In A Completely Randomized Design: 10 Experimental Units Wer
In A Completely Randomized Design 10 Experimental Units Were Used For
In a completely randomized design, 10 experimental units were used for the first treatment, 12 for the second treatment, and 19 for the third treatment. Sum of Squares due to Treatments and Sum of Squares Total is computed as 1100 and 1700 respectively. Prepare the ANOVA table and complete the same (fill out all the cells). State the Hypotheses. At a .05 level of significance, is there a significant difference between the treatments? Use both p-Value and Critical-Value approaches. 1. Imagine yourself at a fair playing one of the midway games. Pick a game and calculate the expected value and post your results along with how you calculated them. For example, you may decide to throw a basketball to try to win a $10 bear. You paid $2.00 for three shots. What is the expected value? (Please do not use this example in your answer) 2. In one sentence, summarize what you understand of the Central Limit Theorem. 4. Use the mean and the standard deviation obtained from the last module and test the claim that the mean age of all books in the library is greater than 2005. Share your results with the class.
Paper For Above instruction
The analysis of variance (ANOVA) is a statistical method used to determine whether there are significant differences among the means of three or more groups. In the context of a completely randomized design, the ANOVA helps in assessing whether observed variations in data can be attributed to the treatments applied or are merely due to random chance. This paper aims to construct a comprehensive ANOVA table based on provided sums of squares, state the corresponding hypotheses, and perform significance testing using both p-value and critical-value approaches. Additionally, the paper explores applying probabilistic expectations at a fair game, provides a brief understanding of the Central Limit Theorem, and conducts a hypothesis test concerning the average age of books in a library.
Constructing the ANOVA Table
Given data includes three treatments with different numbers of experimental units, sums of squares due to treatments (SST) = 1100, and total sum of squares (SSTotal) = 1700. First, we identify the degrees of freedom (df) for each component:
- Number of treatments (k) = 3
- Number of experimental units per treatment:
- Trial 1: 10 units
- Trial 2: 12 units
- Trial 3: 19 units
The total number of observations (N) = 10 + 12 + 19 = 41
Degrees of freedom for treatments (dftreatments) = k - 1 = 2
Degrees of freedom for total (dftotal) = N - 1 = 40
Sum of Squares for Error (SSE) can be calculated as:
SSE = SSTotal - SST = 1700 - 1100 = 600
Mean Squares (MS) are calculated as:
- MStreatments = SST / dftreatments = 1100 / 2 = 550
- MSError = SSE / dferror = 600 / (N - k) = 600 / (41 - 3) = 600 / 38 ≈ 15.79
Now, the ANOVA table is as follows:
| Source of Variation | Sum of Squares (SS) | Degrees of Freedom (df) | Mean Square (MS) | F-Value |
|---|---|---|---|---|
| Treatments | 1100 | 2 | 550 | F = MStreatments / MSerror = 550 / 15.79 ≈ 34.84 |
| Error | 600 | 38 | 15.79 | |
| Total | 1700 | 40 |
Thus, the calculated F-statistic is approximately 34.84. To determine the significance, we compare this with the critical F-value at α = 0.05 with degrees of freedom df1=2 and df2=38.
Hypotheses Statement
Null hypothesis (H0): There is no significant difference among the treatment means.
Alternative hypothesis (Ha): At least one treatment mean differs significantly from the others.
Significance Testing
The critical F-value for df1=2 and df2=38 at α=0.05 is approximately 3.23 (from F-distribution tables). Since our calculated F-value (~34.84) exceeds this critical value, we reject H0, indicating significant differences among treatment means.
Using the p-value approach, the p-value associated with F(2,38)=34.84 is much less than 0.05, reaffirming the conclusion to reject H0.
Expected Value in a Fair Game
Consider a game where you pay $3 for a chance to win different prizes. Suppose there are three outcomes: winning a $15 prize, a $5 prize, or no prize. The probabilities for each are 1/5, 2/5, and 2/5 respectively. The expected value (EV) is computed as:
EV = (1/5 $15) + (2/5 $5) + (2/5 * $0) = $3 + $2 + $0 = $5
This means, on average, a player can expect to win $5 per game, which is higher than the cost to play, indicating a favorable game in the long run.
Summary of the Central Limit Theorem
The Central Limit Theorem states that the sampling distribution of the sample mean approaches a normal distribution as the sample size increases, regardless of the population's distribution, provided the samples are independent and identically distributed.
Testing the Mean Age of Library Books
Suppose the mean and standard deviation from previous data are 2020 years and 15 years, respectively. To test whether the true mean age exceeds 2005, we set up:
- Null hypothesis (H0): μ = 2005
- Alternative hypothesis (Ha): μ > 2005
Using a one-sample t-test with sample mean = 2020, standard deviation = 15, and sample size N, we calculate the t-statistic:
t = (2020 - 2005) / (15 / √N)
If, for example, N=30, then:
t = (15) / (15 / √30) = √30 ≈ 5.48
This t-value is compared with the critical t-value at α=0.05 and df=29, which is approximately 2.045. Since 5.48 > 2.045, we reject H0, concluding that the average age of books is significantly greater than 2005.
This analysis demonstrates that the library's books are, on average, older than 2005, with statistical significance.
Conclusion
Applying ANOVA in experiments allows us to determine whether differences in treatment effects are statistically significant. By calculating the F-statistic and comparing it with critical values or p-values, we can draw valid inferences. Furthermore, using expected value calculations at fairs provides insights into game fairness and potential profitability. The Central Limit Theorem underpins many statistical procedures by assuring normality in sampling distributions for sufficiently large samples. Lastly, hypothesis testing concerning the mean age of books exemplifies the application of inferential statistics in real-world contexts, enabling data-driven decisions in library management and beyond.
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