
(a)
Complete the F table.
(a)

Answer to Problem 22CAP
The complete F table is,
Source of Variation | SS | df | MS | |
Between groups | 30 | 2 | 15 | 11.72 |
Within groups (error) | 50 | 39 | 1.28 | |
Total | 80 | 41 |
Explanation of Solution
Calculation:
From the given data, there are three groups highly experienced, moderately experienced and inexperienced athletes. The
The formulas for computing the F table are,
Source of Variation | SS | df | MS | |
Between groups | ||||
Within groups (error) | ||||
Total | 80 |
Where
Source of Variation | SS | df | MS | |
Between groups | ||||
Within groups (error) | 50 | |||
Total | 80 |
Thus, the ANOVA table is,
Source of Variation | SS | df | MS | |
Between groups | 30 | 2 | 15 | 11.72 |
Within groups (error) | 50 | 39 | 1.28 | |
Total | 80 | 41 |
(b)
Find the value of eta-squared
(b)

Answer to Problem 22CAP
The value of eta-squared
Explanation of Solution
Calculation:
The ANOVA table suggests that the value of
Eta-squared
One of the measure for effect size is Eta-squared
Formula is given by,
Where,
Description of effect size using eta-squared:
- If the eta-squared value is less than 0.01, then the effect size is small.
- If the eta-squared value lies between 0.01 and 0.09, then the effect size is medium.
- If the eta-squared value is greater than 0.25, then the effect size is larger.
Substitute the value of
The eta-squared value is,
Justification: The value of eta-squared is 0.38, which is greater than 0.25. Hence, the effect size is larger.
Hence, the value of eta-squared
(c)
Observe whether the decision is to retain or reject the null hypothesis.
(c)

