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Chapter 11: Problem 22

The following data refers to yield of tomatoes \((\mathrm{kg} /\) plot) for four different levels of salinity; salinity level here refers to electrical conductivity (EC), where the chosen levels were \(\mathrm{EC}=1.6,3.8,6.0\), and \(10.2 \mathrm{nmhos} / \mathrm{cm}\) : \(\begin{array}{rrrrrr}1.6: & 59.5 & 53.3 & 56.8 & 63.1 & 58.7 \\ 3.8: & 55.2 & 59.1 & 52.8 & 54.5 & \\ 6.0: & 51.7 & 48.8 & 53.9 & 49.0 & \\ 10.2: & 44.6 & 48.5 & 41.0 & 47.3 & 46.1\end{array}\) Use the \(F\) test at level \(\alpha=.05\) to test for any differences in true average yield due to the different salinity levels.

Short Answer

Expert verified
There are significant differences in tomato yields at different salinity levels.

Step by step solution

01

Organize the Data

Write down the data for each salinity level:- EC 1.6: \(59.5, 53.3, 56.8, 63.1, 58.7\)- EC 3.8: \(55.2, 59.1, 52.8, 54.5\)- EC 6.0: \(51.7, 48.8, 53.9, 49.0\)- EC 10.2: \(44.6, 48.5, 41.0, 47.3, 46.1\).
02

Calculate Group Means

Compute the mean yield for each salinity level:- Mean(EC 1.6) = \(\frac{59.5 + 53.3 + 56.8 + 63.1 + 58.7}{5} = 58.28\)- Mean(EC 3.8) = \(\frac{55.2 + 59.1 + 52.8 + 54.5}{4} = 55.4\)- Mean(EC 6.0) = \(\frac{51.7 + 48.8 + 53.9 + 49.0}{4} = 50.85\)- Mean(EC 10.2) = \(\frac{44.6 + 48.5 + 41.0 + 47.3 + 46.1}{5} = 45.5\).
03

Calculate the Overall Mean

Find the overall mean yield by averaging all data points:\[ \text{Overall Mean} = \frac{59.5 + 53.3 + 56.8 + 63.1 + 58.7 + 55.2 + 59.1 + 52.8 + 54.5 + 51.7 + 48.8 + 53.9 + 49.0 + 44.6 + 48.5 + 41.0 + 47.3 + 46.1}{18} \approx 53.43 \].
04

Compute Sum of Squares for Treatment (SST)

SST is calculated using the formula:\[ SST = n \times \sum (\text{Group Mean} - \text{Overall Mean})^2 \]\[ SST = 5 \times (58.28 - 53.43)^2 + 4 \times (55.4 - 53.43)^2 + 4 \times (50.85 - 53.43)^2 + 5 \times (45.5 - 53.43)^2 \approx 337.455 \].
05

Compute Sum of Squares for Error (SSE)

Find the SSE, which is the sum of the variances within each group:\[ SSE = \sum (x_{ij} - \text{Group Mean})^2 \].After calculations, SSE \approx 226.82.
06

Determine Mean Squares

Calculate the mean square for treatment (MST) and error (MSE):- MST = \(\frac{SST}{k-1} = \frac{337.455}{3} \approx 112.485\)- MSE = \(\frac{SSE}{N-k} = \frac{226.82}{14} \approx 16.201\).
07

Calculate F Statistic

The F statistic is given by:\[ F = \frac{MST}{MSE} = \frac{112.485}{16.201} \approx 6.945 \].
08

Compare With Critical Value

Using an F-table, find the critical value for \(\alpha = 0.05\), \(df_1 = 3\), and \(df_2 = 14\). The critical value is approximately 3.34. Since 6.945 > 3.34, we 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 Hypothesis Testing
At the heart of statistical hypothesis testing lies the concept of making informed decisions about a population based on sample data. Imagine you are conducting an experiment and you want to know if a treatment leads to different outcomes compared to no treatment. You'd start with a null hypothesis (H鈧�), which typically states that there is no effect or difference, and an alternative hypothesis (H鈧�), which suggests that there is an effect or difference.

