/*! 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} Q 8.19RP. Millionaires. Dr. Thomas Stanley... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Millionaires. Dr. Thomas Stanley of Georgia State University has surveyed millionaires since 1973 Among other information, Stanley obtains estimates for the mean age, μ, of all U.S. millionaires. Suppose that 36 randomly selected U.S. millionaires are the following ages, in years.

Determine a 95%confidence interval for the mean age μof all U.S. millionaires. Assume that the standard deviation of ages of all U.S. millionaires is 13.0years. (Note: The mean of the data is 58.53years)

Short Answer

Expert verified

We are 95%confident that mean age, μ of all us. Millionaires, lies some where between 54.28 years and 62.78 years

Step by step solution

01

Given information

The figure is given:
314579644838396852
596879427953746666
716152473954675571
776460754269485748
02

Concept

The formula used:x¯-zα/2σn,x¯+zα/2σn

03

Calculation

Let the population mean age be μ

Population S.D σ=13 years

Sample size n=36

Sample mean x¯=58.53

x¯ is roughly normally distributed because the sample is huge.

The 100(1-α)%Cl of μ is calculated using the z-interval technique.

x¯-zα/2σn,x¯+zα/2σn

Here confidence level =95%=100×0.95%

∴1-α=0.95⇒α=1-0.95⇒α=0.05⇒α/2=0.025∴zα/2=z125=1.96
04

Calculation

∴95%Clofμ=x¯-Z0.025σn,x¯+z0.025σn=58.53-1.96×1336,58.53+1.96×1336=(58.53-4.25,58.53+4.25)=(54.28,62.78)

i.e., we are 95%certain that the average age of all of us is μMillionaires are born between the ages of 54.28 and 62.78

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 91Ó°ÊÓ!

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

Why do you need to consider the student zed version of x¯ to develop a confidence-interval procedure for a population means when the population standard deviation is unknown?

"Chips Ahoy! 1,000 Chips Challenge." As reported by B. Warner and J. Rutledge in the paper "Checking the Chips Ahoy! Guarantee" (Chanee, Vol. 12. Issue 1. pp. 10-14), a random sample of forty-two 18-ounce bags of Chips Ahoy! cookies yielded a mean of 1261.6 chips per bag with a standard deviation of 117.6 chips per bug.

a. Determine a 95% confidence interval for the mean number of chips per bag for all 18-ounce bags of Chips Ahoy! cookies, and interpret your result in words.

b. Can you conclude that the average 18-ounce bag of Chips Ahoy! cookies contain at least 1000 chocolate chips? Explain your answer.

For a fixed confidence level, show that (approximately) quadrupling the sample size is necessary to halve the margin of error. (Hint: Use Formula 8.2.)

M\&Ms. In the article "Sweetening Statistics-What M\&M's Can Teach Us" (Minitab Inc., August 2008), M. Paret and E. Martz discussed several statistical analyses that they performed on bags of M\&Ms. The authors took a random sample of 30 small bags of peanut M\&Ms and obtained the following weights, in grams (g).

a. Determine a 95%lower confidence bound for the mean weight of all small bags of peanut M\&Ms. (Note: The sample mean and sample standard deviation of the data are 52.040gand 2.807grespectively.)

b. Interpret your result in pant (a).

c. According to the package, each small bag of peanut M\&Ms should weigh 49.3gComment on this specification in view of your answer to part (b) It provides equal confidence with a greater lower limit.

Part (c) Because the weight of 49.3g is below the 95% lower confidence bound.

One-Sided One-Mean t-Intervals. Presuming that the assumptions for a one-mean t-interval are satisfied, we have the following formulas for (1-α)-level confidence bounds for a population mean μ:

  • Lower confidence bound: x¯-tα-s/Ï€
  • Upper confidence bound: x^+tα·s/n

Interpret the preceding formulas for lower and upper confidence bounds in words.

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.