Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Q 8.8 (page 320)

A simple random sample is taken from a population and yields the following data for a variable of the population:

Find a point estimate for the population mean (i.e., the mean of the variable).

Short Answer

Expert verified

The point estimatate for the population mean is 10

Step by step solution

01

Step 1:Gien information

02

Step 2:Formulae used

The fomulae to find the sample mean is,

x¯=∑xin

Where,∑xiis the sum of all obserations andn is the total number of observations.

03

Step 3:Simplification

The sample mean is

x¯=∑xin=20+2+6+2+12+6+9+8+16+8+2111=11011=10

Thus,The point estimatate for the population mean is10

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Toxic Mushrooms?Cadmium, a heavy metal, is toxic to animals. Mushrooms, however, are able to absorb and accumulate cadmium at high concentrations. The Czech and Slovak governments have set a safety limit for cadmium in dry vegetables at 0.5part per million (ppm). M. Melgar et al. measured the cadmium levels in a random sample of the edible mushroom Bolefus pinicola and published the results in the paper "Influence of Some Factors in Toxicity and Accumulation of Cd from Edible Wild Macrofungi in NW Spain (Journal of Envionmental Science and Health. Here are the data obtained by the researchers.

Find and interpret a 99%confidence interval for the mean cadmium level of all Boletus pinicola mushrooms. Assume a population standard deviation of cadmium levels in Bolefus pinicola mushrooms of 0.37ppm.

Find the confidence level and ∝ for

a. 90%confidence interval.

b. 94%confidence interval.

Suppose that you take 1000simple random samples from a population and that, for each sample, you obtain a 95% confidence interval for an unknown parameter. Approximately how many of those confidence intervals will contain the value of the unknown parameter?

Class Project: Gestation Periods of Humans. This exercise can be done individually or, better yet, as a class project. Gestation periods of humans are normally distributed with a mean of 266 days and a standard deviation of 16 days.

a. Simulate 100 samples of nine human gestation periods each.

b. For each sample in part (a), obtain a 95% confidence interval for the population mean gestation period.

c. For the 100 confidence intervals that you obtained in part (b), roughly how many would you expect to contain the population mean gestation period of 266 days?

d. For the 100 confidence intervals that you obtained in part (b), determine the number that contain the population mean gestation period of 266 days.

e. Compare your answers from parts (c) and (d), and comment on any observed difference.

Venture-Capital Investments. In Exercise 8.69, you found a 95%confidence interval for the mean amount of all venture-capital investments in the fiber optics business sector to be from \(5.389million to\)7.274million. Obtain the margin of error by

a. taking half the length of the confidence interval.

b. using Formula 8.1on page 325. (Recall that and that σ=$2.04million.)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation
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.

Sign-up for free