In our tomato yield example, the null hypothesis is that different salinity levels do not affect the yield. Conversely, the alternative hypothesis posits that there is a difference in yields across different salinity levels.

When performing hypothesis testing, you must choose a significance level, denoted by \(ackslash\alpha\), usually set at 0.05. This level tells us how much risk we are willing to take in making an incorrect decision by rejecting the null hypothesis. The decision-making process involves computing a test statistic, which is then compared to a critical value derived from statistical tables. If the test statistic exceeds the critical value, the null hypothesis is rejected, leading to the conclusion that there is a significant effect or difference.
ANOVA (Analysis of Variance)
Analysis of Variance, or ANOVA, is a powerful statistical method used to compare means between three or more groups. It checks if at least one of the group means is significantly different from the others.

Think of ANOVA as a tool that partitions the data's total variability into variability "between" and "within" groups. The "between" groups variability assesses how the group means differ from the overall mean, implying that a significant difference could be due to the treatment (like salinity levels). The "within" groups variability measures how much each observation within a group diverges from the group's mean, which isn't attributed to the treatment.

In our example with tomato yields, ANOVA helps us determine whether the differences in yield at various salinity levels are statistically significant. By computing the F statistic, which is the ratio of mean square due to treatment (MST) to mean square error (MSE), ANOVA concludes whether the observed variability between the group means is more than expected by chance.
Experimental Design
A good experimental design is crucial for reliable statistical analysis. It involves carefully planning how to collect data and manage variables to ensure unbiased and valid results.

In the case of assessing tomato yields under different salinity levels, a well-structured experimental design means organizing experiments so that each salinity level is tested independently and variously. This includes ensuring an even distribution of sample sizes across levels to maintain balance. Randomization might also be used to prevent bias, ensuring that results are due to changes in salinity and not other factors.

The essence of a robust experimental design is to maximize the accuracy of findings and reduce error variance, thus making it possible to apply statistical tests confidently. By adhering to a thoughtful design, the outcomes from statistical methods like ANOVA can be trusted to reflect true differences or effects stemming from experimental treatments.

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Most popular questions from this chapter

When both factors are random in a two-way ANOVA experiment with \(K\) replications per combination of factor levels, the expected mean squares are \(E(\mathrm{MSE})=\sigma^{2}, E(\mathrm{MSA})=\sigma^{2}+\) \(K \sigma_{G}^{2}+J K \sigma_{A}^{2}, E(\mathrm{MSB})=\sigma^{2}+K \sigma_{G}^{2}+I K \sigma_{B}^{2}\), and \(E(\mathrm{MSAB})=\sigma^{2}+K \sigma_{G}^{2}\) a. What \(F\) ratio is appropriate for testing \(H_{0 G}: \sigma_{G}^{2}=0\) versus \(H_{\mathrm{a} G}: \sigma_{G}^{2}>0\) ? b. Answer part (a) for testing \(H_{0 A}: \sigma_{A}^{2}=0\) versus \(H_{\mathrm{aA}}: \sigma_{A}^{2}>0\) and \(H_{0 B}: \sigma_{B}^{2}=0\) versus \(H_{\mathrm{a} B}: \sigma_{B}^{2}>0\)

In a study to assess the effects of malaria infection on mosquito hosts ("Plasmodium cynomolgi: Effects of Malaria Infection on Laboratory Flight Performance of Anopheles stephensi Mosquitos," Exp. Parasitol., 1977: 397-404), mosquitoes were fed on either infective or noninfective rhesus monkeys. Subsequently the distance they flew during a 24 -h period was measured using a flight mill. The mosquitoes were divided into four groups of eight mosquitoes each: infective rhesus and sporozites present (IRS), infective rhesus and oocysts present (IRD), infective rhesus and no infection developed (IRN), and noninfective (C). The summary data values are \(\bar{x}_{1}=4.39(\mathrm{IRS}), \quad \bar{x}_{2}=4.52(\mathrm{IRD}), \quad \bar{x}_{3}=\) \(5.49(\mathrm{IRN}), \quad \bar{x}_{4}=6.36(\mathrm{C}), \quad \bar{x}_{. .}=5.19, \quad\) and \(\sum \sum x_{i j}^{2}=911.91\). Use the ANOVA \(F\) test at level \(.05\) to decide whether there are any differences between true average flight times for the four treatments.

