \). Using a t-distribution table, the critical value for a two-tailed test at \( df = 20 \) is approximately \( \pm 2.086 \).
06

Compare Test Statistic with Critical Value

Since the calculated \( t \) value of \( -0.14 \) is within the range of \( -2.086 \) to \( 2.086 \), we fail to reject the null hypothesis.
07

Conclusion: Draw Results

At the 5% significance level, there is insufficient evidence to conclude that there is a difference in yields between organic farming (Method A) and conventional farming (Method B).

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Key Concepts

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

Two-Sample t-Test
A two-sample t-test is a statistical method used to determine if there is a significant difference between the means of two independent groups. In this context, we are looking at organic and conventional farming yields to see if they produce significantly different results. The test begins with the formulation of two hypotheses:
  • Null Hypothesis ( H_0 ): There is no difference in the mean yields of the two groups ( \( \mu_A = \mu_B \)
  • Alternative Hypothesis ( H_1 ): There is a difference in the mean yields ( \( \mu_A eq \mu_B \)
The two-sample t-test assumes that the data samples are independent and normally distributed, although it can accommodate unequal variances with the Welch's t-test variant. This variant is especially useful when sample sizes or variances differ between the groups, as is the case with our organic and conventional farming yield data. Here, the t-test helps us to understand if any observed difference in sample means is statistically meaningful or if it's just a random variation.
To conduct a test, we calculate the t-statistic, which measures the difference between group means standardized by the data's variability. Once we have the t-statistic, we compare it to a critical value from the t-distribution to decide whether to accept or reject the null hypothesis. A t-statistic falling within the range of critical values means there is not enough evidence to suggest a significant difference between the two group means, as was concluded in the farming yields scenario.
Organic vs Conventional Farming Yields
When evaluating farming practices, yield is a crucial metric. In this exercise, we compared lima bean yields from organic (Method A) and conventional (Method B) farming practices. Organic farming emphasizes natural processes, often using fewer synthetic inputs than conventional methods. Thus, it's valuable to analyze if there's a statistical difference in yields because this can influence agricultural practices and economic decisions.
Looking at the sample data, Method A (organic) and Method B (conventional) yields vary slightly. Organic yields had a mean (\( \bar{x}_A \)) of 1.76 tons per acre, while conventional farming showed a mean (\( \bar{x}_B \)) of 1.78 tons per acre.
In practical terms, such a small difference might not appear significant to a farmer considering switching methods. However, using statistical analysis, we gain insight into whether this observed difference could stem from natural variability in yields across farms or from the farming methods themselves.
Statistically, the two-sample t-test results suggested that the difference in means (about 0.02 tons/acre) was not statistically significant at the 5% level. This means that from a statistical standpoint, we do not have enough evidence to claim that organic or conventional methods result in different yields.
Sample Means Calculation
Calculating sample means is a foundational step in statistical analysis that provides an average value representing the central tendency of data. This calculation is crucial when comparing two data sets, such as the organic and conventional farming yields.
To find the mean yield for each farming method, we sum all individual observations for each method and divide by the number of observations. For example, for organic farming (Method A) with samples, we calculated:\[\bar{x}_A = \frac{1.83 + 2.34 + 1.61 + 1.99 + 1.78 + 2.01 + 2.12 + 1.15 + 1.41 + 1.95 + 1.25}{11} = 1.76\text{ tons/acre}\]Similarly, for conventional farming (Method B), the calculation is:\[\bar{x}_B = \frac{2.15 + 2.17 + 2.11 + 1.89 + 1.34 + 1.88 + 1.96 + 1.10 + 1.75 + 1.80 + 1.53 + 2.21}{12} = 1.78\text{ tons/acre}\]By doing this, we're able to utilize these mean values in further statistical tests, like the two-sample t-test, to explore the differences between the groups. Mean comparison helps us determine if observed changes or differences between farming methods are statistically meaningful or not. In this case, the close mean yields suggest similarities, yet statistical testing is needed to assess the significance of such closeness in means.

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Most popular questions from this chapter

Is there a relation between police protection and fire protection? A random sample of large population areas gave the following information about the number of local police and the number of local firefighters (units in thousands). $$ \begin{array}{l|rrrrrrrrrrrrr} \hline \text { Area } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \\\ \hline \text { Police } & 11.1 & 6.6 & 8.5 & 4.2 & 3.5 & 2.8 & 5.9 & 7.9 & 2.9 & 18.0 & 9.7 & 7.4 & 1.8 \\ \text { Firefighters } & 5.5 & 2.4 & 4.5 & 1.6 & 1.7 & 1.0 & 1.7 & 5.1 & 1.3 & 12.6 & 2.1 & 3.1 & 0.6 \\ \hline \end{array} $$ (i) Rank-order police using 1 as the largest data value. Also rank-order firefighters using 1 as the largest data value. Then construct a table of ranks to be used for a Spearman rank correlation test. (ii) Use a \(5 \%\) level of significance to test the claim that there is a monotone relationship (either way) between the ranks of number of police and number of firefighters.

To apply the sign test, do you need independent or dependent (matched pair) data?

A psychology professor is studying the relation between overcrowding and violent behavior in a rat colony. Eight colonies with different degrees of overcrowding are being studied. By using a television monitor, lab assistants record incidents of violence. Each colony has been ranked for crowding and violence. A rank of 1 means most crowded or most violent. The results for the eight colonies are given in the following table, with \(x\) being the population density rank and \(y\) the violence rank. $$ \begin{array}{l|rrrrrrrr} \hline \text { Colony } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline x \text { rank } & 3 & 5 & 6 & 1 & 8 & 7 & 4 & 2 \\ y \text { rank } & 1 & 3 & 5 & 2 & 8 & 6 & 4 & 7 \\ \hline \end{array} $$ Using a \(0.05\) level of significance, test the claim that lower crowding ranks mean lower violence ranks (i.e., the variables have a monotone-increasing relationship).

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When applying the rank-sum test, do you need independent or dependent samples?

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