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Bromeliads are tropical flowering plants. Many are epiphytes that attach to trees and obtain moisture and nutrients from air and rain. Their leaf bases form cups that collect water and are home to the larvae of many insects. 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: 18 $$ \begin{array}{l|llllllll} \hline \text { Control } & 11 & 13 & 16 & 15 & 15 & 11 & 12 & \\ \hline \text { Nitrogen } & 15 & 14 & 15 & 16 & 17 & 18 & 17 & 13 \\ \hline \end{array} $$ Is there evidence that adding nitrogen increases the mean number of new leaves formed?

Step by step solution

01

Define Hypotheses

For this statistical test, we need to set up null and alternative hypotheses. The null hypothesis (\(H_0\)) states that adding nitrogen does not increase the mean number of new leaves, while the alternative hypothesis (\(H_a\)) suggests that adding nitrogen does increase the mean number.\[H_0: \mu_{control} = \mu_{nitrogen} \H_a: \mu_{control} < \mu_{nitrogen}\]

02

Collect and Summarize Data

We need to summarize the given data for control and nitrogen groups. First, calculate the mean and standard deviation for each group: - Control Group: 11, 13, 16, 15, 15, 11, 12 - Nitrogen Group: 15, 14, 15, 16, 17, 18, 17, 13

03

Calculate Sample Means and Standard Deviations

Calculate the sample mean for each group:- Mean of control group (\(\bar{x}_{control}\)) = \(\frac{11 + 13 + 16 + 15 + 15 + 11 + 12}{7} = 13.29\)- Mean of nitrogen group (\(\bar{x}_{nitrogen}\)) = \(\frac{15 + 14 + 15 + 16 + 17 + 18 + 17 + 13}{8} = 15.63\)Next, calculate the standard deviations (\(s\)) for each group.

04

Conduct a t-Test

Use a two-sample t-test for unequal variances (Welch's t-test). The test statistic is calculated as follows:\[t = \frac{\bar{x}_{nitrogen} - \bar{x}_{control}}{\sqrt{\frac{s_{control}^2}{n_{control}} + \frac{s_{nitrogen}^2}{n_{nitrogen}}}}\]where \(n\) is the sample size for each group.

05

Determine the Critical Value and Conclusion

Determine the critical value for a one-tailed test at a chosen level of significance (let's assume 0.05). Compare the test statistic to the critical value: - If the test statistic is more significant than the critical value, reject the null hypothesis. - If the test statistic is less than or equal to the critical value, do not reject the null hypothesis. Calculate the critical value using a t-distribution table or software.

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Key Concepts

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

Null Hypothesis

When conducting a statistical test, the null hypothesis is the statement that assumes no effect or no difference. It's the default or initial assumption that we try to test. In our bromeliad study exercise, the null hypothesis (\(H_0\)) claims that adding nitrogen to the bromeliad plants does not increase the mean number of new leaves formed. This hypothesis is crucial because we need a baseline to challenge with our collected data.
Here are some key features of the null hypothesis:

• It is typically denoted as \(H_0\).
• The null hypothesis assumes equality or no difference between groups.
• In our example, \(H_0: \mu_{control} = \mu_{nitrogen}\).

The null hypothesis helps us focus on the effects of the treatment (in this case, nitrogen) by providing a basis for comparison. In hypothesis testing, failing to reject the null hypothesis means that the data doesn't provide enough evidence to support the alternative hypothesis, under the set significance level.

Alternative Hypothesis

The alternative hypothesis (\(H_a\)) is the statement we aim to support through our statistical analysis. It suggests that there is an effect or a difference due to the treatment or condition being tested. Referring to the exercise, the alternative hypothesis posits that adding nitrogen to bromeliad plants increases the mean number of new leaves.A few important characteristics of the alternative hypothesis include:

• It is typically represented as \(H_a\).
• It contradicts the null hypothesis by suggesting a significant difference or effect.
• In our scenario, \(H_a: \mu_{control} < \mu_{nitrogen}\) implies a one-tailed test because it suggests an increase.

By proposing the alternative hypothesis, researchers seek to demonstrate that observed data differs significantly from what the null hypothesis predicts. If statistical evidence favors \(H_a\), the null can be rejected, supporting the notion that nitrogen does increase bromeliad leaf production.

t-Test

A t-Test is a statistical test used to determine if there are significant differences between the means of two groups, which may be related in certain features. In this exercise, a two-sample t-test is used to compare the control and nitrogen-treated groups of bromeliads. The specific test employed here is Welch's t-test, which is applicable when the variances of two samples are not assumed to be equal.Some key elements of performing a t-Test include:

• Calculate the sample means and standard deviations for each group.
• Use the t-statistic formula: \[t = \frac{\bar{x}_{nitrogen} - \bar{x}_{control}}{\sqrt{\frac{s_{control}^2}{n_{control}} + \frac{s_{nitrogen}^2}{n_{nitrogen}}}}\]
• Compare the calculated t-value against a critical value from a t-distribution table to make inferences.

Conducting a t-Test allows us to statistically infer whether observed variations in our bromeliad groups are due to nitrogen. Through this method, it helps in either rejecting or not rejecting our initial null hypothesis, indicating a potential influence of the treatment.

Sample Mean

The sample mean is a statistical measure representing the average value in a set of numbers. We use this to summarize the data collected from samples. In the case of the bromeliad exercise, calculating the sample means for both control and nitrogen groups helps us compare their average leaf production.Here is how we deal with sample means:

• For a given sample, sum all observed values and divide by the number of observations.
• The formula for the sample mean (\(\bar{x}\)) is: \[\bar{x} = \frac{\sum_{i=1}^{n}x_i}{n}\] where \(x_i\) represents each individual observation, and \(n\) is the total number of observations.
• In the exercise, \(\bar{x}_{control} = 13.29\) and \(\bar{x}_{nitrogen} = 15.63\).

Comparing sample means provides insight into differences between groups. This difference forms the basis of our t-Test analysis, indicating whether additional nitrogen affects the bromeliads' growth substantially.