/*! 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. 10. Attitudes Refer to Exercise 4. I... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Q. 10. (page 564)

Attitudes Refer to Exercise 4. In the study of older students鈥� attitudes, the sample mean SSHA score was 125.7 and the sample standard deviation was 29.8. A significance test yields a P-value of 0.0101.

a. Explain what it would mean for the null hypothesis to be true in this setting.

b. Interpret the P-value.

Short Answer

Expert verified

Part a) The correct mean score for the students who are at least 30years of age is 115

Part b) The P-value is if the population mean is equal to115, then there is the possibility of1.01%of getting a random sample with a sample mean of125.7or more.

Step by step solution

01

Part a) Step 1: Given information

From the previous exercise,

H0:渭=115Ha:渭>115

渭is the population mean score of students who are at least 30 years of age.

02

Part a) Step 2: The objective is to explain the mean for the null hypothesis to be true in this setting

If the null hypothesis H0:渭=115is correct, the true mean score for students who are at least 30years old is115

03

Part b) Step 1: Given information

n=45x炉=125.7s=29.8P=0.0101=1.01%

04

Part b) Step 2: The objective is to explain the p value

Result of the previous exercise:

H0:渭=115Ha:渭>115

If the population means is115then there is a 1.01%chance of getting a random sample with a sample means of125.7 or higher.

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

How much juice? Refer to Exercise 3. The mean amount of liquid in the bottles is 179.6ml and the standard deviation is 1.3ml. A significance test yields a P-value of 0.0589. Interpret the P-value.

Tests and confidence intervals The P-value for a two-sided test of the null hypothesis H0:渭=15is0.03

a. Does the 99% confidence interval for 渭 include 15? Why or why not?

b. Does the 95% confidence interval for 渭 include 15? Why or why not?

Candy! A machine is supposed to fill bags with an average of 19.2 ounces of candy. The manager of the candy factory wants to be sure that the machine does not consistently underfill or overfill the bags. So the manager plans to conduct a significance test at the 伪=0.10significance level of

H0:渭=19.2Ha:渭notequalto19.2

where 渭=the true mean amount of candy (in ounces) that the machine put in all bags filled that day. The manager takes a random sample of 75 bags of candy produced that day and weighs each bag. Check if the conditions for performing the test are met.

Calculations and conclusions Refer to Exercise R9.1. Find the standardized test statistic and P-value in each setting, and make an appropriate conclusion.

Fast connection? How long does it take for a chunk of information to travel

from one server to another and back on the Internet? According to the site

internettrafficreport.com, the average response time is 200 milliseconds (about one-fifth of a second). Researchers wonder if this claim is true, so they collect data on response times (in milliseconds) for a random sample of 14 servers in Europe. A graph of the data reveals no strong skewness or outliers.

a. State an appropriate pair of hypotheses for a significance test in this setting. Be sure to define the parameter of interest.

b. Check conditions for performing the test in part (a).

c. The 95% confidence interval for the mean response time is 158.22 to 189.64

milliseconds. Based on this interval, what conclusion would you make for a test of the hypotheses in part (a) at the 5% significance level?

d. Do we have convincing evidence that the mean response time of servers in the United States is different from 200 milliseconds? Justify your answer.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Probability and Statistics

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.