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The calcium content of a powdered mineral substance was analyzed five times by each of three methods, with similar standard deviations: $$ \begin{array}{llllll} \text { Method } & {\text { Percent Calcium }} \\ \hline 1 && .0279 & .0276 & .0270 & .0275 & .0281 \\ 2 && .0268 & .0274 & .0267 & .0263 & .0267 \\ 3 && .0280 & .0279 & .0282 & .0278 & .0283 \end{array} $$ Use an appropriate test to compare the three methods of measurement. Comment on the validity of any assumptions you need to make.

Short Answer

Expert verified
Answer: A one-way ANOVA (analysis of variance) test is used to compare the calcium content determined by different methods. The assumptions made to validate the conclusions are: 1) the data are normally distributed, 2) the population variances are equal, and 3) the samples are independent of one another.

Step by step solution

01

Compute the mean for each method

Calculate the mean for each method by summing the test results for each method and dividing by the number of tests (5). Let's label the means as \(M_1\), \(M_2\), and \(M_3\) respectively for Methods 1, 2, and 3. \(M_1 = \frac{0.0279+0.0276+0.0270+0.0275+0.0281}{5} = 0.02762\) \(M_2 = \frac{0.0268+0.0274+0.0267+0.0263+0.0267}{5} = 0.02678\) \(M_3 = \frac{0.0280+0.0279+0.0282+0.0278+0.0283}{5} = 0.02804\)
02

Compute the grand mean

Calculate the grand mean (\(GM\)), which is the mean of all the test results regardless of the method. \(GM = \frac{0.0279+0.0276+0.0270+0.0275+0.0281+0.0268+0.0274+0.0267+0.0263+0.0267+0.0280+0.0279+0.0282+0.0278+0.0283}{15} = 0.02748\)
03

Compute the sum of squares for each method and the total sum of squares

Calculate the sum of squares within each method (SSW), which measures the variation between the test results for each method. SSW = \(\sum_{i=1}^3[(Method_i - M_i)^2]\) For Method 1: SSW1 = \((0.0279-0.02762)^2+(0.0276-0.02762)^2+(0.0270-0.02762)^2+(0.0275-0.02762)^2+(0.0281-0.02762)^2 = 0.00018\) For Method 2: SSW2 = \((0.0268-0.02678)^2+(0.0274-0.02678)^2+(0.0267-0.02678)^2+(0.0263-0.02678)^2+(0.0267-0.02678)^2 = 0.00018\) For Method 3: SSW3 = \((0.0280-0.02804)^2+(0.0279-0.02804)^2+(0.0282-0.02804)^2+(0.0278-0.02804)^2+(0.0283-0.02804)^2 = 0.00012\) Now, calculate the total sum of squares within all methods (SSW_total): SSW_total = SSW1 + SSW2 + SSW3 = 0.00018 + 0.00018 + 0.00012 = 0.00048
04

Compute the sum of squares between methods

Calculate the sum of squares between the methods (SSB), which measures the variation between the methods. SSB = \(5[(M_1-GM)^2+(M_2-GM)^2+(M_3-GM)^2]\) SSB = \(5[(0.02762-0.02748)^2+(0.02678-0.02748)^2+(0.02804-0.02748)^2] = 0.00034\)
05

Compute the F statistic

Calculate the F statistic by using the following formula: \(F = \frac{\frac{SSB}{(3-1)}}{\frac{SSW_{total}}{(15-3)}}\) \(F = \frac{\frac{0.00034}{2}}{\frac{0.00048}{12}} = 4.25\)
06

Find the critical F value and draw conclusions

Find the critical F value for the given degrees of freedom (2, 12) and a significance level of 0.05. The critical F value from the F-distribution table is approximately 3.89. Since the calculated F statistic (4.25) is greater than the critical F value (3.89), we reject the null hypothesis that there is no significant difference between the methods. As a result, there is a significant difference between at least two of the three methods for measuring calcium content in the powdered mineral substance. However, to determine which methods are significantly different, a post-hoc test such as Tukey's HSD test would be required. As for the assumptions, we must assume the data is normally distributed, population variances are equal, and the samples are independent of one another. One way to verify these assumptions would be through graphical methods like histograms, box plots, and Q-Q plots, or by performing specific tests like Shapiro-Wilk test for normality and Levene's test for equal variances.

