/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 Growing tomatoes An agricultural... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Growing tomatoes An agricultural field trial compares the yield of two varieties of tomatoes for commercial use. Researchers randomly select 10 Variety A and 10 Variety 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 B plant goes in the other half. After harvest, they compare the yield in pounds for the plants at each location. The 10 differences (Variety \(A-\) Variety \(B )\) give \(\overline{x}=0.34\) and \(s_{x}=0.83\) A graph of the differences looks roughly symmetric and single- peaked with no outliers. Is there convincing evidence that Variety A has the higher mean yield? Perform a significance test using \(\alpha=0.05\) to answer the question.

Short Answer

Expert verified
There is no convincing evidence that Variety A has a higher mean yield.

Step by step solution

01

State the Hypotheses

We need to determine whether there is sufficient evidence that the mean yield of Variety A tomatoes is higher than that of Variety B. Set up the null hypothesis as \( H_0: \mu_A - \mu_B = 0 \), indicating no difference in mean yield. The alternative hypothesis is \( H_a: \mu_A - \mu_B > 0 \), suggesting that Variety A has a higher mean yield than Variety B.
02

Check Assumptions

Since the data comes from a normal distribution as suggested by the symmetric and single-peaked graph with no outliers, and the sample size is 10, it is appropriate to use a t-test for the mean difference.
03

Calculate the Test Statistic

Use the formula for the t-statistic: \[ t = \frac{\overline{x} - 0}{s_x / \sqrt{n}} \] where \( \overline{x} = 0.34 \), \( s_x = 0.83 \), and \( n = 10 \). Substituting in these values gives \[ t = \frac{0.34 - 0}{0.83 / \sqrt{10}} = \frac{0.34}{0.2627} \approx 1.29 \].
04

Determine the Critical Value

For a one-tailed test at \( \alpha = 0.05 \) with \( df = n - 1 = 9 \), use the t-distribution table to find the critical value. The critical value for \( t \) is approximately 1.833.
05

Make a Decision

Compare the calculated t-statistic (1.29) with the critical value (1.833). Since 1.29 is less than 1.833, we fail to reject the null hypothesis \( H_0 \).
06

Conclusion

There is not enough statistical evidence to support the claim that Variety A has a higher mean yield than Variety B at the 0.05 significance level. Therefore, we cannot conclude that Variety A is superior in yield based on this trial.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Hypothesis Testing
Hypothesis testing is a fundamental method in statistics used to determine whether there is enough evidence to support a particular belief or hypothesis about a data set. It typically involves two hypotheses: the null hypothesis and the alternative hypothesis.
- The **null hypothesis** (\(H_0\)) is a statement of no effect or no difference; it is what you assume is true until you have evidence to suggest otherwise. In this exercise, it states that the mean yield of Variety A is equal to that of Variety B.
- The **alternative hypothesis** (\(H_a\)) represents what you want to test for. It hints that there is an effect or a difference, such as Variety A yielding more than Variety B.
Once the hypotheses are set, the next step is to decide on a significance level (\(\alpha\)), which in this exercise is 0.05. This level is the threshold for deciding whether to reject the null hypothesis. A lower p-value than \(\alpha\) suggests that there is enough evidence against the null hypothesis, whereas a higher p-value indicates failure to reject it.
t-Test
A t-test is a statistical test used to compare the means of two groups. In our case, it's used to see if the difference in yield between Variety A and Variety B is statistically significant.
For the t-test to be valid, a few assumptions need to be met: the data should be normally distributed, data points should be independent, and the data should come from populations with equal variances.
In this example, we check for normal distribution, and it's confirmed by the graph showing symmetric, single-peaked data with no outliers.
The t-statistic is calculated using the formula: \[ t = \frac{\overline{x} - 0}{s_x / \sqrt{n}} \]where \(\overline{x}=0.34\), \(s_x=0.83\), and \(n=10\).
The t-statistic helps in understanding how many standard deviations the sample mean is away from the null hypothesis mean (which is zero in this case). If it's far enough, we might consider the results statistically significant.
Critical Value
The critical value is a cutoff point that helps determine whether to reject the null hypothesis. It's obtained from a statistical table based on the significance level and degrees of freedom (\(df\)).
For a one-tailed t-test as done here, the significance level, \(\alpha\), is 0.05, and the degrees of freedom is calculated as \(n - 1\), which is 9.
You look up the t-distribution table to find the critical value matching your confidence level and \(df\). In this situation, the critical value is approximately 1.833.
The critical value helps set a boundary; if your test statistic falls beyond this point, you have statistically significant results and can reject the null hypothesis. If it's within, like our example's t-statistic of 1.29, there's not enough evidence to do so.
P-Value
A p-value is a probability measure that helps to determine the strength of the evidence against the null hypothesis. It signifies the likelihood of obtaining a result at least as extreme as the one observed, under the assumption that the null hypothesis is true.
In hypothesis testing, the p-value is compared to the pre-determined significance level (\(\alpha\)). If the p-value is smaller than \(\alpha\), you reject the null hypothesis, indicating that the results are statistically significant. Conversely, if the p-value is larger, you fail to reject it.
While the p-value calculation wasn't directly shown in this solution, understanding it is crucial. It complements the t-test by providing a probability concerned with the observed data under specific assumptions. In this example, since the calculated t-statistic of 1.29 was less than the critical value, it implies that the p-value is greater than 0.05, endorsing our decision not to reject the null hypothesis that there's no significant difference in mean yields.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Explain why we aren’t safe carrying out a one-sample z test for the population proportion p. No test You toss a coin 10 times to test the hypothesis \(H_{0} : p=0.5\) that the coin is balanced.

Tests and CIs The P-value for a two-sided test of the null hypothesis \(H_{0} : \mu=10\) is \(0.06 .\) (a) Does the 95\(\%\) confidence interval for \(\mu\) include 10 ? Why or why not? (b) Does the 90\(\%\) confidence interval for \(\mu\) include 10? Why or why not?

Side effects A drug manufacturer claims that less than 10% of patients who take its new drug for treating Alzheimer’s disease will experience nausea. To test this claim, researchers conduct an experiment. They give the new drug to a random sample of 300 out of 5000 Alzheimer’s patients whose families have given informed consent for the patients to participate in the study. In all, 25 of the subjects experience nausea. Use these data to perform a test of the drug manufacturer’s claim at the \(\alpha=0.05\) significance level.

You are testing \(H_{0} : \mu=10\) against \(H_{a} : \mu \neq 10\) based on an SRS of 15 observations from a Normal population. What values of the \(t\) statistic are statistically significant at the \(\alpha=0.005\) level? $$ \begin{array}{ll}{\text { (a) } t>3.326} & {\text { (d) } t<-3.326 \text { or } t>3.326} \\ {\text { (b) } t>3.286} & {\text { (e) } t<-3.286 \text { or } t>3.286}\end{array} $$ (c) \(t > 2.977\)

One-sided test Suppose you carry out a significance test of \(H_{0} : \mu=5\) versus \(H_{a} : \mu>5\) based on a sample of size \(n=20\) and obtain \(t=1.81\) (a) Find the P-value for this test using (i) Table B and (ii) your calculator. What conclusion would you draw at the 5% significance level? At the 1% significance level? (b) Redo part (a) using an alternative hypothesis of \(H_{a} : \mu \neq 5\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.