/*! 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 9 Six samples of each of four type... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \((\mu \mathrm{g} / \mathrm{g})\) : $$ \begin{array}{lllllll} \text { Wheat } & 5.2 & 4.5 & 6.0 & 6.1 & 6.7 & 5.8 \\ \text { Barley } & 6.5 & 8.0 & 6.1 & 7.5 & 5.9 & 5.6 \\ \text { Maize } & 5.8 & 4.7 & 6.4 & 4.9 & 6.0 & 5.2 \\ \text { Oats } & 8.3 & 6.1 & 7.8 & 7.0 & 5.5 & 7.2 \end{array} $$ Does this data suggest that at least two of the grains differ with respect to true average thiamin content? Use a level \(\alpha=.05\) test based on the \(P\)-value method.

Short Answer

Expert verified
Use ANOVA to test the means; if P-value ≤ 0.05, reject the null hypothesis.

Step by step solution

01

Set up the hypotheses

We are asked to determine if at least two of the grains differ in their true average thiamin content. Therefore, we set up the null hypothesis as follows: \( H_0: \) All four grain types have the same thiamin content averages, \( \mu_1 = \mu_2 = \mu_3 = \mu_4 \). The alternative hypothesis is: \( H_a: \) At least two grains have different average thiamin contents.
02

Choose the test and calculate the test statistic

Since we are comparing more than two means, we use ANOVA (Analysis of Variance). We compute the ANOVA F-statistic based on the variation between sample means and within the samples themselves. We start by calculating the following values:- **Grand mean (\( \bar{x}_{grand} \))**: The mean of all thiamin content values across all grain types.- **Between-group variation (SSB)**: Measures how much the sample means vary from the grand mean.- **Within-group variation (SSW)**: Measures how much individual observations vary within their own group means.Then, compute the F-statistic using:\[ F = \frac{MSB}{MSW} \]where MSB is the mean square between (\( MSB = \frac{SSB}{dfB} \)) and MSW is the mean square within (\( MSW = \frac{SSW}{dfW} \)), with \( dfB = k - 1 \) and \( dfW = N - k \), where \( k \) is the number of groups and \( N \) is the total number of observations.
03

Calculate the P-value

Using the calculated F-statistic and the degrees of freedom (\( dfB \) for the numerator and \( dfW \) for the denominator), find the P-value from the F-distribution table. The P-value tells us the probability of observing an F-statistic at least as extreme as the one calculated, given that the null hypothesis is true.
04

Make a decision

Compare the P-value to the significance level \( \alpha = 0.05 \). If the P-value is less than or equal to 0.05, reject the null hypothesis, indicating there is enough evidence to suggest at least two grain types differ in their average thiamin content. If the P-value is greater than 0.05, we fail to reject the null hypothesis, which means there is not sufficient evidence to suggest a difference in means.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a method used in statistics to test an assumption about a population parameter. In this exercise, we aim to determine if there are differences in thiamin contents among four types of cereal grains — wheat, barley, maize, and oats.
The first step in hypothesis testing involves setting up two hypotheses: the null hypothesis \( H_0 \), and the alternative hypothesis \( H_a \). The null hypothesis is a statement we look to find evidence against. Here, it suggests that all grain types have the same average thiamin content \( \mu_1 = \mu_2 = \mu_3 = \mu_4 \). The alternative hypothesis asserts that at least two types of grains differ in their average thiamin content.
This comparison is made using the ANOVA test (Analysis of Variance), which helps us understand if the means across different groups (grain types in this case) are significantly different.
F-statistic
The F-statistic is a crucial part of the ANOVA test, helping us determine if the means across multiple groups are significantly different.
To compute the F-statistic, we first calculate several components:
  • Grand Mean \( \bar{x}_{grand} \): This is the overall mean of thiamin content values from all grain types combined.
  • Between-group variation (SSB): This measures how much the sample means differ from the grand mean.
  • Within-group variation (SSW): This assesses how much variability there is within each group.
With these values, the F-statistic formula is:\[F = \frac{MSB}{MSW}\]where MSB (Mean Square Between) is calculated as \( \frac{SSB}{dfB} \) and MSW (Mean Square Within) as \( \frac{SSW}{dfW} \). Here, \( dfB \) is the degrees of freedom between groups (number of groups - 1) and \( dfW \) is the degrees of freedom within groups (total observations - number of groups). The F-statistic is a ratio that indicates whether the means of the various groups are statistically different.
P-value Method
The P-value method is an essential part of hypothesis testing in statistics. It provides a way to measure the strength of the evidence against the null hypothesis.
After computing the F-statistic, we use it to find the P-value from the F-distribution. The P-value signifies the probability of obtaining an F-statistic as extreme as, or more extreme than, the observed value, assuming the null hypothesis is true. A lower P-value suggests stronger evidence against the null hypothesis.
When comparing the P-value to a predetermined significance level \( \alpha \), in this exercise \( \alpha = 0.05 \), we decide whether to reject the null hypothesis:
  • If the P-value is less than or equal to \( \alpha \), reject the null hypothesis, indicating significant differences among group means.
  • If the P-value is greater than \( \alpha \), we fail to reject the null hypothesis, suggesting no significant differences among group means.
Statistical Significance
Statistical significance is the cornerstone of hypothesis testing. It quantifies how likely a result is not just due to random chance. In this context, statistical significance helps determine if the variation in thiamin content among different grains is noteworthy.
Achieving statistical significance is reliant on the P-value and the chosen significance level \( \alpha \). A result is statistically significant if the P-value is less than the significance level, typically set at 0.05. This threshold is somewhat arbitrary but widely accepted in statistical analyses.
When we find statistical significance, it implies that the observed data would be unlikely if the null hypothesis were true. Therefore, we have enough evidence to support the alternative hypothesis. In simpler terms, if our results are statistically significant, it suggests that the difference in mean thiamin contents is likely real, not just an outcome of random sampling variations.

