/*! 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 9 Determine the F-test statistic b... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 9

Determine the F-test statistic based on the given summary statistics. $$ \begin{array}{cccc} \text { Population } & \text { Sample Size } & \text { Sample Mean } & \text { Sample Variance } \\ \hline 1 & 10 & 40 & 48 \\ \hline 2 & 10 & 42 & 31 \\ \hline 3 & 10 & 44 & 25 \end{array} $$

Short Answer

Expert verified
The F-test statistic \( F \approx 1.548 \).

Step by step solution

01

- Identify the formula for the F-test statistic

The F-test statistic is calculated using the formula: \[ F = \frac{s_1^2}{s_2^2} \] where \( s_1^2 \) and \( s_2^2 \) are the sample variances of the populations being compared.
02

- Identify the populations for comparison

From the given data, choose two populations to compare. For this example, let's compare Population 1 and Population 2.
03

- Insert the sample variances into the formula

For Population 1, the sample variance \( s_1^2 = 48 \). For Population 2, the sample variance \( s_2^2 = 31 \). The formula for the F-test statistic becomes: \[ F = \frac{48}{31} \]
04

- Calculate the F-test statistic

Perform the division to get the F-test statistic: \[ F = \frac{48}{31} \approx 1.548 \]

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Variance
To understand the F-test statistic, it's important to know about sample variance. Sample variance measures how spread out the values in a sample are around the sample mean. It tells us how much the values differ from the average value. The formula for sample variance is: \[ s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} \] where \(x_i\) are the values in the sample, \(\bar{x}\) is the sample mean, and \(n\) is the sample size. Calculating sample variance helps in understanding the dispersion in your data.
Sample Size
Sample size is the number of observations in your sample. In the context of statistical tests, knowing the sample size is crucial since it impacts the reliability of your results. Larger sample sizes yield more reliable estimates because they better approximate the population parameters. In our example, each population has a sample size of 10. This consistency helps simplify the F-test calculation.
Statistical Comparison
Statistical comparison involves comparing datasets to determine if there are significant differences between them. In an F-test, we compare the variances of two populations to see if one is significantly larger than the other. This can help determine if two datasets come from populations with different variability. Efficient statistical comparison allows us to make informed decisions based on data.
F-distribution
The F-distribution is a probability distribution that arises frequently in the context of variance analysis. It's skewed to the right and is used in F-tests to compare variances. The shape of the F-distribution depends on two degrees of freedom parameters, \(d_1\) and \(d_2\). These parameters are determined by the sample sizes of the populations being compared. Understanding the F-distribution is key to interpreting the results of an F-test correctly.

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

Given the following ANOVA output, answer the questions that follow. \(\begin{array}{lrrrrr}\text { Source } & \text { df } & \text { SS } & \text { MS } & F & P \\ \text { Factor A } & 1 & 531.2 & 531.2 & 11.73 & 0.003 \\\ \text { Factor B } & 2 & 3018.0 & 1509.0 & 33.33 & 0.000 \\ \text { Interaction } & 2 & 16.3 & 8.2 & 0.18 & 0.836 \\ \text { Error } & 18 & 814.9 & 45.3 & & \end{array}\) (a) Is there evidence of an interaction effect? Why or why not? (b) Based on the \(P\) -value, is there evidence of a difference in the means from factor A? Based on the \(P\) -value, is there evidence of a difference in the means from factor \(\mathrm{B} ?\) (c) What is the mean square error?

Researchers Olivia Carrillo-Gamboa and Richard Gunst presented the following data in their article, "Measurement-Error-Model Collinearities." The data represent the compressive strength (in pounds per square inch) of a random sample of concrete 2 days, 7 days, and 28 days after pouring. $$ \begin{array}{cccc} \text { Concrete } & \text { 2 Days } & \text { 7 Days } & \text { 28 Days } \\\ \hline \mathbf{1} & 2830 & 3505 & 4470 \\ \hline \mathbf{2} & 3295 & 3430 & 4740 \\ \hline \mathbf{3} & 2710 & 3670 & 5115 \\ \hline \mathbf{4} & 2855 & 3355 & 4880 \\ \hline \mathbf{5} & 2980 & 3985 & 4445 \\ \hline \mathbf{6} & 3065 & 3630 & 4080 \\ \hline \mathbf{7} & 3765 & 4570 & 5390 \end{array} $$ (a) Normal probability plots for each treatment indicate that the requirement of normality is satisfied. Verify that the requirement of equal population variances for each treatment is satisfied. (b) Is there sufficient evidence that the mean strength is different among the three days at the \(\alpha=0.05\) level of significance? (c) If the null hypothesis from part (b) was rejected, use Tukey's test to determine which pairwise means differ using a familywise error rate of \(\alpha=0.05 .\)

