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Chapter 10: Q.1 (page 664)

A study of road rage asked separate random samples of 596men and 523women about their behavior while driving. Based on their answers, each re-spondent was assigned a road rage score on a scale of 0to 20. Are the conditions for performing a two-sample t test satisfied?

a) Maybe; we have independent random samples, but we need to look at the data to check Normality.

(b) No; road rage scores in a range between 0 and 20 can’t be Normal.

(c) No; we don’t know the population standard deviations.

(d) Yes; the large sample sizes guarantee that the corresponding population distributions will be Normal.

(e) Yes; we have two independent random samples and large sample sizes.

Short Answer

Expert verified

The correct answer is (e) Yes; we have two independent random samples and large sample sizes.

Step by step solution

01

Given Information

A study of road rage asked separate random samples of 596 men and 523 women about their behavior while driving.

02

Explanation

Conditions for performing a two-sample t-test: Random, Normal and Independent.

Random: Satisfied, because the samples have been given to be random samples.

Normal: Can be assumed to be satisfied, because the sample size of each sample is at least 30.

Independent: Satisfied. The random samples have been given to be separate and thus are independent. Moreover, the sample sizes are less than 10% of the population size.

Thus all three conditions have been satisfied and thus answer (e) is correct.

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Most popular questions from this chapter

According to sleep researchers, if you are between the ages of 12and18years old, you need 9hours of sleep to be fully functional. A simple random sample of 28students was chosen from a large high school, and these students were asked how much sleep they got the previous night. The mean of the responses was 7.9hours, with a standard deviation of 2.1hours.If we are interested in whether students at this high school are getting too little sleep, which of the following represents the appropriate null and alternative hypotheses?

(a) H0:μ=7.9andHa:μ<7.9

(b) H0:μ=7.9andHa:μ≠7.9

(c) H0:μ=9andHa:μ≠9

(d)H0:μ=9andHa:μ<9

(e)H0:μ≤9andHa:μ≥9

A driving school wants to find out which of its two instructors is more effective at preparing students to pass the state’s driver’s license exam. An incoming class of 100students is randomly assigned to two groups, each of size 50. One group is taught by Instructor A; the other is taught by Instructor B. At the end of the course, 30of Instructor A’s students and 22of Instructor B’s students pass the state exam. Do these results give convincing evidence that Instructor A is more effective?

Min Jae carried out the significance test shown below to answer this question. Unfortunately, he made some mistakes along the way. Identify as many mistakes as you can, and tell how to correct each one.

State: I want to perform a test of

H0:p1-p2=0

Ha:p1-p2>0

where p1=the proportion of Instructor A's students that passed the state exam and p2=the proportion of Instructor B's students that passed the state exam. Since no significance level was stated, I'll use σ=0.05

Plan: If conditions are met, I’ll do a two-sample ztest for comparing two proportions.

Random The data came from two random samples of 50students.

- Normal The counts of successes and failures in the two groups -30,20,22, and 28-are all at least 10.

- Independent There are at least 1000 students who take this driving school's class.

Do: From the data, p^1=2050=0.40and p^2=3050=0.60. So the pooled proportion of successes is

p^C=22+3050+50=0.52

- Test statistic

localid="1650450621864" z=(0.40-0.60)-00.52(0.48)100+0.52(0.48)100=-2.83

- p-value From Table A, localid="1650450641188" P(z≥-2.83)=1-0.0023=0.9977.

Conclude: The p-value, 0.9977, is greater than α=0.05, so we fail to reject the null hypothesis. There is no convincing evidence that Instructor A's pass rate is higher than Instructor B's.

42. Long words Mary was interested in comparing the mean word length in articles from a medical journal and an airline’s in-flight magazine. She counted the number of letters in the first 200 words of an article in
the medical journal and in the first 100 words of an article in the airline magazine. Mary then used Minitab statistical software to produce the histograms shown.

Construct and interpret a 95%confidence interval for p1-p2 in Exercise 24. Explain what additional information the confidence interval provides.

Coaching and SAT scores (10.2) What we really want to know is whether coached students improve more than uncoached students and whether any advantage is large enough to be worth paying for. Use the information above to answer these questions:

(a) Is there good evidence that coached students gained more on average than uncoached students? Carry out a significance test to answer this question.

(b) How much more do coach students gain on the average? Construct and interpret a 99% confidence interval.

(c) Based on your work, what is your opinion: do you think coaching courses are worth paying for?
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