Shrubs and fire Fire is a serious threat to shrubs in dry climates. Some shrubs can

resprout from their roots after their tops are destroyed. Researchers wondered if fire would help with resprouting. One study of resprouting took place in a dry area of Mexico. The researchers randomly assigned shrubs to treatment and control groups. They clipped the tops of all the shrubs. They then applied a propane torch to the stumps of the treatment group to simulate a fire. All 12of the shrubs in the treatment group resprouted. Only 8of the 12shrubs in the control group resprouted.

a. State appropriate hypotheses for performing a significance test. Be sure to define the parameters of interest.

b. Check if the conditions for performing the test are met.

Short Answer

Expert verified

a. The hypothesis are H0:p1=p2and Ha:p1>p2

b. All conditions are not met.

Step by step solution

01

Given Information

It is given that the researchers want to know that if there is a difference in the proportion of shrubs that received the treatment and resprouted and the shrubs that did not received the treatment and resprouted.

x1=12

x2=8

n1=12

n2=12

02

Appropriate hypothesis

Claim is that proportion is greater than shrubs in treatment group.

The appropriate hypothesis is:

Null: H0:p1=p2

Alternative: Ha:p1>p2

p1is proportion of shrubs that received the treatment and resprouted.

p2is proportion of shrubs that did not received the treatment and resprouted.

03

Conditions

Three conditions are:

Random: Independent random samples assigned to shrubs.

Independent: 12shrubs is less than 10%of all shrubs.

Normal: There are 12success and zero failures, zero is less than ten.

All conditions are not satisfied. We cannot use hypothesis for testing a claim.

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Most popular questions from this chapter

The P-value for the stated hypotheses is 0.002Interpret this value in the context of this study.

a. Assuming that the true mean road rage score is the same for males and females, there is a 0.002probability of getting a difference in sample means equal to the one observed in this study.

b. Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability of getting a difference in sample means at least as large in either direction as the one observed in this study.

c. Assuming that the true mean road rage score is different for males and females, there is a 0.002 probability of getting a difference in sample means at least as large in either direction as the one observed in this study.

d. Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability that the null hypothesis is true.

e. Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability that the alternative hypothesis is true.

A scatterplot and a least-squares regression line are shown in the figure. What effect does point P have on the slope of the regression line and the correlation?

a. Point P increases the slope and increases the correlation.

b. Point P increases the slope and decreases the correlation.

c. Point P decreases the slope and decreases the correlation.

d. Point P decreases the slope and increases the correlation .

e. No conclusion can be drawn because the other coordinates are unknown.

Friday the 13thRefer to Exercise 88.

a. Construct and interpret a 90%confidence interval for the true mean difference. If you already defined parameters and checked conditions in Exercise 88, you don’t need to do them again here.

b. Explain how the confidence interval provides more information than the test in Exercise 88.

Suppose the probability that a softball player gets a hit in any single at-bat is 0.300. Assuming that her chance of getting a hit on a particular time at bat is independent of her other times at bat, what is the probability that she will not get a hit until her fourth time at bat in a game?

a.(43)(0.3)1(0.7)33051526=0.200=20.0%43(0.3)1(0.7)3

b.(43)(0.3)3(0.7)13051526=0.200=20.0%43(0.3)3(0.7)1

C.(41)(0.3)3(0.7)13051526=0.200=20.0%41(0.3)3(0.7)1

d.(0.3)3(0.7)13051526=0.200=20.0%(0.3)3(0.7)1

e.(0.3)1(0.7)33051526=0.200=20.0%(0.3)1(0.7)3

Coaching and SAT scores What we really want to know is whether coached students improve more than uncoached students, on average, and whether any advantage is large enough to be worth paying for. Use the information above to answer these questions:

a. How much more do coached students gain, on average, compared to uncoached students? Construct and interpret a 99%confidence interval.

b. Does the interval in part (a) give convincing evidence that coached students gain more, on average, than uncoached students? Explain your answer.

c. Based on your work, what is your opinion: Do you think coaching courses are worth paying for?

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