An F-curve has df=(24,30). In each case, find the F-value having the specified area to its right.

a.0.05

b.0.01

c.0.025

Short Answer

Expert verified

(a) The required f-valueF0.05=1.89

(b)The required f-valueF0.01=2.47

(c) The required f-value F0.025=2.14

Step by step solution

01

Part (a) Step 1: Given information

Given in the question that , AnF-curve has df=(24,30). we need to find the F-value having the area0.05to its right.

02

Part (a) Step 2: Explanation

Given that,

The for an F-curvedf=(24,30)

To the right of the F-value=0.05

The numerator's degrees of freedom are =24.

The denominator's degrees of freedom are=30.

Statistical significance α=0.05

The critical values for the above supplied values are F0.05=1.89using the F-distribution tables.

The critical value and probability are shown in the shaded zone of the graph below.

03

Part (b) Step 1: Given information

Given in the question that, An F-curve has df=(24,30). we need to flnd the F-value having the area0.01 to its right.

04

Part (b) Step 2: Explanation

Given: df=(24,30)on an F-curve

The area immediately to the right of the F-value =0.01.

The degrees of freedom for the numerator are =24.

The denominator's degrees of freedom are=30.

α=0.01significance level

The critical values for the above supplied values areF0.01=2.47using the F-distribution tables.

The critical value and probability are shown in the shaded zone of the graph below.

05

Part (c) Step 1: Given information

Given in the question that, An F-curve has df=(24,30). we need to flnd theF-value having the area 0.025 to its right.

06

Part (b) Step 2: Explanation

given,

The for an F-curvedf=(24,30)

To the right of F-value =0.025is the area.

The degrees of freedom for the numerator are =24.

The denominator's degrees of freedom are=30.

α=0.025significance level

The critical values for the above supplied values are F0.025=2.14when using the F-distribution tables.

The critical value and probability are shown in the shaded zone of the graph below.

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 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

For an F-curve with df=(6,10), find

a. F0.05-

b. F0.01.

c. F002s-

Breast Milk and IQ. Considerable controversy exists over whether long-term neurodevelopment is affected by nutritional factors in early life. A. Lucas and R. Morley summarized their findings on that question for preterm babies in the publication "Breast Milk and Subsequent Intelligence Quotient in Children Born Preterm (The Lancet, Vol. 339, Issue 8788, Pp, 261-264). The researchers analyzed IQ data on children at age 712-8years. The mothers of the children in the study had chosen whether to provide their infants with breast milk within 72hours of delivery. The researchers used the following designations. Group I: mothers declined to provide breast milk; Group ll a: mothers had chosen but were unable to provide breast milk and Group Il b; mothers had chosen and were able to provide breast milk. Here are the summary statistics on IQ.

At the 1%significance level, do the data provide sufficient evidence to conclude that a difference exists in mean IQ at age 712-8years for preterm children among the three groups? Note: For the degrees of freedom in this exercise:

SAMPLE 1
SAMPLE 2
SAMPLE 13
824
413
636


3

Permeation Sampling. Permeation sampling is a method o sampling air in buildings for pollutants. It can be used over a long period of time and is not affected by humidity, air currents, or temperature. In the paper "Calibration of Permeation Passive Samplers With Silicone Membranes Based on Physicochemical Properties of the Analytes" (Analytical Chemistry, Vol. 75, No, 13, pp. 3182-3192). B. Zabiegata et al. obtained calibration constants experimentally for samples of compounds in each of four compound groups. The following data summarize their results.

At the 5%significance level, do the data provide sufficient evidence to conclude that a difference exists in the mean calibration constant among the four compound groups? (Note:,T3=0.870,T4=0.499,Σxi2=0.456919)

Staph Infections. In the article "Using EDE, ANOVA and Regression to Optimize Some Microbiology Data" (Journal of Statistics Education, Vol. 12, No. 2, online), N. Binnie analyzed bacteria culture data collected by G. Cooper at the Auckland University of Technology. Five strains of cultured Staphylococcus aureus bacteria that cause staph infections were observed for 24hours at 27oC. The following table reports bacteria counts, in millions, for different cases from each of the five strains.

At the 5%significance level, do the data provide sufficient evidence to conclude that a difference exists in mean bacteria counts among the five strains of Staphylococcus aureus? (Note: T1=104,T2=129, T3=185,T4=98,T5=194,Σx2i=25,424.)

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.

Sign-up for free