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Researchers suspect that Variety A tomato plants have a higher average yield than Variety \(\mathrm{B}\) tomato plants. To find out, researchers randomly select 10 Variety A and 10 Variety \(\mathrm{B}\) tomato plants. Then the researchers divide in half each of 10 small plots of land in different locations. For each plot, a coin toss determines which half of the plot gets a Variety A plant; a Variety \(\mathrm{B}\) plant goes in the other half. After harvest, they compare the yield in pounds for the plants at each location. The 10 differences in yield (Variety \(\mathrm{A}\) - Variety \(\mathrm{B}\) ) are recorded. A graph of the differences looks roughly symmetric and single-peaked with no outliers. A paired \(t\) test on the differences yields \(t=1.295\) and \(P\) -value \(=0.1138\). (a) State appropriate hypotheses for the paired \(t\) test. Be sure to define your parameter. (b) What are the degrees of freedom for the paired \(t\) test? (c) Interpret the \(P\) -value in context. What conclusion should the researchers draw? (d) Describe a Type I error and a Type II error in this setting. Which mistake could researchers have made based on your answer to part (c)?

Step by step solution

01

State Hypotheses

For a paired \( t \) test, we test the difference between two related samples. Here, we define \( \mu_d \) as the true mean difference in yields between Variety A and B plants. The null hypothesis \( H_0 \) is that there is no difference in mean yields, implying \( \mu_d = 0 \). The alternative hypothesis \( H_a \) is that Variety A yields more than Variety B on average, implying \( \mu_d > 0 \). Thus, we have:\[H_0: \mu_d = 0 \H_a: \mu_d > 0\]

02

Determine Degrees of Freedom

The degrees of freedom for a paired \( t \) test is calculated as the number of differences minus one. Since there are 10 differences in yield, the degrees of freedom are \( 10 - 1 = 9 \).

03

Interpret the P-value

The \( P \)-value of 0.1138 indicates the probability of observing a test statistic as extreme as \( t = 1.295 \), assuming the null hypothesis is true. Since the \( P \)-value (0.1138) is greater than a typical significance level (such as 0.05), we do not have enough evidence to reject the null hypothesis. Therefore, the researchers should conclude that there is not enough evidence to support the claim that Variety A has a higher yield than Variety B.

04

Describe Potential Errors

A Type I error would occur if we incorrectly reject the null hypothesis, concluding that Variety A has a higher yield when it does not. A Type II error occurs when we fail to reject the null hypothesis, not concluding that Variety A has a higher yield when it actually does. Based on our conclusion in Step 3, the researchers could have made a Type II error if there indeed is a mean difference, but the test failed to detect it.

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Key Concepts

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

Null Hypothesis

The null hypothesis is a fundamental idea in hypothesis testing. In the context of the paired t-test, the null hypothesis establishes a baseline that there is no effect or difference. For the tomato plant example, the null hypothesis (\( H_0 \)) suggests that there is no difference in the mean yields between Variety A and Variety B tomato plants. This is mathematically expressed as:

• \( H_0: \mu_d = 0 \), where \( \mu_d \) is the true mean difference in yields between the two varieties.

To "test" this hypothesis, researchers use data to assess whether there is enough statistical evidence to reject it. By defining a null hypothesis, we set a reference point for evaluating the observed data against what we would expect to observe if there was in fact no difference between the two tomato varieties.

Degrees of Freedom

Understanding degrees of freedom is crucial for interpreting statistical tests. In a paired t-test, the degrees of freedom is calculated as one less than the number of paired observations. It helps adjust the test to the sample size and influences the shape of the t-distribution used in the test.
For the tomato plant experiment, researchers collected 10 observations of the yield differences. Therefore, the degrees of freedom (\( df \)) are:

• \( df = 10 - 1 = 9 \)

The degrees of freedom are significant because they characterize the variability in the data and impact the statistical conclusions we draw. Different degrees of freedom alter the critical t-values, which in turn influence how researchers determine statistical significance.

P-value Interpretation

The P-value is a metric used to gauge the strength of evidence against the null hypothesis. It represents the probability of obtaining results at least as extreme as those observed, presuming the null hypothesis is true.
In this case study, a P-value of 0.1138 indicates the likelihood of seeing a test statistic like the calculated t-value of 1.295 if Variety A and Variety B truly had no difference in yield. Generally, if the P-value is greater than a typical significance level like 0.05, researchers do not reject the null hypothesis.

• Here, since the P-value (\( 0.1138 \)) is greater than 0.05, there's insufficient evidence to conclude that Variety A has a higher yield.

While interpreting P-values, remember that a smaller value means stronger evidence against the null hypothesis, and a larger value means weaker evidence.

Type I and Type II Errors

Type I and Type II errors are potential pitfalls in hypothesis testing. Each represents a different kind of mistake that can occur.
A Type I error occurs when we incorrectly reject the null hypothesis, which would mean incorrectly concluding that Variety A has a different yield when it does not truly have one.

• This error is serious because we claim a difference where none exists.

A Type II error, on the other hand, happens when we fail to reject the null hypothesis, even though there is, in fact, a genuine difference.

• In this scenario, researchers might miss out on detecting that Variety A does have a higher yield.

Based on the conclusion from the P-value interpretation, the researchers might have made a Type II error if there is indeed a yield difference that wasn't detected by the test.