Abstract
We need to consider several important collateral issues that complement our discussion in Chap. 2. We first examine the standard assumptions typically made about the probability distribution of the ε’s in our statistical model. Next, we discuss a nonparametric test that is appropriate if the assumption of normality, one of the standard assumptions, is seriously violated. We then review hypothesis testing, a technique that was briefly discussed in the previous chapter and is an essential part of the ANOVA and that we heavily rely on throughout the text. This leads us to a discussion of the notion of statistical power and its determination in an ANOVA. Finally, we find a confidence interval for the true mean of a column and for the difference between two true column means.
This is a preview of subscription content, log in via an institution to check access.
Notes
- 1.
There are different ways to assess the normality of the random error component. The most common types include histograms, normal probability plots, and dot plots.
- 2.
We note the word homoscedasticity solely to prepare readers for it, should they see it in other texts or treatises on the subject. It means “constant variance,” or something close to that, in Greek. It is sometimes spelled homoskedasticity.
- 3.
The Hartley F max test is described, for example, in R. Ott and M. Longnecker (2010), An Introduction to Statistical Methods and Data Analysis, 6th edition, p. 376. This test assumes independent samples of a normally-distributed population with equal sample sizes.
- 4.
In JMP , the Kruskal-Wallis test is found under Nonparametric, Wilcoxon Test. The Wilcoxon test – also known as Mann-Whitney, Mann-Whitney-Wilcoxon, or Wilcoxon rank-sum test – is similar to the Kruskal-Wallis test; however, the latter can accommodate more than two groups (or columns). In Excel , there is no function for the Kruskal-Wallis test.
- 5.
In fact, there are close relationships among the F distribution , the chi-square distribution (χ2), the Student t distribution (t), and the standard normal, Z distribution; here we relate each to the F distribution:
$$ {\displaystyle \begin{array}{c}{\chi}^2\left({df}_1\right)/{df}_1=F\left({df}_1,\infty \right)\\ {}{t}^2\left({df}_2\right)=F\left(1,{df}_2\right)\\ {}\ {Z}^2=F\left(1,\infty \right)\end{array}} $$ - 6.
Alternatively, certain commands in Excel can be used to obtain table values for a t distribution (TINV) and F distribution (FINV). In Chap. 2, we provided details of the FINV command. The p-values, which we will see soon, can also be obtained in Excel, considering a χ2 distribution (CHIDIST), t distribution (TDIST), and F distribution (FDIST).
- 7.
Although not necessarily the case from a mathematical perspective, the two hypotheses are collectively exhaustive for all practical purposes, and certainly in the applications we undertake. This means that the hypotheses represent all potential states of the world.
- 8.
In the criminal courts, H 0 is the presumption of innocence; it is rejected only if the evidence (data) is judged to indicate guilt beyond a reasonable doubt (that is, the evidence against innocence is substantial).
- 9.
In most cases, the standard deviation is not known. Indeed, we have encountered a known standard deviation only when (1) the process being studied had a standard deviation that historically has remained constant, and the issue was whether the process was properly calibrated or was off-center, or (2) the quantity being tested is a proportion, in which case the standard deviation is treated as if known, as a direct function of the hypothesized value of the proportion. However, the assumption of known standard deviation is useful in this presentation. Our goal at this point does not directly include distinctions between the Z and the t distributions; that changes in Chap. 4, where the Student t distribution is discussed.
- 10.
Notice that, in this example, logic suggests a critical (rejection) region that is two-sided (more formally called two-tailed, and the test is said to be a two-tailed test ). After all, common sense says that we should reject H 0: μ = 160 if the \( \overline{X} \) is either too small (that is, a lot below 160) or too large (that is, a lot above 160). Because α is whatever it is (here, .05) in total, it must be split between the upper and lower tails. It is traditional, when there are two tails, to split the area equally between the tails.
- 11.
Obviously, we don’t know for sure the exact values of the μ’s, or we would not have a need to test. However, in very rare cases we are not certain whether the μ’s are equal or not, but we do know what they are if they are not all equal.
- 12.
These terms arise from the notion that the producer is hurt economically by rejecting good-quality products, whereas the consumer is hurt economically by accepting bad-quality products. Interestingly, consumer risk is often more costly to the producer than producer’s risk; indeed, it is seldom true that a Type II error leaves the producer unscathed.
- 13.
It may be possible to obtain these values, and those of the next section, via software. However, there is insight to be gained, here, as elsewhere throughout the text, through seeing the entire tables in print.
- 14.
Usually, when degrees of freedom for the t distribution exceed 30, we simply use the corresponding z value, which here is 1.96. However, for 120 degrees of freedom, the value of the t is 1.98. For 116 degrees of freedom, it is very close to 1.98.
- 15.
Potential causes and issues associated with these discrepancies are discussed by D. B. Merrington and J. A. Thompson (2011), “Problematic Standard Errors and Confidence Intervals for Skewness and Kurtosis .” Behavior Research Methods, vol. 43, pp. 8–17.
Table 3.5 Confidence interval summary in SPSS
Author information
Authors and Affiliations
Appendix
Appendix
Example 3.5 Study of Battery Lifetime with Kruskal-Wallis test using SPSS
The Kruskal-Wallis test can be performed in SPSS using the following procedure. The data in Table 3.1 are entered in a way similar to that which we used in Chap. 2, using 24 rows and two columns. Using the K Independent Samples… command under Analyze > Nonparametric Tests > Legacy Dialogs, we obtain an output shown in Table 3.4, which indicates we do not have evidence sufficient to reject the null hypothesis .
