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Chapter 8: Q51E (page 488)

Patron amenability to supply biomass. Relate to the Biomass and Energy (Vol. 36, 2012) study of the amenability of directors to supply biomass products similar to fat hay, Exercise8.20 (p. 469). Recall that independent samples of Missouri directors and Illinois directors were surveyed. Another aspect of the study concentrated on the service directors who were willing to supply. One essential service involves windrowing (mowing and piling) hay. Of the 558 Missouri directors surveyed, 187 were willing to offer windrowing. Of the 940 Illinois directors surveyed, 380 were willing to offer windrowing services. The experimenters want to know if the proportion of directors willing to offer windrowing services to the biomass request differs for the two areas, Missouri and Illinois.

a. Specify the parameter of interest to the experimenters.

b. Set up the null and indispensable suppositions for testing whether the proportion of directors willing to offer windrowing services differs in Missouri and Illinois.

c. A Minitab analysis of the data is given below. Detect the test statistic on the printout.

d. provide the rejection region for the test using a = .01.

e. Detect the p- the value of the test on the printout.

f. Make the applicable conclusion using both the p-value and rejection region approach. Your conclusions should agree.

Short Answer

Expert verified

A service provides value to consumers by enabling desired results without the ownership of particular potential costs.

Step by step solution

01

Step-by-Step Solution Step 1: Specify the parameter of interest

The parameter of interest is the difference between the proportions of producers who were willing to offer windrowing services to the biomass market area Missouri (p1) and the proportions of producers who were willing to offer windrowing services to the biomass market area Ilinois (p2).

That is, the parameter of interest is P1鈥� P2.

02

Write the null and alternative hypotheses

Null hypothesis:

H0: P鈧�-P鈧�=0

There is no difference between the proportions of producers who were willing to offer windrowing services to the biomass market area in Missouri and Illinois.

Alternative hypothesis:

Ha: P鈧�-P鈧佲墵0

There is a difference between the proportions of producers who were willing to offer windrowing services to the biomass market area in Missouri and Illinois.

03

from the MINITAB, the test statistic

from the MINITAB output, the test statistic (z) value is -2.67.

04

state is a region of rejection

Let the confidence level be 0.99.

1 鈥� 伪 = 0.99

伪 = 1 鈥� 0.99

= 0.01

a2=0.005

05

From the appendix D table-2

From appendix D- Table2, the value of za/2is given below.

za2= z0.005

= 2.58

So, the value of Za/2 is 2.58 .

Rejection region :

If z > za/2(=2.58), then reject the null hypothesis H0.

If z > za/2(= - 2.58), then reject the null hypothesis H0.

06

Minitab

From the MINITAB output, the p-value is 0.008.

07

conclusion for rejection region approach

The critical value is 2.58, and the value of z is 2.67.

Here, the value is greater than the value of Za/2.

That is, z(= 2.67) > Za/2, (= 2.58).

so,by the rejection rule, reject the null hypothesis (H0).

Thus, it can be concluded that there is evidence to reject the null hypothesis (H0) at 伪=0.01.

Hence, there is a difference between the proportions of producers willing to supply windrowing services to the biomass market area in Missouri and Illinois.

08

Conclusion for the p-value approach

Use the significance level, 伪 =0.01.

Here p-value is 0.008, which is lesser than the level of significance.

That is, (p-value = 0.008) <(伪=0.01).

Therefore, by the condition, If the p-value < 伪, reject the null hypothesis.

Hence, reject the null hypothesis H0.

Thus, it can be concluded that there is evidence to reject the null hypothesis at a=0.01.

There is a difference between the proportions of producers who were willing to supply windrowing services to the biomass market area in Missouri and Illinois.

Hence, the conclusions obtained from both approaches are the same.

