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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 sweet corn (in tons/acre) (Reference: Agricultural Statistics, U.S. Department of Agriculture). $$ \begin{array}{l|llllllllllll} \hline \text { Method A } & 6.88 & 6.86 & 7.12 & 5.91 & 6.80 & 6.92 & 6.25 & 6.98 & 7.21 & 7.33 & 5.85 & 6.72 \\ \hline \text { Method B } & 5.71 & 6.93 & 7.05 & 7.15 & 6.79 & 6.87 & 6.45 & 7.34 & 5.68 & 6.78 & 6.95 & \\ \hline \end{array} $$ Use a \(5 \%\) level of significance to test the claim that there is no difference between the yield distributions.

Short Answer

Expert verified
There is no significant difference in yields between organic and conventional farming at \(5\%\) significance.

Step by step solution

01

Set Up Hypotheses

To determine if there is a difference in yields, we formulate the null hypothesis \( H_0 \) and the alternative hypothesis \( H_a \). The null hypothesis is that there is no difference in the mean yields of the two methods: \( H_0: \mu_A = \mu_B \). The alternative hypothesis is that there is a difference: \( H_a: \mu_A eq \mu_B \).
02

Collect Sample Data

Identify the data provided for both methods. For Method A (organic), the yields are 6.88, 6.86, 7.12, 5.91, 6.80, 6.92, 6.25, 6.98, 7.21, 7.33, 5.85, 6.72. For Method B (conventional), the yields are 5.71, 6.93, 7.05, 7.15, 6.79, 6.87, 6.45, 7.34, 5.68, 6.78, 6.95.
03

Calculate Sample Means and Standard Deviations

Calculate the sample means: \( \bar{x}_A = 6.75 \) and \( \bar{x}_B = 6.74 \). Compute the sample standard deviations: \( s_A = 0.47 \) and \( s_B = 0.53 \).
04

Determine the Test Statistic

Since the sample sizes are small, use a t-test for two independent samples. The formula for the t-statistic is \( t = \frac{\bar{x}_A - \bar{x}_B}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}} \), where \( n_A \) and \( n_B \) are the sample sizes of A and B, respectively.
05

Compute the Test Statistic

Plug in the values: \( t = \frac{6.75 - 6.74}{\sqrt{\frac{0.47^2}{12} + \frac{0.53^2}{11}}} \). Calculate: \( t \approx 0.06 \).
06

Determine the Critical Value

For a two-tailed test with \( \alpha = 0.05 \), look up the t-distribution table with \( df = 21 \) (approximated by \( df = \min(n_A - 1, n_B - 1) = 11 \)). The critical t-value is approximately 2.201 at \( 5\% \) significance level.
07

Make a Decision

Compare the test statistic to the critical value: \( |0.06| < 2.201 \). Since the absolute value of the calculated t-statistic is less than the critical value, we do not reject the null hypothesis.
08

Draw a Conclusion

Based on the test, there is no statistically significant difference in the sweet corn yields between organic and conventional farming methods at the \( 5\% \) significance level.

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Key Concepts

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

Organic Farming
Organic farming is an agricultural method focused on growing crops without the use of synthetic fertilizers and pesticides. Instead, it relies on natural sources of nutrients and pest control. This approach not only aims to produce healthier crops but also seeks to minimize environmental impact. Farmers employing organic methods often emphasize soil health, using crop rotation, compost, and green manure as integral practices.

Here are some benefits of organic farming:
  • Environmental sustainability
  • Biodiversity preservation
  • Reduced pollution
  • Potentially healthier produce
However, organic farming can sometimes result in lower yields compared to conventional methods. This variance in yields prompted the exercise to analyze data using statistical methods.
Conventional Farming
Conventional farming involves the use of advanced technology, synthetic fertilizers, and pesticides to enhance crop production. The primary goal is to maximize yields and efficiency to meet the growing demand for food. Conventional farming practices include the use of genetically modified organisms (GMOs) and chemical control of pests and weeds.

Some common features of conventional farming include:
  • Higher yield potential
  • Technology-driven solutions
  • Efficient use of land resources
  • More reliance on chemical inputs
While effective in increasing food availability, conventional farming raises concerns about environmental impact and food safety, leading to comparisons with organic practices.
Two-Sample t-Test
The two-sample t-test is a statistical method used to compare the means of two independent samples to see if there is a statistically significant difference between them. It's crucial in situations where we want to determine if one method or treatment is superior to another.

For our exercise, this test helps assess whether the yields of sweet corn from organic farming differ significantly from those of conventional farming. The steps involve formulating hypotheses, calculating sample means and standard deviations, and ultimately calculating the t-statistic. The t-statistic, calculated as:\[ t = \frac{\bar{x}_A - \bar{x}_B}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}} \]allows researchers to understand the extent of difference between the samples. If the calculated t-statistic is more extreme than a critical value from the t-distribution, we could claim a significant difference.
Confidence Interval
A confidence interval provides a range of values that is likely to contain the population parameter with a certain level of confidence. In hypothesis testing, it serves as an alternative approach to using a critical value or p-value, offering insight into reliability and precision of estimation. For the current test examining farming yield differences, calculating a confidence interval around the mean difference can help verify findings. If the interval includes zero, it suggests no significant difference, aligning with the decision not to reject the null hypothesis. Confidence levels are usually set at 95% or 99%, indicating the proportion of intervals that would contain the true parameter in repeated sampling. Variability and sample size significantly influences the width of confidence intervals, affecting certainty in the estimates.

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Most popular questions from this chapter

To compare two elementary schools regarding teaching of reading skills, 12 sets of identical twins were used. In each case, one child was selected at random and sent to school A, and his or her twin was sent to school B. Near the end of fifth grade, an achievement test was given to each child. The results follow: $$ \begin{array}{l|cccccc} \hline \text { Twin Pair } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text { School A } & 177 & 150 & 112 & 95 & 120 & 117 \\ \hline \text { School B } & 86 & 135 & 115 & 110 & 116 & 84 \\ \hline \end{array} $$ $$ \begin{array}{l|cccccc} \hline \text { Twin Pair } & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \text { School A } & 86 & 111 & 110 & 142 & 125 & 89 \\ \hline \text { School B } & 93 & 77 & 96 & 130 & 147 & 101 \\ \hline \end{array} $$ Use a \(0.05\) level of significance to test the hypothesis that the two schools have the same effectiveness in teaching reading skills against the alternate hypothesis that the schools are not equally effective.

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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.

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