/*! 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 35 To study the effect of temperatu... [FREE SOLUTION] | 91Ó°ÊÓ

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To study the effect of temperature on yield in a chemical process, five batches were produced at each of three temperature levels. The results follow. Construct an analysis of variance table. Use a .05 level of significance to test whether the temperature level has an effect on the mean yield of the process. $$\begin{array}{ccc} & \text { Temperature } & \\ \mathbf{5 0}^{\circ} \mathbf{C} & \mathbf{6 0}^{\circ} \mathbf{C} & \mathbf{7 0}^{\circ} \mathbf{C} \\ 34 & 30 & 23 \\ 24 & 31 & 28 \\ 36 & 34 & 28 \\ 39 & 23 & 30 \\ 32 & 27 & 31 \end{array}$$

Short Answer

Expert verified
There is no significant effect of temperature on yield at the 0.05 level.

Step by step solution

01

Organize the Data

First, we need to list out all the yields for each temperature level: - At 50°C: 34, 24, 36, 39, 32 - At 60°C: 30, 31, 34, 23, 27 - At 70°C: 23, 28, 28, 30, 31.
02

Calculate Overall Mean Yield

Calculate the mean yield of all 15 batches regardless of temperature. The total yield is given as \[ 34 + 24 + 36 + 39 + 32 + 30 + 31 + 34 + 23 + 27 + 23 + 28 + 28 + 30 + 31 = 420 \]Therefore, the overall mean yield \( \bar{Y} \) is \[ \bar{Y} = \frac{420}{15} = 28 \].
03

Calculate Group Means

Determine the mean yield for each temperature level. For 50°C:\[ \bar{Y}_{50} = \frac{34 + 24 + 36 + 39 + 32}{5} = \frac{165}{5} = 33 \]For 60°C:\[ \bar{Y}_{60} = \frac{30 + 31 + 34 + 23 + 27}{5} = \frac{145}{5} = 29 \]For 70°C:\[ \bar{Y}_{70} = \frac{23 + 28 + 28 + 30 + 31}{5} = \frac{140}{5} = 28 \]
04

Calculate Sum of Squares Between Groups (SSB)

Use the formula \[ SSB = n(\bar{Y}_{50} - \bar{Y})^2 + n(\bar{Y}_{60} - \bar{Y})^2 + n(\bar{Y}_{70} - \bar{Y})^2 \]where \( n = 5 \):- For 50°C: \[ 5(33 - 28)^2 = 5 \times 25 = 125 \]- For 60°C: \[ 5(29 - 28)^2 = 5 \times 1 = 5 \]- For 70°C: \[ 5(28 - 28)^2 = 5 \times 0 = 0 \]Hence, \[ SSB = 125 + 5 + 0 = 130 \].
05

Calculate Sum of Squares Within Groups (SSW)

Calculate the sum of squares for each group:- For 50°C: \[ (34-33)^2 + (24-33)^2 + (36-33)^2 + (39-33)^2 + (32-33)^2 = 1 + 81 + 9 + 36 + 1 = 128 \]- For 60°C: \[ (30-29)^2 + (31-29)^2 + (34-29)^2 + (23-29)^2 + (27-29)^2 = 1 + 4 + 25 + 36 + 4 = 70 \]- For 70°C: \[ (23-28)^2 + (28-28)^2 + (28-28)^2 + (30-28)^2 + (31-28)^2 = 25 + 0 + 0 + 4 + 9 = 38 \]Summing them, \[ SSW = 128 + 70 + 38 = 236 \].
06

Calculate Total Sum of Squares (SST) and Validate

Confirm calculations with total sum of squares formula: \[ SST = \sum (Y_{ij} - \bar{Y})^2 \].This should equate to \[ SST = SSB + SSW \].Calculating directly:\[ (34-28)^2+(24-28)^2+(36-28)^2+(39-28)^2+(32-28)^2+(30-28)^2+(31-28)^2+(34-28)^2+(23-28)^2+(27-28)^2+(23-28)^2+(28-28)^2+(28-28)^2+(30-28)^2+(31-28)^2 = 366 \].Verify that \[ 366 = 130 + 236 \].
07

