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Fertilizing a Tropical Plant. Bromeliads are tropical flowering plants. Many are epiphytes that attach to trees and obtain moist ure and nutrients from air and rain. In an experiment in Costa Rica, Jacqueline Ngai and Diane Srivastava studied whether added nitrogen increases the productivity of bromeliad plants. Bromeliads were randomly assigned to nitrogen or control groups. Here are data on the number of new leaves produced over a seven-month period: 19 III FERT \begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|} \hline Control & 11 & 13 & 16 & 15 & 15 & 11 & 12 & \\ \hline Nitrogen & 15 & 14 & 15 & 16 & 17 & 18 & 17 & 13 \\ \hline \end{tabular} Is there evidence that adding nitrogen increases the mean number of new leaves formed?

Step by step solution

01

Understand the Problem

We need to determine if adding nitrogen increases the productivity (number of new leaves) of bromeliad plants. This involves comparing the mean number of new leaves between plants given nitrogen and those that were not.

02

Organize the Data

The control group (no nitrogen) data is [11, 13, 16, 15, 15, 11, 12], and the nitrogen group data is [15, 14, 15, 16, 17, 18, 17, 13]. We need to compute the means for these groups.

03

Calculate Means

Calculate the mean number of leaves for each group. Control Mean: \( \frac{11 + 13 + 16 + 15 + 15 + 11 + 12}{7} = 13 \)Nitrogen Mean: \( \frac{15 + 14 + 15 + 16 + 17 + 18 + 17 + 13}{8} = 15.625 \)

04

State Hypotheses

Set up the null and alternative hypotheses. Null Hypothesis (\( H_0 \)): Adding nitrogen does not increase the mean number of new leaves.Alternative Hypothesis (\( H_a \)): Adding nitrogen increases the mean number of new leaves.

05

Conduct Statistical Test

Perform a two-sample t-test for the difference between two means to see if there's a significant difference. This requires the standard deviation and sample size of each group.

06

Calculate Standard Deviations

Compute the standard deviations for each group.Control: \( s_{control} = \sqrt{\frac{(11-13)^2+(13-13)^2+(16-13)^2+(15-13)^2+(15-13)^2+(11-13)^2+(12-13)^2}{7-1}} \approx 2.16 \)Nitrogen: \( s_{nitrogen} = \sqrt{\frac{(15-15.625)^2+(14-15.625)^2+(15-15.625)^2+(16-15.625)^2+(17-15.625)^2+(18-15.625)^2+(17-15.625)^2+(13-15.625)^2}{8-1}} \approx 1.60 \)

07

Perform T-Test Calculation

Use the formula for the t-test statistic for two independent samples: \[ t = \frac{\overline{X}_1 - \overline{X}_2}{\sqrt{\frac{s_{1}^2}{n_1} + \frac{s_{2}^2}{n_2}}} \]Substitute the calculated means, standard deviations, and sample sizes into the formula:

08

Interpret the Results

Compare the calculated t-value to the critical t-value from the t-distribution table or calculate the p-value to determine significance. If the p-value is less than the significance level (commonly 0.05), we reject the null hypothesis.

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Key Concepts

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

Null Hypothesis

In statistical testing, the null hypothesis is a fundamental concept. It sets an initial claim that there is no effect or difference between groups. In our bromeliad experiment, the null hypothesis (\( H_0 \)) states that adding nitrogen does not increase the mean number of new leaves in bromeliad plants.

• This serves as a starting point for testing that any observed difference is due to chance.
• We assume the null hypothesis is true until evidence suggests otherwise.

The null hypothesis forms the basis of statistical tests, including the t-test, by providing a benchmark against which we compare our sample data.

Alternative Hypothesis

When conducting a statistical analysis, formulating an alternative hypothesis is crucial. The alternative hypothesis (\( H_a \)) proposes that there is an actual effect or difference. For the bromeliad plants, the alternative hypothesis suggests that adding nitrogen increases the mean number of new leaves.

• If our test results show a significant effect, we consider the alternative hypothesis more likely.
• Rejecting the null hypothesis in favor of the alternative hypothesis indicates that the treatment has an effect.

It's what researchers aim to prove through their experiments.

Standard Deviation

Standard deviation is a statistical measure that reflects the degree of variation or dispersion in a set of values. In our bromeliad experiment, we calculate the standard deviation for both groups (control and nitrogen) to understand how their leaf production varies.

• The formula involves the differences between each data point and the mean, squared, summed, and then averaged.
• A smaller standard deviation means that the data points are close to the mean, indicating consistency.
• For the control group, the standard deviation is approximately 2.16, and for the nitrogen group, it is about 1.60.

Understanding standard deviation helps interpret the reliability and variability of experimental data.

Mean Difference

The mean difference is another vital concept in hypothesis testing. It reflects the difference between the average outcomes of two groups. In the bromeliad study, this refers to the contrast in the mean number of new leaves between the nitrogen-increased group and the control group.

• The control group has a mean of 13 leaves, while the nitrogen group has a mean of 15.625 leaves.
• The mean difference assists in evaluating the effectiveness of the treatment.
• A larger mean difference suggests a stronger effect from the treatment.

Mean difference plays a key role in the calculation of t-test statistics, helping to determine if observed differences are statistically significant.