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Key Concepts

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

ANOVA
Analysis of Variance, often abbreviated as ANOVA, is a statistical technique that is used to determine if there are any statistically significant differences between the means of three or more independent groups. It is particularly useful in situations where you want to compare more than two groups at once. The essence of ANOVA is to test the null hypothesis that all group means are equal.
The key advantage of using ANOVA over multiple individual tests is that it helps control the type I error rate. Conducting multiple tests separately increases the chance of making a false positive error. ANOVA provides a robust method to make this comparison in a single test with a controlled error rate.
In practice, ANOVA involves calculating a statistic called the "F" ratio. It compares the variance between group means to the variance within the groups. If the ratio is significantly large, it suggests at least one group mean is different from the others.
F-test
The F-test, central to ANOVA, is a statistical test used to compare variances and evaluate if there is more variance between groups than within groups. Essentially, the F-test is used to test the null hypothesis that the group means are all very similar, thus indicating that the groups do not differ significantly from one another.
To perform an F-test, an F statistic is calculated. This is the ratio of two variances: the variance between the group means (sum of squares between) and the variance within the groups (sum of squares within). The formula is:\[F = \frac{\text{MSB}}{\text{MSW}}\]where MSB is the mean square between the groups, and MSW is the mean square within the groups.
The F statistic is then compared to a critical value from the F distribution table, which is determined by the degrees of freedom and the level of significance. If the F statistic is larger than this critical value, we reject the null hypothesis, suggesting significant differences between at least some group means.
experimental design
Experimental design is the process and planning of how to conduct a study to ensure that valid and reliable results can be obtained. With respect to ANOVA, the design involves determining how the data will be collected across different groups.
A good experimental design ensures:
  • Randomization: Allocating subjects randomly to different groups to avoid bias.
  • Replication: Performing the experiment more than once to ensure results can be duplicated and are not due to chance.
  • Control: Managing external factors so any observed effects can be attributed to the experimental treatment rather than other variables.
In the context of an ANOVA, each method of measurement provides the structure for the groups being compared. The experimental design assists in determining how these different methods may affect the outcomes, and whether the observed differences in means are statistically significant or not.
statistical assumptions
When using ANOVA and the F-test, several statistical assumptions need to be considered to ensure valid results. These assumptions include:
  • Normality: The data within each group should be approximately normally distributed. This can often be assessed using visual tools like Q-Q plots and histograms or statistical tests such as the Shapiro-Wilk test.
  • Homogeneity of variance: The variances among the groups should be approximately equal, making them comparable. Tests like Levene's test can be conducted to verify this assumption.
  • Independence: Observations must be independent of each other, implying that the data from one group do not influence the data from another group.
Failure to meet these assumptions can lead to incorrect conclusions. Therefore, examining these assumptions before performing an ANOVA is crucial to ensure the reliability of the results. Visualization tools and diagnostic tests provide insights into whether these assumptions hold true.

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Most popular questions from this chapter

Refer to Exercise \(11.54 .\) The data for this experiment are shown in the table. $$ \begin{array}{lccc} && {\text { Training (A) }} \\ \hline \text { Situation (B) } & \text { Trained } & \text { Not Trained } & \text { Totals } \\ \hline \text { Standard } & 85 & 53 & 519 \\ & 91 & 49 & \\ & 80 & 38 & \\ & 78 & 45 & \\ \hline \text { Emergency } & 76 & 40 & 473 \\ & 67 & 52 & \\ & 82 & 46 & \\ & 71 & 39 & \\ \hline \text { Totals } & 630 & 362 & 992 \end{array} $$ a. Construct the ANOVA table for this experiment, b. Is there a significant interaction between the presence or absence of training and the type of decision-making situation? Test at the \(5 \%\) level of significance. c. Do the data indicate a significant difference in behavior ratings for the two types of situations at the \(5 \%\) level of significance? d. Do behavior ratings differ significantly for the two types of training categories at the \(5 \%\) level of significance. e. Plot the average scores using an interaction plot. How would you describe the effect of training and emergency situation on the decision-making abilities of the supervisors?

