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Credit cards and incentives A bank wants to know which of two incentive plans will most increase the use of its credit cards. It offers each incentive to a group of current credit card customers, determined at random, and compares the amount charged during the following six months.

(a) Is this a problem with comparing means or comparing proportions? Explain.

(b) What type of study design is being used to produce data?

Short Answer

Expert verified

From the given information,

a) It is about comparing means.

b)Experimental type of study design is being used to produce data.

Step by step solution

01

Part (a) Step 1: Given Information

It is given in the question that, a bank needs to determine which of two incentive strategies will increase Visa usage the greatest. It gives every incentive to a randomly selected group of current Visa clients and considers the amount charged over the preceding six months.

02

Part (a) Step 2: Explanation

In this situation, two scores are involved, and scores are numerical numbers. Mean scores can be calculated because the data is numerical. As a result, the problem can be characterised as a comparison of means..

03

Part (b) Step 1: Given Information

It is given in the question that, a bank needs to determine which of two incentive strategies will increase Visa usage the greatest. It gives every incentive to a randomly selected group of current Visa clients and considers the amount charged over the preceding six months. What type of study design is being used to produce data?

04

Part (b) Step 2: Explanation

The volunteers in an experimental study are given treatment and their responses are tracked. In this experiment, subjects were utilized to track the response.

As a result, it's merely a test.

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Most popular questions from this chapter

Prayer and pregnancy Two hundred women who were about to undergo IVF served as subjects in an experiment. Each subject was randomly assigned to cither a treatment group or at control group. Women in the treatment group were intentionally prayed for by several people (called intercessors) who did not know them, a process known as intercessory prayer. 'The praying continued for three weeks following VT. The intercessors did not pray for the women in the control group. Here are the results: 44of the 88women in the treatment group got pregnant, compared to 21out of 81in the control group.

Is the pregnancy rate significantly higher for women who received intercessory prayer? To find out, researchers perform a test of H0:P1=P2versus H:P1>P2, where P1&P2are the actual pregnancy rates for women like those in the study who do and don't receive intercessory prayer, respectively,

(a) Name the appropriate test and check that the conditions for carrying out this test are met.

(b) The appropriate test from part (a) yields a P-value of 0.0007. Interpret this P-value in context.

(c) What conclusion should researchers draw at the =0.05significance level? Explain.

(d) The women in the study did not know if they were being prayed for. Explain why this is important.

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.

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鈥檛 be Normal.

(c) No; we don鈥檛 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.

A large university is considering the establishment of a schoolwide recycling program. To gauge interest in the program by means of a questionnaire, the university takes separate random samples of undergraduate students, graduate students, faculty, and staff. This is an example of what type of sampling design?

(a) Simple random sample

(b) Stratified random sample

(c) Convenience sample

(d) Cluster sample

(e) Systematic sample

A study of road rage asked samples of 596men and 523women about their behaviour while driving. Based on their answers, each person was assigned a road rage score on a scale of 0to20. The participants were chosen by random digit dialling of telephone numbers. We suspect that men are more prone to road rage than women. To see if this is true, test these hypotheses for the mean road rage scores of all male and female drivers

(a) H0:M=FversusH0:M>F

(b) H0:M=FversusH0:MF

(c) role="math" localid="1650363760526" H0:M=FversusH0:M<F

(d) H0:xM=xFversusH0:xM>xF

(e)H0:xM=xFversusH0:xM<xF

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