Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Coffee markets that conform to organic standards focus on the environmental aspects of coffee growing, such as the use of shade trees and a reduced reliance on chemical pesticides. A study of organic coffee growers was published in Food Policy (Vol. 36, 2010). In a representative sample of 845 coffee growers from southern Mexico, 417 growers were certified to sell to organic coffee markets while 77 growers were transitioning to become organic certified. In the United States, 60% of coffee growers are organic certified. Is there evidence to indicate that fewer than 60% of the coffee growers in southern Mexico are either organic certified or transitioning to become organic certified? State your conclusion so that there is only a 5% chance of making a Type I error.

Short Answer

Expert verified

At a 5% significance level, we do not have sufficient evidence to conclude that fewer than 60% of the coffee growers in southern Mexico are either got an organic certification or transitioning to become organically certified.

Step by step solution

01

Given information

In the United States, the proportion of coffee growers who got an organic certification is 60%.

As per an article published in Food Policy (Vol. 36, 2010), out of 845 coffee growers from southern Mexico, 494 (=417+77) coffee growers either got an organic certification or transitioning to become organically certified.

That is

The size of the sample isn=845

The sample proportion is

p^=494845=0.585

Where is the sample proportion of coffee growers who are either got an organic certification or transitioning to become organically certified.

02

Setting up the hypotheses

We have to test whether coffee growers have either got organic certification or are transitioning to become certified organic is lesser than 60%.

The null and alternative hypotheses are given as

H0:p=0.60

That is, in southern Mexico, the true proportion of coffee growers who are either got an organic certification or transitioning to become organically certified is not less than 60%.

And

Ha:p<0.60

That is, in southern Mexico, the true proportion of coffee growers who are either got an organic certification or transitioning to become organically certified is fewer than 60%.

03

Calculating the test statistic

The test statistic for testing these hypotheses is

Z=p^-pp1-pn=0.585-0.600.601-0.60845=-0.0150.000284024=-0.89

04

Calculating the critical value

Here

α:The level of significance (chance of making a type I error)

α=.05

Using the standard normal table, the critical value at the 5% significance level is -1.645.

We can see that

Z=-0.89>-1.645

Hence, we failed to reject the null hypothesis.

05

Conclusion

At a 5% significance level, we do not have sufficient evidence to conclude that fewer than 60% of the coffee growers in southern Mexico are either got an organic certification or transitioning to become organically certified.

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 Vaia!

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

If the rejection of the null hypothesis of a particular test would cause your firm to go out of business, would you want ato be small or large? Explain

The National Association of Realtors (NAR) reported the results of an April 2015 survey of home buyers. In a random sample of 1,971 residential properties purchased during the year, 414 were purchased as a vacation home. Five years ago, 10% of residential properties were vacation homes.

a. Do the survey results allow the NAR to conclude (at α=.01) that the percentage of all residential properties purchased for vacation homes is greater than 10%?

b. In a previous year, the NAR sent the survey questionnaire to a nationwide sample of 45,000 new home owners, of which 1,982 responded to the survey. How might this bias the results? [Note: In the most recent survey, the NAR used a more valid sampling method.

Performance of stock screeners. Recall, from Exercise 6.36 (p. 350), that stock screeners are automated tools used by investment companies to help clients select a portfolio of stocks to invest in. The data on the annualized percentage return on investment (as compared to the Standard & Poor’s 500 Index) for 13 randomly selected stock screeners provided by the American Association of Individual Investors (AAII) are repeated in the accompanying table. You want to determine whether \(\mu \) , the average annualized return for all AAII stock screeners, is positive (which implies that the stock screeners perform better, on average, than the S&P 500). An XLSTAT printout of the analysis is shown on the top of page 418.

9.0 -.1 -1.6 14.6 16.0 7.7 19.9 9.8 3.2 24.8 17.6 10.7 9.1

  1. State \({H_0}\,and\,{H_a}\) for this test

Stability of compounds in new drugs. Refer to the ACS Medicinal Chemistry Letters (Vol. 1, 2010) study of the metabolic stability of drugs, Exercise 2.22 (p. 83). Recall that two important values computed from the testing phase are the fraction of compound unbound to plasma (fup) and the fraction of compound unbound to microsomes (fumic). A key formula for assessing stability assumes that the fup/fumic ratio is 1:1. Pharmacologists at Pfizer Global Research and Development tested 416 drugs and reported the fup/fumic ratio for each. These data are saved in the FUP file, and summary statistics are provided in the accompanying Minitab printout. Suppose the pharmacologists want to determine if the true mean ratio, μ, differs from 1.

a. Specify the null and alternative hypotheses for this test.

b. Descriptive statistics for the sample ratios are provided in the Minitab printout on page 410. Note that the sample mean,\(\overline x = .327\)is less than 1. Consequently, a pharmacologist wants to reject the null hypothesis. What are the problems with using such a decision rule?

c. Locate values of the test statistic and corresponding p-value on the printout.

d. Select a value of\(\alpha \)the probability of a Type I error. Interpret this value in the words of the problem.

e. Give the appropriate conclusion based on the results of parts c and d.

f. What conditions must be satisfied for the test results to be valid?

What is the level of significance of a test of hypothesis?

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.

Sign-up for free