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Most popular questions from this chapter

Although tea is the world's most widely consumed beverage after water, little is known about its nutritional value. Folacin is the only B vitamin present in any significant amount in tea, and recent advances in assay methods have made accurate determination of folacin content feasible. Consider the accompanying data on folacin content for randomly selected specimens of the four leading brands of green tea. $$ \begin{array}{cccccccc} \text { Brand } & \multicolumn{8}{c}{\text { Observations }} \\ \hline 1 & 7.9 & 6.2 & 6.6 & 8.6 & 8.9 & 10.1 & 9.6 \\ 2 & 5.7 & 7.5 & 9.8 & 6.1 & 8.4 & & \\ 3 & 6.8 & 7.5 & 5.0 & 7.4 & 5.3 & 6.1 & \\ 4 & 6.4 & 7.1 & 7.9 & 4.5 & 5.0 & 4.0 & \\ \hline \end{array} $$ (Data is based on "Folacin Content of Tea," J. Amer: Dietetic Assoc., 1983: 627-632.) Does this data suggest that true average folacin content is the same for all brands? a. Carry out a test using \(\alpha=.05\) via the \(P\)-value method. b. Assess the plausibility of any assumptions required for your analysis in part (a). c. Perform a multiple comparisons analysis to identify significant differences among brands.

The following data refers to yield of tomatoes ( \(\mathrm{kg} / \mathrm{plot}\) ) for four different levels of salinity; salinity level here refers to electrical conductivity \((\mathrm{EC})\), 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.

In single-factor ANOVA with \(I\) treatments and \(J\) observations per treatment, let \(\mu=(1 / I) \sum \mu_{i}\). a. Express \(E\left(\bar{X}_{\underline{.}}\right)\) in terms of \(\mu\). [Hint: \(\bar{X}_{. .}=(1 / I) \sum \bar{X}_{i \cdot}\) ] b. Compute \(E\left(\bar{X}_{i .}^{2}\right)\). [Hint: For any rv \(Y, E\left(Y^{2}\right)=V(Y)+\) \(\left.[E(Y)]^{2} .\right]\) c. Compute \(E\left(\bar{X}_{. .}^{2}\right)\). d. Compute \(E(\mathrm{SSTr})\) and then show that $$ E(\mathrm{MSTr})=\sigma^{2}+\frac{J}{I-1} \sum\left(\mu_{i}-\mu\right)^{2} $$ e. Using the result of part (d), what is E(MSTr) when \(H_{0}\) is true? When \(H_{0}\) is false, how does \(E(\mathrm{MSTr})\) compare to \(\sigma^{2} ?\)

The article "Computer-Assisted Instruction Augmented with Planned Teacher/Student Contacts" (J. Exp. Educ., Winter, 1980-1981: 120-126) compared five different methods for teaching descriptive statistics. The five methods were traditional lecture and discussion (L/D), programmed textbook instruction \((\mathrm{R})\), programmed text with lectures (R/L), computer instruction (C), and computer instruction with lectures (C/L). Forty-five students were randomly assigned, 9 to each method. After completing the course, the students took a 1-hour exam. In addition, a 10-minute retention test was administered 6 weeks later. Summary quantities are given. $$ \begin{array}{lcccc} & {}{}{\text { Exam }} && {}{}{\text { Retention Test }} \\ \hline \text { Method } & \overline{\boldsymbol{x}}_{\boldsymbol{i} \cdot} & \boldsymbol{s}_{i} & \overline{\boldsymbol{x}}_{i \cdot} & \boldsymbol{s}_{i} \\\ \hline \mathrm{L} / \mathrm{D} & 29.3 & 4.99 & 30.20 & 3.82 \\ \mathrm{R} & 28.0 & 5.33 & 28.80 & 5.26 \\ \mathrm{R} / \mathrm{L} & 30.2 & 3.33 & 26.20 & 4.66 \\ \mathrm{C} & 32.4 & 2.94 & 31.10 & 4.91 \\ \mathrm{C} / \mathrm{L} & 34.2 & 2.74 & 30.20 & 3.53 \\ \hline \end{array} $$ The grand mean for the exam was \(30.82\), and the grand mean for the retention test was \(29.30\). a. Does the data suggest that there is a difference among the five teaching methods with respect to true mean exam score? Use \(\alpha=.05\). b. Using a .05 significance level, test the null hypothesis of no difference among the true mean retention test scores for the five different teaching methods.

The article "Origin of Precambrian Iron Formations" (Econ. Geology, 1964: 1025-1057) reports the following data on total Fe for four types of iron formation ( \(1=\) carbonate, \(2=\) silicate, \(3=\) magnetite, \(4=\) hematite). $$ \begin{array}{llllll} 1: & 20.5 & 28.1 & 27.8 & 27.0 & 28.0 \\ & 25.2 & 25.3 & 27.1 & 20.5 & 31.3 \\ 2: & 26.3 & 24.0 & 26.2 & 20.2 & 23.7 \\ & 34.0 & 17.1 & 26.8 & 23.7 & 24.9 \\ 3: & 29.5 & 34.0 & 27.5 & 29.4 & 27.9 \\ & 26.2 & 29.9 & 29.5 & 30.0 & 35.6 \\ 4: & 36.5 & 44.2 & 34.1 & 30.3 & 31.4 \\ & 33.1 & 34.1 & 32.9 & 36.3 & 25.5 \end{array} $$ Carry out an analysis of variance \(F\) test at significance level .01, and summarize the results in an ANOVA table.

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