Answer to Problem 22CAP
The decision is rejecting the null hypothesis.
Explanation of Solution
Calculation:
The given information suggest that the degrees of freedom for between groups is 2 and the degrees of freedom for error is 39.
Rejection Rule:
If the value of the test statistic is greater than the critical value then reject the null hypothesis.
The given ANOVA table suggests that the degrees of freedom for between groups is 2 and the degrees of freedom for error is 39. That is, the degrees of freedom for numerator is 2 because the between group degrees of freedom represents the numerator degrees of freedom and the degrees of freedom for denominator is 39 because the error group degrees of freedom represents the denominator degrees of freedom.
Critical value:
Assume that the significance level is
The numerator degrees of freedom are 2, the denominator degrees of freedom as 39 and the alpha level is 0.05.
From the Appendix B: Table B.3 the F Distribution:
- Locate the value 2 in the numerator degrees of freedom row.
- Locate the value 39 in the denominator degrees of freedom column.
- Locate the 0.05 in level of significance.
- The intersecting value that corresponds to the numerator degrees of freedom 2, the denominator degrees of freedom 39 with level of significance 0.05 is 3.24.
Thus, the critical value for the numerator degrees of freedom 2, the denominator degrees of freedom 39 with level of significance 0.05 is 3.24.
The ANOVA gives that the test statistic value is
Conclusion:
The value of test statistic is 11.72
The critical value is 3.24.
The value of test statistic is greater than the critical value.
The test statistic value falls under critical region.
By the decision rule, the conclusion is rejecting the null hypothesis.
Hence, the decision is rejecting the null hypothesis.
Want to see more full solutions like this?
Chapter 12 Solutions
Statistics for the Behavioral Sciences
- Client 1 Weight before diet (pounds) Weight after diet (pounds) 128 120 2 131 123 3 140 141 4 178 170 5 121 118 6 136 136 7 118 121 8 136 127arrow_forwardClient 1 Weight before diet (pounds) Weight after diet (pounds) 128 120 2 131 123 3 140 141 4 178 170 5 121 118 6 136 136 7 118 121 8 136 127 a) Determine the mean change in patient weight from before to after the diet (after – before). What is the 95% confidence interval of this mean difference?arrow_forwardIn order to find probability, you can use this formula in Microsoft Excel: The best way to understand and solve these problems is by first drawing a bell curve and marking key points such as x, the mean, and the areas of interest. Once marked on the bell curve, figure out what calculations are needed to find the area of interest. =NORM.DIST(x, Mean, Standard Dev., TRUE). When the question mentions “greater than” you may have to subtract your answer from 1. When the question mentions “between (two values)”, you need to do separate calculation for both values and then subtract their results to get the answer. 1. Compute the probability of a value between 44.0 and 55.0. (The question requires finding probability value between 44 and 55. Solve it in 3 steps. In the first step, use the above formula and x = 44, calculate probability value. In the second step repeat the first step with the only difference that x=55. In the third step, subtract the answer of the first part from the…arrow_forward
- If a uniform distribution is defined over the interval from 6 to 10, then answer the followings: What is the mean of this uniform distribution? Show that the probability of any value between 6 and 10 is equal to 1.0 Find the probability of a value more than 7. Find the probability of a value between 7 and 9. The closing price of Schnur Sporting Goods Inc. common stock is uniformly distributed between $20 and $30 per share. What is the probability that the stock price will be: More than $27? Less than or equal to $24? The April rainfall in Flagstaff, Arizona, follows a uniform distribution between 0.5 and 3.00 inches. What is the mean amount of rainfall for the month? What is the probability of less than an inch of rain for the month? What is the probability of exactly 1.00 inch of rain? What is the probability of more than 1.50 inches of rain for the month? The best way to solve this problem is begin by creating a chart. Clearly mark the range, identifying the lower and upper…arrow_forwardProblem 1: The mean hourly pay of an American Airlines flight attendant is normally distributed with a mean of 40 per hour and a standard deviation of 3.00 per hour. What is the probability that the hourly pay of a randomly selected flight attendant is: Between the mean and $45 per hour? More than $45 per hour? Less than $32 per hour? Problem 2: The mean of a normal probability distribution is 400 pounds. The standard deviation is 10 pounds. What is the area between 415 pounds and the mean of 400 pounds? What is the area between the mean and 395 pounds? What is the probability of randomly selecting a value less than 395 pounds? Problem 3: In New York State, the mean salary for high school teachers in 2022 was 81,410 with a standard deviation of 9,500. Only Alaska’s mean salary was higher. Assume New York’s state salaries follow a normal distribution. What percent of New York State high school teachers earn between 70,000 and 75,000? What percent of New York State high school…arrow_forwardPls help asaparrow_forward
- Solve the following LP problem using the Extreme Point Theorem: Subject to: Maximize Z-6+4y 2+y≤8 2x + y ≤10 2,y20 Solve it using the graphical method. Guidelines for preparation for the teacher's questions: Understand the basics of Linear Programming (LP) 1. Know how to formulate an LP model. 2. Be able to identify decision variables, objective functions, and constraints. Be comfortable with graphical solutions 3. Know how to plot feasible regions and find extreme points. 4. Understand how constraints affect the solution space. Understand the Extreme Point Theorem 5. Know why solutions always occur at extreme points. 6. Be able to explain how optimization changes with different constraints. Think about real-world implications 7. Consider how removing or modifying constraints affects the solution. 8. Be prepared to explain why LP problems are used in business, economics, and operations research.arrow_forwardged the variance for group 1) Different groups of male stalk-eyed flies were raised on different diets: a high nutrient corn diet vs. a low nutrient cotton wool diet. Investigators wanted to see if diet quality influenced eye-stalk length. They obtained the following data: d Diet Sample Mean Eye-stalk Length Variance in Eye-stalk d size, n (mm) Length (mm²) Corn (group 1) 21 2.05 0.0558 Cotton (group 2) 24 1.54 0.0812 =205-1.54-05T a) Construct a 95% confidence interval for the difference in mean eye-stalk length between the two diets (e.g., use group 1 - group 2).arrow_forwardAn article in Business Week discussed the large spread between the federal funds rate and the average credit card rate. The table below is a frequency distribution of the credit card rate charged by the top 100 issuers. Credit Card Rates Credit Card Rate Frequency 18% -23% 19 17% -17.9% 16 16% -16.9% 31 15% -15.9% 26 14% -14.9% Copy Data 8 Step 1 of 2: Calculate the average credit card rate charged by the top 100 issuers based on the frequency distribution. Round your answer to two decimal places.arrow_forward
- Please could you check my answersarrow_forwardLet Y₁, Y2,, Yy be random variables from an Exponential distribution with unknown mean 0. Let Ô be the maximum likelihood estimates for 0. The probability density function of y; is given by P(Yi; 0) = 0, yi≥ 0. The maximum likelihood estimate is given as follows: Select one: = n Σ19 1 Σ19 n-1 Σ19: n² Σ1arrow_forwardPlease could you help me answer parts d and e. Thanksarrow_forward
- Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw HillHolt Mcdougal Larson Pre-algebra: Student Edition...AlgebraISBN:9780547587776Author:HOLT MCDOUGALPublisher:HOLT MCDOUGAL