Construct ANOVA Table

Construct the ANOVA table:\[\begin{array}{lccc}\text{Source of Variation} & \text{Sum of Squares (SS)} & \text{Degrees of Freedom (df)} & \text{Mean Square (MS)} & \text{F} \ \hline\text{Between Groups} & 130 & 2 & 65 & \frac{65}{23.6} \approx 2.75 \ \text{Within Groups} & 236 & 12 & 19.67 & \\text{Total} & 366 & 14 & & \ \end{array}\]The calculated F-value is approximately 2.75, with degrees of freedom 2 for between groups and 12 for within groups.
08

Make Decision Based on F-Value

With an F-statistic of approximately 2.75 and the critical F-value at a 0.05 significance level is calculated (or obtained from tables of F-distribution) as a value from F(2,12), approximately 3.88: Since 2.75 < 3.88, we do not reject the null hypothesis.

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Key Concepts

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

Statistical Significance
In the context of ANOVA, statistical significance is crucial in determining whether the observed differences between group means are likely due to a specific factor—here, the temperature—or merely by some random chance. Imagine you run an experiment multiple times under different circumstances, and you notice different outcomes each time. Statistical significance helps you determine if these different outcomes are truly caused by variations in the conditions you set (such as different temperatures) or just because of random fluctuations.

In this exercise, we want to see if temperature changes have a significant impact on the chemical process yield. We compare the F-statistic, derived from comparing the variance between the groups (different temperatures) against the variance within the groups (individual differences within each temperature batch), to the critical F-value from the F-distribution table at a 0.05 significance level. If the calculated F-value exceeds the critical F-value, it indicates significant differences among group means, suggesting that temperature indeed affects yield. In our case, the F-value of 2.75 is not significant (since it's less than 3.88), pointing out that temperature may not have a considerable effect on yield.
Sum of Squares
The sum of squares in ANOVA is all about variance and how it can explain differences in data. It's split into two parts: the sum of squares between groups (SSB) and the sum of squares within groups (SSW).

**Sum of Squares Between Groups (SSB)**
SSB measures how much the group means differ from the overall mean, providing evidence of potential systematic differences due to the treatment or condition applied—in this case, temperature. Larger SSB values indicate greater variability between group means, suggesting that the treatment might have a significant effect.

**Sum of Squares Within Groups (SSW)**
SSW captures the variation within each group, which is attributed to random error or intrinsic factors unrelated to the primary factor under study. This reflects the natural spread of scores within groups without any treatment effect.

For this exercise, we calculated SSB as 130 and SSW as 236. Together, they form the total sum of squares (SST), which quantifies overall variance in the dataset. Verifying their sum provides insights for constructing the ANOVA table and calculating the mean square values needed to find the F-statistic.
Degrees of Freedom
Degrees of freedom (df) in statistics relate to the number of values in a calculation that are free to vary while estimating statistical parameters. In ANOVA, degrees of freedom are vital for computing the mean squares and determining statistical significance of results.

ANOVA primarily involves two degrees of freedom:
  • **Degrees of Freedom Between Groups (dfB):** This is determined by the number of groups minus one. In our scenario with three temperature levels, dfB is calculated as 3 - 1 = 2. These two degrees of freedom account for the comparison of mean yields across the different temperature levels.
  • **Degrees of Freedom Within Groups (dfW):** Calculated as the total number of observations minus the number of groups. With three groups and five observations per group, the dfW becomes 15 - 3 = 12. This number indicates the variance attributed to differences within each temperature treatment group.
The total degrees of freedom (dfT) adds up to 14 (dfB + dfW), providing a sense of the total size of the dataset across all observations. Understanding degrees of freedom is crucial, as they influence the accuracy and reliability of the F-statistic computed for testing the hypothesis about the mean differences.

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Most popular questions from this chapter

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