/*! 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 1 Give as much information as you ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 1

Give as much information as you can about the \(P\) -value for an upper-tailed \(F\) test in each of the following situations. a. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=15, F=5.37\) b. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=15, F=1.90\) c. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=15, F=4.89\) d. \(\mathrm{df}_{1}=3, \mathrm{df}_{2}=20, F=14.48\) e. \(\mathrm{df}_{1}=3, \mathrm{df}_{2}=20, F=2.69\) f. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=50, F=3.24\)

Short Answer

Expert verified
This problem involves analyzing each case by comparing the observed F-value with the critical F-value. The P-value provides a measure of the strength of evidence against the null hypothesis. If the observed F-value is larger than the critical value, the P-value will be small, prompting the rejection of the null hypothesis. Note: Actual P-values cannot be calculated without utilising a statistical software or an F-distribution table.

Step by step solution

01

Identify the given parameters

For each situation, clearly identify the degrees of freedom for the numerator \(\mathrm{df}_1\), degrees of freedom for the denominator \(\mathrm{df}_2\), and the observed F-value \(F\).
02

Use a F-distribution table or statistical software

Use an F-distribution table or statistical software to find the upper critical value of F for the given degrees of freedom. Comparing this value with the observed F will let one know whether or not to reject the null hypothesis.
03

Compute and Compare for P-value

Calculate the P-value associated with the observed F-value, using the formula for P-value in an F-distribution with the given degrees of freedom. If the observed F-value is greater than the critical value, the P-value is small (<0.05) and the null hypothesis is rejected. Otherwise, it is not rejected. Remember the P-value is the probability of seeing an F-value as extreme or more extreme (in the direction of the alternative hypothesis) than the observed value.

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

Suppose that a random sample of size \(n=5\) was selected from the vineyard properties for sale in Sonoma County, California, in each of three years. The following data are consistent with summary information on price per acre (in dollars, rounded to the nearest thousand) for disease-resistant grape vineyards in Sonoma County (Wines and Vines, November 1999). $$ \begin{array}{llllll} 1996 & 30,000 & 34,000 & 36,000 & 38,000 & 40,000 \\ 1997 & 30,000 & 35,000 & 37,000 & 38,000 & 40,000 \\ 1998 & 40,000 & 41,000 & 43,000 & 44,000 & 50,000 \end{array} $$ a. Construct boxplots for each of the three years on a common axis, and label each by year. Comment on the similarities and differences. b. Carry out an ANOVA to determine whether there is evidence to support the claim that the mean price per acre for vineyard land in Sonoma County was not the same for the three years considered. Use a significance level of \(.05\) for your test.

Exercise \(15.8\) presented the accompanying summary information on satisfaction levels for employees on three different work schedules: $$ \begin{array}{lll} n_{1}=24 & n_{2}=24 & n_{3}=20 \\ \bar{x}_{1}=6.60 & \bar{x}_{2}=5.37 & \bar{x}_{3}=5.20 \end{array} $$ MSE was computed to be 2.028. Calculate the \(95 \%\) T-K intervals, and identify any significant differences.

\(\mathbf{1 5 . 9} \nabla\) High productivity and carbohydrate storage ability of the Jerusalem artichoke make it a promising agricultural crop. The article "Leaf Gas Exchange and Tuber Yield in Jerusalem Artichoke Cultivars" (Field Crops Research [1991]: 241-252) reported on various plant characteristics. Consider the accompanying data on chlorophyll concentration \(\left(\mathrm{gm} / \mathrm{m}^{2}\right)\) for four varieties of Jerusalem artichoke: \(\begin{array}{lccll}\text { Variety } & \text { BI } & \text { RO } & \text { WA } & \text { TO } \\ \text { Sample mean } & .30 & .24 & .41 & .33\end{array}\) Suppose that the sample sizes were \(5,5,4\), and 6, respectively, and also that \(\mathrm{MSE}=.0130 .\) Do the data suggest that true average chlorophyll concentration depends on the variety? State and test the appropriate hypotheses using a significance level of \(.05\).

15.19 The Gunning Fog index is a measure of reading difficulty based on the average number of words per sentence and the percentage of words with three or more syllables. High values of the Gunning Fog index are associated with difficult reading levels. Independent random samples of six advertisements were taken from three different magazines, and Gunning Fog indices were computed to obtain the data given in the accompanying table ("Readability Levels of Magazine Advertisements"'Journal of Advertising Research [1981]: 45-50). Construct an ANOVA table, and then use a significance level of .01 to test the null hypothesis of no difference between the mean Gunning Fog index levels for advertisements appearing in the three magazines. $$ \begin{array}{lrrrrrr} \text { Scientific American } & 15.75 & 11.55 & 11.16 & 9.92 & 9.23 & 8.20 \\ \text { Fortune } & 12.63 & 11.46 & 10.77 & 9.93 & 9.87 & 9.42 \\ \text { New Yorker } & 9.27 & 8.28 & 8.15 & 6.37 & 6.37 & 5.66 \end{array} $$

The article "Utilizing Feedback and Goal Setting to Increase Performance Skills of Managers" (Academy of Management Journal \([1979]: 516-526)\) reported the results of an experiment to compare three different interviewing techniques for employee evaluations. One method allowed the employee being evaluated to discuss previous evaluations, the second involved setting goals for the employee, and the third did not allow either feedback or goal setting. After the interviews were concluded, the evaluated employee was asked to indicate how satisfied he or she was with the interview. (A numerical scale was used to quantify level of satisfaction.) The authors used ANOVA to compare the three interview techniques. An \(F\) statistic value of \(4.12\) was reported. a. Suppose that a total of 33 subjects were used, with each technique applied to 11 of them. Use this information to conduct a level \(.05\) test of the null hypothesis of no difference in mean satisfaction level for the three interview techniques. b. The actual number of subjects on which each technique was used was \(45 .\) After studying the \(F\) table, explain why the conclusion in Part (a) still holds.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

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.