Example 3.6 Confidence Interval for Clinic Satisfaction Study using SPSS
SPSS can also be used to find the confidence interval for the means in Example 3.4 (Clinic Satisfaction Study). After entering the data as previously shown, we select the Explore command under Analyze > Descriptive Statistics and fill in the dependent and factor lists. A portion of the output summary is shown in Table 3.5. Note that some values are slightly different from what we have calculated by hand and using JMP .Footnote 15
Example 3.7 Study of Battery Lifetime using R
The Kruskal-Wallis test can be done easily in R. After importing the data as we have done previously, we can use the kruskal.test() function and assign it to a new object (lifetime.kw). Here, we show two ways in which the arguments can be written:
# Option 1:
> lifetime.kw1 <- kruskal.test(lifetime$V2, lifetime$V1)
# Option 2:
> lifetime.kw2 <- kruskal.test(V2~V1, data=lifetime)
Although the output summary is slightly different, the results are the same.
# Option 1:
> lifetime.kw1 Kruskal-Wallis rank sum test data: lifetime$V2 and lifetime$V1 Kruskal-Wallis chi-squared = 13.01, df = 7, p-values = 0.07188
# Option 2:
> lifetime.kw2 Kruskal-Wallis rank sum test data: V2 by V1 Kruskal-Wallis chi-squared = 13.01, df = 7, p-values = 0.07188
If necessary, we can find the critical value (c) using the qchisq() function
> qchisq(.95, df=7) [1] 14.06714
where .95 is the 95th percentile of the χ2 distribution.
Example 3.8 Confidence Interval for Clinic Satisfaction Study using R
As we have seen in this chapter, we can find the confidence intervals of a particular mean and of the difference between two means. Both procedures require a series of commands in R. First, let’s find the confidence interval of the mean, taking the data for the Amesbury clinic. We import the data as two columns (V1 – location, V2 – satisfaction level) and find the ANOVA table.
> satisfaction.aov <- aov(V2~V1, data=satisfaction)
# names( ) function can be used to obtain the names of an object
> names(satisfaction.aov)
[1] | “coefficients” | “residuals” | “effects” | “rank” |
[5] | “fitted.values” | “assign” | “qr” | “df.residual” |
[9] | “ contrasts ” | “xlevels” | “call” | “terms” |
[13] | “model” |
In order to find the confidence interval, we will calculate each term of Eq. 3.2 separately. We find the mean of the Amesbury clinic using the mean( ) function:
> amesbury <- mean(satisfaction1$model$V2[satisfaction.aov+$model$V1==“Amesbury”]) > amesbury [1] 67.1
Next, we find the t value, the estimate of the sample standard deviation (s), and the number of observations (n):
> tvalue <- qt(0.025, satisfaction.aov$df.residual, lower.tail=F) > tvalue [1] 1.980626
> s <- sqrt(sum((satisfaction.aov$residuals)^2)/satisfaction1+$df.residual) > s [1] 2.932527
# s is the square root of MSW . The latter is the sum of the square of residuals (sum(satisfaction.aov$residuals)^2) divided by the degrees of freedom of the residuals (satisfaction.aov$df.residual).
> n <- sum(satisfaction.aov$model$V1==“Amesbury”) > n [1] 30
The confidence interval is:
> amesbury–tvalue*(s/sqrt(n)) ; amesbury+tvalue*(s/sqrt(n)) [1] 66.03957 [1] 68.16043
Now, we will find the confidence interval of the difference between two means, Methuen and Andover. We will use the t value and the estimate of the sample standard deviation (s) found previously to find the confidence interval using Eq. 3.3:
> methuen.n <- sum(satisfaction.aov$model$V1=="Methuen") > methuen.n [1] 30 > andover.n <- sum(satisfaction.aov$model$V1=="Andover") > andover.n [1] 30 > methuen <- mean(satisfaction.aov$model$V2[satisfaction.aov+$model$V1==Methuen"]) > methuen [1] 57.56667 > andover <- mean(satisfaction.aov$model$V2[satisfaction.aov$model$V1=="Andover"]) > andover [1] 50.4 > (methuen–andover) – tvalue*(s*sqrt(1/methuen.n+1/andover.n));+(methuen–andover) + tvalue*(s*sqrt(1/methuen.n+1/andover.n)) [1] 5.666986 [1] 8.666347
Since n 1 = n2, we can also find the confidence interval using Eq. 3.4:
> (methuen–andover) – tvalue*(s*sqrt(2/n)) ; (methuen–andover) + + tvalue*(s*sqrt(2/n)) [1] 5.666986 [1] 8.666347
Rights and permissions
Copyright information
© 2018 Springer International Publishing AG
About this chapter
Cite this chapter
Berger, P.D., Maurer, R.E., Celli, G.B. (2018). Some Further Issues in One-Factor Designs and ANOVA. In: Experimental Design. Springer, Cham. https://doi.org/10.1007/978-3-319-64583-4_3
Download citation
DOI: https://doi.org/10.1007/978-3-319-64583-4_3
Published:
Publisher Name: Springer, Cham
Print ISBN: 978-3-319-64582-7
Online ISBN: 978-3-319-64583-4
eBook Packages: Mathematics and StatisticsMathematics and Statistics (R0)