/*! 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. 10.105 Left-Tailed Hypothesis Tests and... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Left-Tailed Hypothesis Tests and CIs. If the assumptions for a nonpooled t-interval are satisfied, the formula for a (1-α) level upper confidence bound for the difference, μ1-μ2. between two population means is

f1-f2+t0·s12/n1+s22/n2

For a left-tailed hypothesis test at the significance level α, the null hypothesis H0:μ1=μ2 will be rejected in favor of the alternative hypothesis H2:μ1<μ2 if and only if the (1-α)-level upper confidence bound for μ1-μ2 is less than or equal to 0. In each case, illustrate the preceding relationship by obtaining the appropriate upper confidence bound and comparing the result to the conclusion of the hypothesis test in the specified exercise.

a. Exercise 10.83

b. Exercise 10.84

Short Answer

Expert verified

a) 95%upper bound is negative.

b)95%upper bound is negative.

Step by step solution

01

Part(a) Step 1: Given Information

To detect a comparison between the hypothesis test result and the confidence interval.

02

Part (a) Step 2: Explanation

Consider the following data,

Consider the confidence interval's upper bound of95%

x¯1-x¯2+ta2·s12n1+s22n2=(7.36-10.50)+(-2.015)1.22214+4.5926

=-6.97256

As a result, the 95% upper bound is negative. As a result, the mean number of acute postoperative days in the hospital is lower than in the dynamic system.

03

Part(b) Step 1: Given Information

To determine the confidence interval and hypothesis test result comparison.

04

Part(b) Step 2: Explanation 

Consider the following data

Consider the confidence interval's upper bound of 95%

x¯1-x¯2+ta2·s12n1+s22n2=(67.90-66.81)+(-1.83311)

5.49210+9.04231

=-3.267

The intervention program lowers the mean heart rate of urban bus drivers since the 95%upper bound is negative.

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

Right-Tailed Hypothesis Tests and CIs. If the assumptions for a pooled t-interval are satisfied, the formula for a (1-α)-level lower confidence bound for the difference, μ1-μ2, between two population means is

x¯1-x^2-ta·Sp1/n1+1/n2

For a right-tailed hypothesis test at the significance level α,

the null hypothesis H0:μ1=μ2will be rejected in favor of the alternative hypothesis Ha:μ1>μ2if and only if the (1-α)-level lower confidence bound for μ1-μ2is greater than or equal to 0. In each case, illustrate the preceding relationship by obtaining the appropriate lower confidence bound and comparing the result to the conclusion of the hypothesis test in the specified exercise.

a. Exercise 10.47

b. Exercise 10.50

Stressed-Out Bus Drivers. An intervention program designed by the Stockholm Transit District was implemented to improve the work conditions of the city's bus drivers. Improvements were evaluated by G. Evans et al., who collected physiological and psychological data for bus drivers who drove on the improved routes (intervention) and for drivers who were assigned the normal routes (control). Their findings were published in the article "Hassles on the Job: A Study of a Job Intervention with Urban Bus Drivers" (Journal of Organizational Behavior, Vol. 20, pp. 199-208). Following are data, based on the results of the study, for the heart rates, in beats per minute, of the intervention and control drivers.

a. At the 5%significance level, do the data provide sufficient evidence to conclude that the intervention program reduces mean heart rate of urban bus drivers in Stockholm? (Note; x1=67.90, s1=5.49,x¯2=66.81and s2=9.04.

b. Can you provide an explanation for the somewhat surprising results of the study?

c. Is the study a designed experiment or an observational study? plain your answer.

Two-Tailed Hypothesis Tests and CIs. As we mentioned on page 413, the following relationship holds between hypothesis tests and confidence intervals: For a two-tailed hypothesis test at the significance level α, the null hypothesis H0:μ1=μ2will be rejected in favor of the alternative hypothesis Ha:μ1≠μ2if and only if the ( 1-α)-level confidence interval for μ1-μ2does not contain 0. In each case, illustrate the preceding relationship by comparing the reults of the hypothesis test and confidence interval in the specified xercises.

a. Exercises 10.48 and 10.54.

b. Exercises 10.49 and 10.55.

State the two conditions required for performing a paired r-procedure. How important are those conditions?

V. Tangpricha et al. did a study to determine whether fortifying orange juice with Vitamin D would result in changes in the blood levels of five biochemical variables. One of those variables was the concentration of parathyroid hormone (PTH), measured in picograms/milliliter ( pg/ml ). The researchers published their results in the paper "Fortification of Orange Juice with Vitamin D: A Novel Approach for Enhancing Vitamin D Nutritional Health" (American Journal of Clinical Nutrition, Vol. 77, pp. 1478-1483). Concentration levels were recorded at the beginning of the experiment and again at the end of 12weeks. The following data, based on the results of the study, provide the decrease (negative values indicate an increase) in PTH levels, in pg/ml, for those drinking the fortified juice and for those drinking the unfortified juice.

At the 5% significance level, do the data provide sufficient evidence to conclude that drinking fortified orange juice reduces PTH level more than drinking unfortified orange juice? (Note: The mean and standard deviation for the data on fortified juice are 9.0pg/mL and 37.4pg/mL, respectively, and for the data on unfortified juice, they are 1.6pg/mLand34.6pg/mL, respectively.)

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.