Researchers (Brian G. Feagan et al. "Erythropoietin with Iron Supplementation to Prevent Allogeneic Blood Transfusion in Total Hip Joint Arthroplasty," Annals of Internal Medicine, Vol. \(133,\) No. 11 ) wanted to determine whether epoetin alfa was effective in increasing the hemoglobin concentration in patients undergoing hip arthroplasty. A complete medical history and physical of the patients was performed for screening purposes and eligible patients were identified. The researchers used a computergenerated schedule to assign the patients to the high-dose epoetin group, low-dose epoetin group, or placebo group. The study was double-blind. Based on ANOVA, it was determined that there were significant differences in the increase in hemoglobin concentration in the three groups with a \(P\) -value less than 0.001 . The mean increase in hemoglobin in the high-dose epoetin group was 19.5 grams per liter \((\mathrm{g} / \mathrm{L}),\) the mean increase in hemoglobin in the low-dose epoetin group was \(17.2 \mathrm{~g} / \mathrm{L},\) and mean increase in hemoglobin in the placebo group was \(1.2 \mathrm{~g} / \mathrm{L}\). (a) Why do you think it was necessary to screen patients for eligibility? (b) Why was a computer-generated schedule used to assign patients to the various treatment groups? (c) What does it mean for a study to be double-blind? Why do you think the researchers desired a double-blind study? (d) Interpret the reported \(P\) -value.

A quality-control manager at an amusement park feels that the amount of time that people spend waiting in line for the American Eagle roller coaster is too long. To determine if a new loading/unloading procedure is effective in reducing wait time in line, he measures the amount of time (in minutes) people are waiting in line for seven days. After implementing the new procedure, he again measures the amount of time (in minutes) people are waiting in line for seven days and obtains the data on the next page. To make a reasonable comparison, he chooses days when weather conditions are alike. Treat each day as a block and the wait times before and after the procedure as the treatment. $$ \begin{array}{lccccccc} \text { Day } & \text { Mon } & \text { Tues } & \text { Wed } & \text { Thurs } & \text { Fri } & \text { Sat } & \text { Sun } \\ \hline \begin{array}{l} \text { Wait time before } \\ \text { new procedure } \end{array} & 11.6 & 25.9 & 20.0 & 38.2 & 57.3 & 32.1 & 81.8 \\ \hline \begin{array}{l} \text { Wait time after } \\ \text { new procedure } \end{array} & 10.7 & 28.3 & 19.2 & 35.9 & 59.2 & 31.8 & 75.3 \\ \hline \end{array} $$ (a) Using the methods introduced in this section, determine whether there is sufficient evidence to conclude that the two loading procedures are resulting in different measurements of the wait time at the \(\alpha=0.05\) level of significance. (b) Using the methods introduced in Section \(11.2,\) determine whether there is sufficient evidence to conclude that the two loading procedures are resulting in different measurements of the wait time at the \(\alpha=0.05\) level of significance. (c) Compare the \(P\) -values of both approaches. Can you conclude that the method presented in this section is a generalization of the matched-pairs \(t\) -test?

Do gender and seating arrangement in college classrooms affect student attitude? In a study at a large public university in the United States, researchers surveyed students to measure their level of feeling at ease in the classroom. Participants were shown different classroom layouts and asked questions regarding their attitude toward each layout. The following data represent feeling-at-ease scores for a random sample of 32 students (four students for each possible treatment). $$ \begin{array}{lcc|cc|cc|cc} \hline && {\text { Tablet-Arm Chairs }} && {\text { U-Shaped }} & {\text { Clusters }} & & {\text { Tables with Chairs }} \\ \hline \text { Female } & 19.8 & 18.4 & 19.2 & 19.2 & 18.1 & 17.5 & 17.3 & 17.1 \\ \hline & 18.1 & 18.5 & 18.6 & 18.7 & 17.8 & 18.3 & 17.7 & 17.6 \\ \hline \text { Male } & 18.8 & 18.2 & 20.6 & 19.2 & 18.4 & 17.7 & 17.7 & 16.9 \\\ \hline & 18.9 & 18.9 & 19.8 & 19.7 & 17.1 & 18.2 & 17.8 & 17.5 \\ \hline \end{array} $$ (a) What is the population of interest? (b) Is this study an experiment or an observational study? Which type? (c) What are the response and explanatory variables? Identify each as qualitative or quantitative. (d) Compute the mean and standard deviation for the scores in the male/U-shaped cell. (e) Assuming that feeling-at-ease scores for males on the U-shaped layout are normally distributed with \(\mu=19.1\) and \(\sigma=0.8,\) what is the probability that you would observe a sample mean as large or larger than actually observed? Would this be unusual? (f) Determine whether the mean feeling-at-ease score is different for males than females using a two-sample \(t\) -test for independent samples. Use the \(\alpha=0.05\) level of significance. (g) Determine whether the mean feeling-at-ease scores for the classroom layouts are different using one-way ANOVA. Use the \(\alpha=0.05\) level of significance. (h) Determine if there is an interaction effect between the two factors. If not, determine if either main effect is significant. (i) Draw an interaction plot of the data. Does the plot support your conclusions in part (h)? (j) In the original study, the researchers sent out e-mails to a random sample of 100 professors at the university asking permission to survey students in their class. Only 32 respondents agreed to allow their students to be surveyed. What type of nonsampling error is this? How might this affect the results of the study?
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Geometry

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.