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Most popular questions from this chapter

Homework assistance for accounting students. How much assistance should accounting professors provide students for completing homework? Is too much assistance counterproductive? These were some of the questions of interest in a Journal of Accounting Education (Vol. 25, 2007) article. A total of 75 junior-level accounting majors who were enrolled in Intermediate Financial Accounting participated in an experiment. All students took a pretest on a topic not covered in class; then, each was given a homework problem to solve on the same topic. However, the students were randomly assigned different levels of assistance on the homework. Some (20 students) were given the completed solution, some (25 students) were given check figures at various steps of the solution, and the rest (30 students) were given no help. After finishing the homework, each student was given a posttest on the subject. One of the variables of interest to the researchers was the knowledge gain (or test score improvement), measured as the difference between the posttest and pretest scores. The sample means knowledge gains for the three groups of students are provided in the table.

a. The researchers theorized that as the level of homework assistance increased, the test score improvement from pretest to post test would decrease. Do the sample means reported in the table support this theory?

b. What is the problem with using only the sample means to make inferences about the population mean knowledge gains for the three groups of students?

c. The researchers conducted a statistical test of the Hypothesis to compare the mean knowledge gain of students in the "no solutions" group with the mean knowledge gain of students in the "check figures" group. Based on the theory, part a sets up the null and alternative hypotheses for the test.

d. The observed significance level of the t-test of the partc was reported as8248 Using 伪=.05, interpret this result.

e. The researchers conducted a statistical test of the hypothesis to compare the mean knowledge gain of students in the "completed solutions" group with the mean knowledge gain of students in the "check figures" group. Based on the theory, part a sets up the null and alternative hypotheses for the test.

f. The observed significance level of the role="math" localid="1652694732458" t-test of part e was reported as 1849. Using 伪=.05, interpret this result.

g. The researchers conducted a statistical test of the Hypothesis to compare the mean knowledge gain of students in the "no solutions" group with the mean knowledge gain of students in the "completed solutions" group. Based on the theory, part a sets up the null and alternative hypotheses for the test.

h. The observed significance level of the t-test of the part wasg reported as2726. Using role="math" localid="1652694677616" 伪=.05, interpret this result.

Refer to the Archives of Paediatrics and Adolescent Medicine (Dec. 2007) study of honey as a children鈥檚 cough remedy, Exercise 2.31 (p. 86). Children who were ill with an upper respiratory tract infection and their parents participated in the study. Parents were instructed to give their sick child dosage of liquid 鈥渕edicine鈥� before bedtime. Unknown to the parents, some were given a dosage of dextromethorphan (DM)鈥攁n over-the-counter cough medicine鈥攚hile others were given a similar dose of honey. (Note: A third group gave their children no medicine.) Parents then rated their children鈥檚 cough symptoms, and the improvement in total cough symptoms score was determined for each child. The data (improvement scores) for the 35 children in the DM dosage group and the 35 in the honey dosage group are reproduced in the next table. Do you agree with the statement (extracted from the article), 鈥淗oney may be a preferable treatment for the cough and sleep difficulty associated with childhood upper respiratory tract infection鈥�? Use the comparison of the two means methodology presented in this section to answer the question.

The data is given below:

Honey Dosage:

12111511101310415169141061081112128129111510159138121089512

DM Dosage:

469477791210116349781212412137101394410159126

Consider the discrete probability distribution shown here.

x

10

12

18

20

p

.2

.3

.1

.4

a. Calculate渭,蟽2 and蟽 .

b. What isP(x<15) ?

c. Calculate 渭卤2蟽 .

d. What is the probability that xis in the interval 渭卤2蟽 ?

Question: Independent random samples n1 =233 and n2=312 are selected from two populations and used to test the hypothesis Ha:(渭1-渭)2=0against the alternative Ha:(渭1-渭)2鈮�0

.a. The two-tailed p-value of the test is 0.1150 . Interpret this result.b. If the alternative hypothesis had been Ha:(渭1-渭)2<0 , how would the p-value change? Interpret the p-value for this one-tailed test.

Gonzaga University professors conducted a study of television commercials and published their results in the Journal of Sociology, Social Work and Social Welfare (Vol. 2, 2008). The key research question was as follows: 鈥淒o television advertisers use religious symbolism to sell goods and services?鈥� In a sample of 797 TV commercials collected ten years earlier, only 16 commercials used religious symbolism. Of the sample of 1,499 TV commercials examined in the more recent study, 51 commercials used religious symbolism. Conduct an analysis to determine if the percentage of TV commercials that use religious symbolism has changed over time. If you detect a change, estimate the magnitude of the difference and attach a measure of reliability to the estimate.
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