/*! 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 68 Increased Yield An agronomist ha... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 68

Increased Yield An agronomist has shown experimentally that a new irrigation/fertilization regimen produces an increase of 2 bushels per quadrat (significant at the \(1 \%\) level) when compared with the regimen currently in use. The cost of implementing and using the new regimen will not be a factor if the increase in yield exceeds 3 bushels per quadrat. Is statistical significance the same as practical importance in this situation? Explain.

Short Answer

Expert verified
Question: Explain the difference between statistical significance and practical importance in the context of the agronomist's experiment. Answer: Statistical significance refers to the low probability that the observed difference in yield between the current and new regimen is due to random chance, while practical importance refers to whether the increase in yield justifies the costs of implementing the new regimen. In the agronomist's experiment, the new irrigation/fertilization regimen is statistically significant, as it increases the yield by 2 bushels per quadrat with a 1% error rate. However, it does not meet the practical importance threshold, because the increase in yield needs to exceed 3 bushels per quadrat to justify the costs of implementation.

Step by step solution

01

Understand Statistical Significance

In this case, the statistical significance at the 1% level means that there is a 99% chance that the new irrigation/fertilization regimen does indeed increase the yield when compared to the current regimen. It indicates that there is a low probability (1%) that the observed difference in yield is due to random chance and not due to the effects of the new regimen.
02

Understand Practical Importance

Practical importance depends on whether the increase in yield due to the new regimen is enough to justify the cost of implementing it. In this situation, practical importance implies an increase in yield of 3 bushels per quadrat or more. If the increase in yield does not exceed 3 bushels per quadrat, the cost of implementing the new regimen will not be justified, and it will not be considered practically important.
03

Comparing Statistical Significance and Practical Importance

The increase in yield from the new regimen is statistically significant, but it may not necessarily be practically important. Here, the new regimen increases yield by 2 bushels per quadrat, which is statistically significant at the 1% level. However, for practical importance, the increase in yield needs to exceed 3 bushels per quadrat. Since the observed increase is only 2 bushels per quadrat, the new regimen鈥檚 practical importance is not met.
04

Conclusion

In this situation, statistical significance and practical importance are not the same. The new irrigation/fertilization regimen has been proven statistically significant in increasing the yield by 2 bushels per quadrat. However, it does not meet the threshold for practical importance, as the increase in yield required to justify the cost of implementation is 3 bushels per quadrat.

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 91影视!

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

Independent random samples of \(n_{1}=140\) and \(n_{2}=140\) observations were randomly selected from binomial populations 1 and \(2,\) respectively. Sample 1 had 74 successes, and sample 2 had 81 successes. a. Suppose you have no preconceived idea as to which parameter, \(p_{1}\) or \(p_{2}\), is the larger, but you want to detect only a difference between the two parameters if one exists. What should you choose as the alternative hypothesis for a statistical test? The null hypothesis? b. Calculate the standard error of the difference in the two sample proportions, \(\left(\hat{p}_{1}-\hat{p}_{2}\right) .\) Make sure to use the pooled estimate for the common value of \(p\). c. Calculate the test statistic that you would use for the test in part a. Based on your knowledge of the standard normal distribution, is this a likely or unlikely observation, assuming that \(H_{0}\) is true and the two population proportions are the same? d. \(p\) -value approach: Find the \(p\) -value for the test. Test for a significant difference in the population proportions at the \(1 \%\) significance level. e. Critical value approach: Find the rejection region when \(\alpha=.01 .\) Do the data provide sufficient evidence to indicate a difference in the population proportions?

Female Models In a study to assess various effects of using a female model in automobile advertising, 100 men were shown photographs of two automobiles matched for price, color, and size, but of different makes. One of the automobiles was shown with a female model to 50 of the men (group \(\mathrm{A}\) ), and both automobiles were shown without the model to th other 50 men (group B). In group A, the automobile shown with the model was judged as more expensive by 37 men; in group \(B\), the same automobile was judged as the more expensive by 23 men. Do these results indicate that using a female model influences the perceived cost of an automobile? Use a one-tailec test with \(\alpha=.05 .\)

Treatment versus Control An experiment was conducted to test the effect of a new drug on a viral infection. The infection was induced in 100 mice, and the mice were randomly split into two groups of 50\. The first group, the control group, received no treatment for the infection. The second group received the drug. After a 30 -day period, the proportions of survivors, \(\hat{p}_{1}\) and \(\hat{p}_{2}\), in the two groups were found to be .36 and \(.60,\) respectively. a. Is there sufficient evidence to indicate that the drug is effective in treating the viral infection? Use \(\alpha=.05 .\) b. Use a \(95 \%\) confidence interval to estimate the actual difference in the cure rates for the treated versus the control groups.

A random sample of 120 observations was selected from a binomial population, and 72 successes were observed. Do the data provide sufficient evidence to indicate that \(p\) is greater than \(.5 ?\) Use one of the two methods of testing presented in this section, and explain your conclusions.

Define the power of a statistical test. As the alternative value of \(\mu\) gets farther from \(\mu_{0}\), how is the power affected?
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation
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.