/*! 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} Q17E Buy-side vs. sell-side analysts'... [FREE SOLUTION] | 91影视

91影视

Buy-side vs. sell-side analysts' earnings forecasts. Refer to the Financial Analysts Journal (Jul. /Aug. 2008) study of financial analysts' forecast earnings, Exercise 2.86 (p. 112). Recall that data were collected from 3,526 buy-side analysts and 58,562forecasts made by sell-side analysts, and the relative absolute forecast error was determined for each. The mean and standard deviation of forecast errors for both types of analysts are given in the table.

a. Construct a 95% confidence interval for the difference between the mean forecast error of buy-side analysts and the mean forecast error of sell-side analysts.

b. Based on the interval, part a, which type of analysis has the greater mean forecast error? Explain.

c. What assumptions about the underlying populations of forecast errors (if any) are necessary for the validity of the inference, part b?

Short Answer

Expert verified

(a) The 95% confidence interval for the difference between the mean forecast error of buy-side analysts and the mean forecast error of sell-side analysts is (0.8359,0.9640).

(b) The buy-side analyst has a meaner forecast error.

(c) The assumptions for validating part b are as follows:

(i) The samples should be randomly selected.

(ii) The sample size should be sufficiently large enough.

Step by step solution

01

Step-by-Step SolutionStep 1: Formula used

After we get the critical value, find the 95% confidence interval with the use of this formula below:

(x1-x2)z/2s12n1+s22n2

Wherex1 is the mean forecast error of the buy-side analystx2 is the mean forecast error of the sell-side analysts12 squared the standard deviation of the buy-side analysts22 squared the standard deviation of the sell-side analystn1 total number of samples - buy-side analyst total n2number of samples - sell-side analystz/2 is the critical values of z-test.

02

To find the level of confidence

Given the below:

One survey was done on financial analysts. The information was gathered from 3,526 buy-side analyst projections and 58,562 sell-side analyst estimates. The information is represented in the table.

n1=3,526,n2=58,562,x1=0.85,x2=-0.05,s1=1.93ands2=0.85

Let 1be the mean forecast error of buy-side analysts and

Let2be the mean forecast error of sell-side analysts.

Critical value:

The level of confidence is 95%.

=0.05

2=0.052

=0.025

Hence, the cumulative area to the left is as follows:

Area to the left - Area to the right

=1-0.025

=0.975

From Table II of the standard normal distribution in Appendix D, the critical value is 1.96

Confidence interval:

CI=(x1-x2)za221n1+21n2=(0.85-(-0.05))1.961.9323,526+0.85258,562=0.900.0640=(.8359,0.9640)

Thus, the 95%confidence interval for the difference between the mean forecast error of buy-side analysts and the mean forecast error of sell-side analysts is(0.8359,0.9640).

03

(b) Based on the confidence interval, which type of analysis has the greater mean forecast error

The95%confidence interval for the difference between the mean forecast error of buy-side analysts and the mean forecast error of sell-side analysts is(.8359,0.9640).

The interval does not contain 0. So, the mean difference is greater than zero. There is evidence to say that the difference is present between the two groups.

All values in the interval are positive; therefore, it indicates that the mean forecast error of buy-side analysts is more than the mean forecast error of sell-side analysts.

04

(c) Assumptions that make this inference valid

The assumptions for this test are the following:

  • Two samples are randomly selected from the two target populations.
  • Sample sizes are both large.

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

History of corporate acquisitions. Refer to the Academy of Management Journal (August 2008) investigation of the performance and timing of corporate acquisitions, Exercise 2.12 (p. 74). Recall that the investigation discovered that in a random sample of 2,778 firms, 748 announced one or more acquisitions during the year 2000. Does the sample provide sufficient evidence to indicate that the true percentage of all firms that announced one or more acquisitions during the year 2000 is less than 30%? Use a=0.05to make your decision.

Given that x is a random variable for which a Poisson probability distribution provides a good approximation, use statistical software to find the following:

a.P(x2) when =1

b.P(x2) when =2

c.P(x2) when =3

d. What happens to the probability of the event {x2} as it increases from 1 to 3? Is this intuitively reasonable?

What are the treatments for a designed experiment with two factors, one qualitative with two levels (A and B) and one quantitative with five levels (50, 60, 70, 80, and 90)?

Refer to the Archives of Paediatrics and Adolescent Medicine (Dec. 2007) study of honey as a children鈥檚 cough remedy, Exercise 2.31 (p. 86). Children who were ill with an upper respiratory tract infection and their parents participated in the study. Parents were instructed to give their sick child dosage of liquid 鈥渕edicine鈥 before bedtime. Unknown to the parents, some were given a dosage of dextromethorphan (DM)鈥攁n over-the-counter cough medicine鈥攚hile others were given a similar dose of honey. (Note: A third group gave their children no medicine.) Parents then rated their children鈥檚 cough symptoms, and the improvement in total cough symptoms score was determined for each child. The data (improvement scores) for the 35 children in the DM dosage group and the 35 in the honey dosage group are reproduced in the next table. Do you agree with the statement (extracted from the article), 鈥淗oney may be a preferable treatment for the cough and sleep difficulty associated with childhood upper respiratory tract infection鈥? Use the comparison of the two means methodology presented in this section to answer the question.

The data is given below:

Honey Dosage:

12111511101310415169141061081112128129111510159138121089512

DM Dosage:

469477791210116349781212412137101394410159126

Shopping vehicle and judgment. Refer to the Journal ofMarketing Research (December 2011) study of shopping cart design, Exercise 2.85 (p. 112). Recall that design engineers want to know whether the mean choice of the vice-over-virtue score is higher when a consumer鈥檚 arm is flexed (as when carrying a shopping basket) than when the consumer鈥檚 arm is extended (as when pushing a shopping cart). The average choice score for the n1 = 11 consumers with a flexed arm was x1= 59, while the average for the n2 = 11
Consumers with an extended arm was x2= 43. In which scenario is the assumption required for a t-test to compare means more likely to be violated, S1= 4 and S2= 2, or, S1= 10 and S2 = 15? Explain.

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.