The two-sample t statistic for the road rage study (male mean minus female mean) is t=3.18. The P-value for testing the hypotheses from the previous exercise satisfies

(a) 0.001<P<0.005.

(b) 0.0005<P<0.001.

(c) 0.001<P<0.002.

(d) 0.002<P<0.01.

(e) P>0.01.

Short Answer

Expert verified

The correct option is (b).

Step by step solution

01

Given Information

Number of men and women are 596and 523respectively.

Test statistic (t)=3.18.

02

Explanation

The degree of freedom can be computed as:

df=minn1,n2

=min(596,523)

=523

Now, the P-value is:

0.0005<P<0.001

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Most popular questions from this chapter

Mrs. Woods and Mrs. Bryan are avid vegetable gardeners. They use different fertilizers, and each claims that hers is the best fertilizer to use when growing tomatoes. Both agree to do a study using the weight of their tomatoes as the response variable. They had each planted the same varieties of tomatoes on the same day and fertilized the plants on the same schedule throughout the growing season. At harvest time, they each randomly select 15tomatoes from their respective gardens and weigh them. After performing a two-sample t-test on the difference in mean weights of tomatoes, they get t=5.24and P=0.0008. Can the gardener with the larger mean claim that her fertilizer caused her tomatoes to be heavier?

(a) No, because the soil conditions in the two gardens is a potential confounding variable.

(b) No, because there was no replication.

(c) Yes, because a different fertilizer was used on each garden.

(d) Yes, because random samples were taken from each garden.

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(a) Construct and interpret a 99%confidence interval for the difference in population proportions (men minus women).

(b) Does your interval from part (a) give convincing evidence of a difference between the population proportions? Explain.

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(a) 0.03±(0.28)(0.72)125+(0.25)(0.75)68

(b) localid="1650367573248" 0.03±1.96(0.28)(0.72)125+(0.25)(0.75)68

(c) 0.03±1.645(0.28)(0.72)125+(0.25)(0.75)68

(d)

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Find the probability that p1^-p2^is less than or equal to 0.02. Show your work.

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