/*! 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} Problem 580 A research worker wishes to esti... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 580

A research worker wishes to estimate the mean of a population using a sample large enough that the probability will be \(.95\) that the sample mean will not differ from the population mean by more than 25 percent of the standard deviation. How large a sample should he take?

Short Answer

Expert verified
The research worker should take a sample of size 62 to ensure there is a \(.95\) probability that the sample mean will not differ from the population mean by more than \(25\%\) of the standard deviation.

Step by step solution

01

Determine the value of Z

For a \(.95\) probability, we will use a \(95\%\) confidence interval. Since the standard normal distribution is symmetrical, we will look for the critical value (\(Z_{\alpha/2}\)) with an area of \(.975\) to the right of it (i.e., the cut-off point of the middle \(95\%\)). Using the standard normal distribution table or calculator functions, we find that: \[Z_{0.975} \approx 1.96\] This is the critical value needed for our calculation.
02

Set up the equation

Now that we have found the value of Z, we can set up the equation. Since the allowable error is \(25\%\) of the standard deviation, we can write it as: \[E = 0.25\sigma\] Now, we can use the formula we mentioned above: \[n = \left(\frac{Z \cdot \sigma}{E}\right)^2\] In our case: \[n = \left(\frac{1.96 \cdot \sigma}{0.25\sigma}\right)^2\]
03

Solve for the sample size

Let us solve for \(n\): \[n = \left(\frac{1.96}{0.25}\right)^2\] \[n = (7.84)^2\] \[n = 61.4656\] Since we cannot have a fraction of a sample, we round up to ensure we meet the desired level of accuracy: \[n = 62\] So, the research worker should take a sample of size 62 to ensure there is a \(.95\) probability that the sample mean will not differ from the population mean by more than \(25\%\) of the standard deviation.

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

Let \(\mathrm{T}\) be distributed with density function \(f(t)=\lambda e^{-\lambda . t} \quad\) for \(t>0\) and \(=0\) otherwise If \(S\) is a new random variable defined as \(S=\) In \(\mathrm{T}\), find the density function of \(\mathrm{S}\).

Let the random variable \(\mathrm{X}\) represent the number of defective radios in a shipment of four radios to a local appliance store. Assume that each radio is equally likely to be defective or non-defective, hence the probability that a radio is defective is \(\mathrm{p}=1 / 2\). Also assume whether or not each radio is defective or non-defective is indipendent of the status of the other radios. Find the expected number of defective radios.

Suppose we want to compare 2 treatments for curing acne (pimples). Suppose, too, that for practical reasons we are obliged to use a presenting sample of patients. We might then decide to alternate the 2 treatments strictly according to the order in which the patients arrive \((\mathrm{A}, \mathrm{B}, \mathrm{A}, \mathrm{B}\), and so on). Let us agree to measure the cure in terms of weeks to reach \(90 \%\) improvement (this may prove more satisfactory than awaiting \(100 \%\) cure, for some patients, may not be completely cured by either treatment, and many patients might not report back for review when they are completely cured). This design would ordinarily call for Wilcoxon's Sum of Ranks Test, but there is one more thing to be considered: severity of the disease. For it could happen that a disproportionate number of mild cases might end up, purely by chance, in one of the treatment groups, which could bias the results in favor of this group, even if there was no difference between the 2 treatments. It would clearly be better to compare the 2 treatments on comparable cases, and this can be done by stratifying the samples. Suppose we decide to group all patients into one or other of 4 categories \- mild, moderate, severe, and very severe. Then all the mild cases would be given the 2 treatments alternatively and likewise with the other groups. Given the results tabulated below (in order of size, not of their actual occurrence), is the evidence sufficient to say that one treatment is better than the other? $$ \begin{array}{|c|c|c|} \hline \text { Category } & \begin{array}{c} \text { Treatment A } \\ \text { Weeks } \end{array} & \begin{array}{c} \text { Treatment B } \\ \text { Weeks } \end{array} \\ \hline \text { (I) Mild } & 2 & 2 \\ & 3 & 4 \\ \hline \text { (II) Moderate } & 3 & 4 \\ & 5 & 6 \\ & 6 & 7 \\ & 10 & 9 \\ \hline \text { (III) Severe } & 6 & 9 \\ & 8 & 14 \\ & 11 & 14 \\ \hline \text { (IV) Very severe } & 8 & 12 \\ & 10 & 14 \\ & 11 & 15 \\ \hline \end{array} $$

Consider the joint distribution of \(\mathrm{X}\) and \(\mathrm{Y}\) given in the form of a table below. The cell (i,j) corresponds to the joint probability that \(\mathrm{X}=\mathrm{i}, \mathrm{Y}=\mathrm{j}\), for \(\mathrm{i}=1,2,3, \mathrm{j}=1,2,3\) $$ \begin{array}{|c|c|c|c|} \hline \mathrm{Y}^{\mathrm{X}} & 1 & 2 & 3 \\ \hline 1 & 0 & 1 / 6 & 1 / 6 \\ \hline 2 & 1 / 6 & 0 & 1 / 6 \\ \hline 3 & 1 / 6 & 1 / 6 & 0 \\ \hline \end{array} $$ Check that this is a proper probability distribution. What is the marginal distribution of \(\mathrm{X} ?\) What is the marginal distribution of \(\mathrm{Y}\) ?

Harvey of Brooklyn surveyed a random sample of 625 students at SUNY-Stony Brook. Being a pre-medical student, he hoped that most students would major in the social sciences rather than the natural sciences, thus provide him with less competition. To Harvey's dismay, \(60 \%\) of the students he surveyed were majoring in the natural sciences. Construct a \(95 \%\) confidence interval for \(p\), the population proportion of students majoring in the natural sciences.
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

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.