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鈥淲ould you marry a person from a lower social class than your own?鈥 Researchers asked this question of a random sample of 385black, never married students at two historically black colleges in the South. Of the 149men in the sample, 91said 鈥淵es.鈥 Among the 236women, 117said 鈥淵es.鈥14Is there reason to think that different proportions of men and women in this student population would be willing to marry beneath their class?

Holly carried out the significance test shown below to answer this question. Unfortunately, she 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

Ha:p1p2

at the 95%confidence level.

Plan: If conditions are met, I鈥檒l do a one-sample ztest for comparing two proportions.

  • Random The data came from a random sample of 385 black, never-married students.
  • Normal One student鈥檚 answer to the question should have no relationship to another student鈥檚 answer.
  • Independent The counts of successes and failures in the two groups91,58,117, and 119are all at least 10

Do: From the data, p^1=91149=0.61and p^2=117236=0.46.

Test statistic

z=(0.61-0.46)-00.61(0.39)149+0.46(0.54)236=2.91

p=value From Table A, role="math" localid="1650292307192" P(z2.91)1-0.39820.0018.

Conclude: The p-value, 0.0018, is less than 0.05, so I鈥檒l reject the null hypothesis. This proves that a higher proportion of men than women are willing to marry someone from a social class lower than their own.

Short Answer

Expert verified

There is sufficient evidence to support the claim of a difference between the population proportions.

Step by step solution

01

Given Information

Given

x1=91

n1=149

x2=117

role="math" n2=236

Determine the hypothesis

H0:p1-p2=0

Ha:p1-p20

02

Explanation

The sample proportion is the number of successes divided by the sample size:

p^1=x1n1=911490.611

p^2=x2n2=1172360.496

p^p=x1+x2n1+n2=91+117385=2083850.540

Determine the value of the test statistic:

localid="1650450515979" z=p^1-p^2p^p1-p^p1n1+1n2=0.611-0.4960.540(1-0.540)1149+12362.21

The p-value is the probability of obtaining the value of the test statistic, or a value more extreme. Determine the p-value using table A:


localid="1650450537687" P=P(Z<-2.21orZ>2.21)=2P(Z<-2.21)=20.0136=0.0272

If the p-value is smaller than the significance level, reject the null hypothesis:

P<0.05RejectH0

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Most popular questions from this chapter

A driving school wants to find out which of its two instructors is more effective at preparing students to pass the state鈥檚 driver鈥檚 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鈥檚 students and 22of Instructor B鈥檚 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 =0.05

Plan: If conditions are met, I鈥檒l 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 =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.

In an experiment to learn whether Substance M can help restore memory, the brains of 20rats were treated to damage their memories. The rats were trained to run a maze. After a day, 10rats (determined at random) were given M and 7of them succeeded in the maze. Only 2of the 10control rats were successful. The two-sample z test for 鈥渘o difference鈥 against 鈥渁 significantly higher proportion of the M group succeeds鈥

(a) gives z=2.25,P<0.02

(b) gives z=2.60,P<0.005

(c) gives z=2.25,P<0.04but not <0.02

(d) should not be used because the Random condition is violated

(e) should not be used because the Normal condition is violated.

Are teens or adults more likely to go online daily? The Pew Internet and American Life Project asked a random sample of 800 teens and a separate random sample of 2253 adults how often they use the Internet. In these two surveys, 63% of teens and 68%of adults said that they go online every day. Construct and interpret a 90% confidence interval for p1-p2?

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.

Coaching and SAT scores (10.1) What proportion of students who take the SAT twice are coached? To answer this question, Jannie decides to construct a 99%confidence interval. Her work is shown below. Explain what鈥檚 wrong with Jannie鈥檚 method.

A 99%CI for p1-p2is

(0.135-0.865)2.5750.135(0.865)3160+0.865(0.135)2733=-0.730.022=(-0.752,-0.708)

We are 99% confident that the proportion of students taking the SAT twice who are coached is between 71 and 75 percentage points lower than students who aren鈥檛 coached.

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