/*! 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 95 A sample with \(n=10, \bar{x}=50... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 95

A sample with \(n=10, \bar{x}=508.5,\) and \(s=21.5\)

Short Answer

Expert verified
The exercise has provided three key statistics: the sample size (10 items), the sample mean (508.5), and the sample standard deviation (21.5). Depending on the specific task required, these can be used to calculate various statistical measures or carry out different types of statistical analysis.

Step by step solution

01

Understand the Given Statistics

The problem provides several pieces of statistical information. \(n=10\) indicates that you have a sample of 10 items or observations from a population. \(\bar{x}=508.5\) signifies that the average (mean) value of these 10 items is 508.5. Finally, \(s=21.5\) means that the standard deviation 鈥� a measure of how much the values in the sample vary from the mean 鈥� is 21.5.
02

Use the Statistics as Needed

Depending upon what you need to do with these statistics, use the appropriate statistical formulas. For instance, if you need to find a confidence interval for the population mean, you can use the formula for the confidence interval with a t-distribution (because the sample size is less than 30), which is \(\bar{x} 卤 t\frac{s}{\sqrt{n}}\). To carry out a hypothesis test on the population mean, you could use the test statistic \(t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}\), where \(\mu_0\) is the hypothesized population mean.

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

We saw in Exercise 6.260 on page 425 that drinking tea appears to offer a strong boost to the immune system. In a study extending the results of the study described in that exercise, \(^{70}\) blood samples were taken on five participants before and after one week of drinking about five cups of tea a day (the participants did not drink tea before the study started). The before and after blood samples were exposed to e.coli bacteria, and production of interferon gamma, a molecule that fights bacteria, viruses, and tumors, was measured. Mean production went from 155 \(\mathrm{pg} / \mathrm{mL}\) before tea drinking to \(448 \mathrm{pg} / \mathrm{mL}\) after tea drinking. The mean difference for the five subjects is \(293 \mathrm{pg} / \mathrm{mL}\) with a standard deviation in the differences of 242 . The paper implies that the use of the t-distribution is appropriate. (a) Why is it appropriate to use paired data in this analysis? (b) Find and interpret a \(90 \%\) confidence interval for the mean increase in production of interferon gamma after drinking tea for one week.

Random samples of the given sizes are drawn from populations with the given means and standard deviations. For each scenario: (a) Find the mean and standard error of the distribution of differences in sample means, \(\bar{x}_{1}-\bar{x}_{2}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 300 from Population 1 with mean 75 and standard deviation 18 and samples of size 500 from Population 2 with mean 83 and standard deviation 22

Involve scores from the high school graduating class of 2010 on the SAT (Scholastic Aptitude Test). The distribution of sample means \(\bar{x}_{m}-\bar{x}_{f},\) where \(\bar{x}_{m}\) represents the mean Critical Reading score for a sample of 50 males and \(\bar{x}_{f}\) represents the mean Critical Reading score for a sample of 50 females, is centered at 5 with a standard deviation of \(22.5 .\) Give notation and define the quantity we are estimating with these sample differences. In the population of all students taking the test, who scored higher on average, males or females?

Susan is in charge of quality control at a small fruit juice bottling plant. Each bottle produced is supposed to contain exactly 12 fluid ounces (fl oz) of juice. Susan decides to test this by randomly sampling 30 filled bottles and carefully measuring the amount of juice inside each. She will recalibrate the machinery if the average amount of juice per bottle differs from 12 fl oz at the \(1 \%\) significance level. The sample of 30 bottles has an average of \(11.92 \mathrm{fl}\) oz per bottle and a standard deviation of \(0.26 \mathrm{fl}\) oz. Should Susan recalibrate the machinery?

Impact of Sample Size on Accuracy Compute the standard error for sample proportions from a population with proportion \(p=0.4\) for sample sizes of \(n=30, n=200,\) and \(n=1000 .\) What effect does increasing the sample size have on the standard error? Using this information about the effect on the standard error, discuss the effect of increasing the sample size on the accuracy of using a sample proportion to estimate a population proportion.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

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.