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

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– P2.

02

Write the null and alternative hypotheses

Null hypothesis:

H0: P₁-P₁=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: P₁-P₁≠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 – α = 0.99

α = 1 – 0.99

= 0.01

= 0.005

05

From the appendix D table-2

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

= 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 α=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, α =0.01.

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

That is, (p-value = 0.008) <(α=0.01).

Therefore, by the condition, If the p-value < α, 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

A random sample of size n = 121 yielded p^ = .88.

a. Is the sample size large enough to use the methods of this section to construct a confidence interval for p? Explain.

b. Construct a 90% confidence interval for p.

c. What assumption is necessary to ensure the validity of this confidence interval?

To use the t-statistic to test for a difference between the means of two populations, what assumptions must be made about the two populations? About the two samples?

Assume that σ12222. Calculate the pooled estimator σ2 for each of the following cases:

a.s12=120,s22=100,n1=n2=25

b.s12=12,s22=20,n1=20,n2=10

c.s12=.15,s22=.20,n1=6,n2=10

d.s12=3000,s22=2500,n1=16,n2=17

Note that the pooled estimate is a weighted average of the sample variances. To which of the variances does the pooled estimate fall nearer in each of the above cases?

Find the following probabilities for the standard normal random variable z:

a.P(0<z<2.25)b.P(-2.25<z<0)b.P(-2.25<z<1.25)d.P(-2.50<z<1.50)e.P(z<-2.33orz>2.33)

The data for a random sample of six paired observations are shown in the next table.

a. Calculate the difference between each pair of observations by subtracting observation two from observation 1. Use the differences to calculate d¯andsd2.

b. If μ1andμ2are the means of populations 1 and 2, respectively, expressed μdin terms of μ1andμ2.

PairSample from Population 1

(Observation 1)

Sample from Population 2(Observation 2)
123456739648417247

c. Form a 95% confidence interval for μd.

d. Test the null hypothesis H0d=0against the alternative hypothesis Had0. Useα=.05 .

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