/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q162S Given that x聽is a hypergeometri... [FREE SOLUTION] | 91影视

91影视

Given that xis a hypergeometric random variable, computep(x)for each of the following cases:

a. N= 8, n= 5, r= 3, x= 2

b. N= 6, n= 2, r= 2, x= 2

c. N= 5, n= 4, r= 4, x= 3

Short Answer

Expert verified

a. The probability distributionpx is 0.5357.

b. The probability distributionpx is 0.2.

c. The probability distributionpx is 0.8.

Step by step solution

01

Given information

X is a hypergeometricrandom variable.

02

Computing the probability p(x) when N = 8, n = 5, r = 3, x = 2

For a hypergeometric random variable, the probability distribution

px=rxN-rn-xNn

a.

HereN=8,n=5,r=3,x=2

px=rxN-rn-xNn=328-35-285=325385=31056=0.5357

Hence, the probability distributionpx is 0.5357.

03

Computing the probability p(x) when N = 6, n = 2, r = 2, x = 2

b.

HereN=5,n=4,r=4,x=3

px=rxN-rn-xNn=226-32-262=324062=3115=0.2

Hence, the probability distributionpx is 0.2.

04

Computing the probability p(x) when N = 5, n = 4, r = 3, x = 3

c.

Here,N=5,n=4,r=4,x=3

px=rxN-rn-xNn=435-44-354=431154=415=0.8

Hence, the probability distributionpx is 0.8.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find a value of the standard normal random variable z, call it z0, such that

a.P(zz0)=0.2090b.P(zz0)=0.7090c.P(-z0z<z0)=0.8472d.P(-z0z<z0)=0.1664e.P(z0zz0)=0.4798f.P(-1<z<z0)

Is honey a cough remedy? Refer to the Archives of Pediatrics and Adolescent Medicine (Dec. 2007) study of honey as a children鈥檚 cough remedy, Exercise 8.23 (p. 470). The data (cough improvement scores) for the 33 children in the DM dosage group and the 35 children in the honey dosage group are reproduced in the table below. In Exercise 8.23, you used a comparison of two means to determine whether 鈥渉oney may be a preferable treatment for the cough and sleep difficulty associated with childhood upper respiratory tract infection.鈥 The researchers also want to know if the variability in coughing improvement scores differs for the two groups. Conduct the appropriate analysis, using =0.10

Homework assistance for accounting students. How much assistance should accounting professors provide students for completing homework? Is too much assistance counterproductive? These were some of the questions of interest in a Journal of Accounting Education (Vol. 25, 2007) article. A total of 75 junior-level accounting majors who were enrolled in Intermediate Financial Accounting participated in an experiment. All students took a pretest on a topic not covered in class; then, each was given a homework problem to solve on the same topic. However, the students were randomly assigned different levels of assistance on the homework. Some (20 students) were given the completed solution, some (25 students) were given check figures at various steps of the solution, and the rest (30 students) were given no help. After finishing the homework, each student was given a posttest on the subject. One of the variables of interest to the researchers was the knowledge gain (or test score improvement), measured as the difference between the posttest and pretest scores. The sample means knowledge gains for the three groups of students are provided in the table.

a. The researchers theorized that as the level of homework assistance increased, the test score improvement from pretest to post test would decrease. Do the sample means reported in the table support this theory?

b. What is the problem with using only the sample means to make inferences about the population mean knowledge gains for the three groups of students?

c. The researchers conducted a statistical test of the Hypothesis to compare the mean knowledge gain of students in the "no solutions" group with the mean knowledge gain of students in the "check figures" group. Based on the theory, part a sets up the null and alternative hypotheses for the test.

d. The observed significance level of the t-test of the partc was reported as8248 Using =.05, interpret this result.

e. The researchers conducted a statistical test of the hypothesis to compare the mean knowledge gain of students in the "completed solutions" group with the mean knowledge gain of students in the "check figures" group. Based on the theory, part a sets up the null and alternative hypotheses for the test.

f. The observed significance level of the role="math" localid="1652694732458" t-test of part e was reported as 1849. Using =.05, interpret this result.

g. The researchers conducted a statistical test of the Hypothesis to compare the mean knowledge gain of students in the "no solutions" group with the mean knowledge gain of students in the "completed solutions" group. Based on the theory, part a sets up the null and alternative hypotheses for the test.

h. The observed significance level of the t-test of the part wasg reported as2726. Using role="math" localid="1652694677616" =.05, interpret this result.

Assume that 12=蟽22=蟽2. Calculate the pooled estimator 2 for each of the following cases:

a.s12=120,s22=100,n1=n2=25

b.s12=12,s22=20,n1=20,n2=10

c.s12=.15,s22=.20,n1=6,n2=10

d.s12=3000,s22=2500,n1=16,n2=17

Note that the pooled estimate is a weighted average of the sample variances. To which of the variances does the pooled estimate fall nearer in each of the above cases?

The gender diversity of a large corporation鈥檚 board of directors was studied in Accounting & Finance (December 2015). In particular, the researchers wanted to know whether firms with a nominating committee would appoint more female directors than firms without a nominating committee. One of the key variables measured at each corporation was the percentage of female board directors. In a sample of 491firms with a nominating committee, the mean percentage was 7.5%; in an independent sample of 501firms without a nominating committee, the mean percentage was role="math" localid="1652702402701" 4.3% .

a. To answer the research question, the researchers compared the mean percentage of female board directors at firms with a nominating committee with the corresponding percentage at firms without a nominating committee using an independent samples test. Set up the null and alternative hypotheses for this test.

b. The test statistic was reported as z=5.1 with a corresponding p-value of 0.0001. Interpret this result if 伪=0.05.

c. Do the population percentages for each type of firm need to be normally distributed for the inference, part b, to be valid? Why or why not?

d. To assess the practical significance of the test, part b, construct a 95% confidence interval for the difference between the true mean percentages at firms with and without a nominating committee. Interpret the result.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.