Numerous factors contribute to the smooth running of an electric motor ("Increasing Market Share Through Improved Product and Process Design: An Experimental Approach," Qual. Engrg., 1991: 361-369). In particular, it is desirable to keep motor noise and vibration to a minimum. To study the effect that the brand of bearing has on motor vibration, five different motor bearing brands were examined by installing each type of bearing on different random samples of six motors. The amount of motor vibration (measured in microns) was recorded when each of the 30 motors was running. The data for this study follows. State and test the relevant hypotheses at significance level .05, and then carry out a multiple comparisons analysis if appropriate. $$ \begin{array}{llllllll} \text { Brand 1: } & 13.1 & 15.0 & 14.0 & 14.4 & 14.0 & 11.6 & 13.68 \\ \text { Brand 2: } & 16.3 & 15.7 & 17.2 & 14.9 & 14.4 & 17.2 & 15.95 \\ \text { Brand 3: } & 13.7 & 13.9 & 12.4 & 13.8 & 14.9 & 13.3 & 13.67 \\ \text { Brand 4: } & 15.7 & 13.7 & 14.4 & 16.0 & 13.9 & 14.7 & 14.73 \\ \text { Brand 5: } & 13.5 & 13.4 & 13.2 & 12.7 & 13.4 & 12.3 & 13.08 \end{array} $$

Six samples of each of four types of cereal grain grown in a certain region were analyzed to determine thiamin content, resulting in the following data \((\mu \mathrm{g} / \mathrm{g})\) : \(\begin{array}{lllllll}\text { Wheat } & 5.2 & 4.5 & 6.0 & 6.1 & 6.7 & 5.8 \\\ \text { Barley } & 6.5 & 8.0 & 6.1 & 7.5 & 5.9 & 5.6 \\ \text { Maize } & 5.8 & 4.7 & 6.4 & 4.9 & 6.0 & 5.2 \\ \text { Oats } & 8.3 & 6.1 & 7.8 & 7.0 & 5.5 & 7.2\end{array}\) a. Check the ANOVA assumptions with a normal probability plot and a test for equal variances. b. Test to see if at least two of the grains differ with respect to true average thiamin content. Use an \(\alpha=.05\) test based on the \(P\)-value method.

The strength of concrete used in commercial construction tends to vary from one batch to another. Consequently, small test cylinders of concrete sampled from a batch are "cured" for periods up to about 28 days in temperature- and moisture-controlled environments before strength measurements are made. Concrete is then "bought and sold on the basis of strength test cylinders" (ASTM C 31 Standard Test Method for Making and Curing Concrete Test Specimens in the Field). The accompanying data resulted from an experiment carried out to compare three different curing methods with respect to compressive strength (MPa). Analyze this data. $$ \begin{array}{lccc} \hline \text { Batch } & \text { Method A } & \text { Method B } & \text { Method C } \\ \hline 1 & 30.7 & 33.7 & 30.5 \\ 2 & 29.1 & 30.6 & 32.6 \\ 3 & 30.0 & 32.2 & 30.5 \\ 4 & 31.9 & 34.6 & 33.5 \\ 5 & 30.5 & 33.0 & 32.4 \\ 6 & 26.9 & 29.3 & 27.8 \\ 7 & 28.2 & 28.4 & 30.7 \\ 8 & 32.4 & 32.4 & 33.6 \\ 9 & 26.6 & 29.5 & 29.2 \\ 10 & 28.6 & 29.4 & 33.2 \end{array} $$
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