These data are observations collected using a completely randomized design: $$ \begin{array}{lll} \text { Sample 1 } & \text { Sample 2 } & \text { Sample 3 } \\ \hline 3 & 4 & 2 \\ 2 & 3 & 0 \\ 4 & 5 & 2 \\ 3 & 2 & 1 \\ 2 & 5 & \end{array} $$ a. Calculate CM and Total SS. b. Calculate SST and MST. c. Calculate SSE and MSE d. Construct an ANOVA table for the data. e. State the null and alternative hypotheses for an analysis of variance \(F\) -test. f. Use the \(p\) -value approach to determine whether there is a difference in the three population means.

Swampy Sites An ecological study was conducted to compare the rates of growth of vegetation at four swampy undeveloped sites and to determine the cause of any differences that might be observed. Part of the study involved measuring the leaf lengths of a particular plant species on a preselected date in May. Six plants were randomly selected at each of the four sites to be used in the comparison. The data in the table are the mean leaf length per plant (in centimeters) for a random sample of ten leaves per plant. The MINITAB analysis of variance computer printout for these data is also provided. $$ \begin{array}{lllllll} \text { Location } & {\text { Mean Leaf Length (cm) }} \\ \hline 1 && 5.7 & 6.3 & 6.1 & 6.0 & 5.8 & 6.2 \\ 2 && 6.2 & 5.3 & 5.7 & 6.0 & 5.2 & 5.5 \\ 3 && 5.4 & 5.0 & 6.0 & 5.6 & 4.9 & 5.2 \\ 4 && 3.7 & 3.2 & 3.9 & 4.0 & 3.5 & 3.6 \end{array} $$ a. You will recall that the test and estimation procedures for an analysis of variance require that the observations be selected from normally distributed (at least, roughly so) populations. Why might you feel reasonably confident that your data satisfy this assumption? b. Do the data provide sufficient evidence to indicate a difference in mean leaf length among the four locations? What is the \(p\) -value for the test? c. Suppose, prior to seeing the data, you decided to compare the mean leaf lengths of locations 1 and \(4 .\) Test the null hypothesis \(\mu_{1}=\mu_{4}\) against the alternative \(\mu_{1} \neq \mu_{4}\) d. Refer to part c. Construct a \(99 \%\) confidence interval for \(\left(\mu_{1}-\mu_{4}\right)\) e. Rather than use an analysis of variance \(F\) -test, it would seem simpler to examine one's data, select the two locations that have the smallest and largest sample mean lengths, and then compare these two means using a Student's \(t\) -test. If there is evidence to indicate a difference in these means, there is clearly evidence of a difference among the four. (If you were to use this logic, there would be no need for the analysis of variance \(F\) -test.) Explain why this procedure is invalid.

Suppose you wish to compare the means of six populations based on independent random samples, each of which contains 10 observations. Insert, in an ANOVA table, the sources of variation and their respective degrees of freedom.

An experiment was conducted to investigate the effect of management training on the decision-making abilities of supervisors in a large corporation. Sixteen supervisors were selected, and eight were randomly chosen to receive managerial training. Four trained and four untrained supervisors were then randomly selected to function in a situation in which a standard problem arose. The other eight supervisors were presented with an emergency situation in which standard procedures could not be used. The response was a management behavior rating for each supervisor as assessed by a rating scheme devised by the experimenter. a. What are the experimental units in this experiment? b. What are the two factors considered in the experiment? c. What are the levels of each factor? \(?\) ? d. How many treatments are there in the experiment? e. What type of experimental design has been used?

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