simple comparative experiments

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• Published: 29 May 2014

Points of significance

Designing comparative experiments

• Martin Krzywinski 1 &
• Naomi Altman 2  

Nature Methods volume  11 ,  pages 597–598 ( 2014 ) Cite this article

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Good experimental designs limit the impact of variability and reduce sample-size requirements.

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In a typical experiment, the effect of different conditions on a biological system is compared. Experimental design is used to identify data-collection schemes that achieve sensitivity and specificity requirements despite biological and technical variability, while keeping time and resource costs low. In the next series of columns we will use statistical concepts introduced so far and discuss design, analysis and reporting in common experimental scenarios.

In experimental design, the researcher-controlled independent variables whose effects are being studied (e.g., growth medium, drug and exposure to light) are called factors. A level is a subdivision of the factor and measures the type (if categorical) or amount (if continuous) of the factor. The goal of the design is to determine the effect and interplay of the factors on the response variable (e.g., cell size). An experiment that considers all combinations of N factors, each with n i levels, is a factorial design of type n 1 × n 2 × ... × n N . For example, a 3 × 4 design has two factors with three and four levels each and examines all 12 combinations of factor levels. We will review statistical methods in the context of a simple experiment to introduce concepts that apply to more complex designs.

Suppose that we wish to measure the cellular response to two different treatments, A and B, measured by fluorescence of an aliquot of cells. This is a single factor (treatment) design with three levels (untreated, A and B). We will assume that the fluorescence (in arbitrary units) of an aliquot of untreated cells has a normal distribution with μ = 10 and that real effect sizes of treatments A and B are d A = 0.6 and d B = 1 (A increases response by 6% to 10.6 and B by 10% to 11). To simulate variability owing to biological variation and measurement uncertainty (e.g., in the number of cells in an aliquot), we will use σ = 1 for the distributions. For all tests and calculations we use α = 0.05.

We start by assigning samples of cell aliquots to each level ( Fig. 1a ). To improve the precision (and power) in measuring the mean of the response, more than one aliquot is needed 1 . One sample will be a control (considered a level) to establish the baseline response, and capture biological and technical variability. The other two samples will be used to measure response to each treatment. Before we can carry out the experiment, we need to decide on the sample size.

( a ) Two treated samples (A and B) with n = 17 are compared to a control (C) with n = 17 and to each other using two-sample t -tests. ( b ) Simulated means and P values for samples in a . Values are drawn from normal populations with σ = 1 and mean response of 10 (C), 10.6 (A) and 11 (B). ( c ) The preferred reporting method of results shown in b , illustrating difference in means with CIs, P values and effect size, d . All error bars show 95% CI.

We can fall back to our discussion about power 1 to suggest n . How large an effect size ( d ) do we wish to detect and at what sensitivity? Arbitrarily small effects can be detected with large enough sample size, but this makes for a very expensive experiment. We will need to balance our decision based on what we consider to be a biologically meaningful response and the resources at our disposal. If we are satisfied with an 80% chance (the lowest power we should accept) of detecting a 10% change in response, which corresponds to the real effect of treatment B ( d B = 1), the two-sample t -test requires n = 17. At this n value, the power to detect d A = 0.6 is 40%. Power calculations are easily computed with software; typically inputs are the difference in means (Δ μ ), standard deviation estimate ( σ ), α and the number of tails (we recommend always using two-tailed calculations).

Based on the design in Figure 1a , we show the simulated samples means and their 95% confidence interval (CI) in Figure 1b . The 95% CI captures the mean of the population 95% of the time; we recommend using it to report precision. Our results show a significant difference between B and control (referred to as B/C, P = 0.009) but not for A/C ( P = 0.18). Paradoxically, testing B/A does not return a significant outcome ( P = 0.15). Whenever we perform more than one test we should adjust the P values 2 . As we only have three tests, the adjusted B/C P value is still significant, P ′ = 3 P = 0.028. Although commonly used, the format used in Figure 1b is inappropriate for reporting our results: sample means, their uncertainty and P values alone do not present the full picture.

A more complete presentation of the results ( Fig. 1c ) combines the magnitude with uncertainty (as CI) in the difference in means. The effect size, d , defined as the difference in means in units of pooled standard deviation, expresses this combination of measurement and precision in a single value. Data in Figure 1c also explain better that the difference between a significant result (B/C, P = 0.009) and a nonsignificant result (A/C, P = 0.18) is not always significant (B/A, P = 0.15) 3 . Significance itself is a hard boundary at P = α , and two arbitrarily close results may straddle it. Thus, neither significance itself nor differences in significance status should ever be used to conclude anything about the magnitude of the underlying differences, which may be very small and not biologically relevant.

CIs explicitly show how close we are to making a positive inference and help assess the benefit of collecting more data. For example, the CIs of A/C and B/C closely overlap, which suggests that at our sample size we cannot reliably distinguish between the response to A and B ( Fig. 1c ). Furthermore, given that the CI of A/C just barely crosses zero, it is possible that A has a real effect that our test failed to detect. More information about our ability to detect an effect can be obtained from a post hoc power analysis, which assumes that the observed effect is the same as the real effect (normally unknown), and uses the observed difference in means and pooled variance. For A/C, the difference in means is 0.48 and the pooled s.d. ( s p ) = 1.03, which yields a post hoc power of 27%; we have little power to detect this difference. Other than increasing sample size, how could we improve our chances of detecting the effect of A?

Our ability to detect the effect of A is limited by variability in the difference between A and C, which has two random components. If we measure the same aliquot twice, we expect variability owing to technical variation inherent in our laboratory equipment and variability of the sample over time ( Fig. 2a ). This is called within-subject variation, σ wit . If we measure two different aliquots with the same factor level, we also expect biological variation, called between-subject variation, σ bet , in addition to the technical variation ( Fig. 2b ). Typically there is more biological than technical variability ( σ bet > σ wit ). In an unpaired design, the use of different aliquots adds both σ wit and σ bet to the measured difference ( Fig. 2c ). In a paired design, which uses the paired t -test 4 , the same aliquot is used and the impact of biological variation ( σ bet ) is mitigated ( Fig. 2c ). If differences in aliquots ( σ bet ) are appreciable, variance is markedly reduced (to within-subject variation) and the paired test has higher power.

( a ) Limits of measurement and technical precision contribute to σ wit (gray circle) observed when the same aliquot is measured more than once. This variability is assumed to be the same in the untreated and treated condition, with effect d on aliquot x and y . ( b ) Biological variation gives rise to σ bet (green circle). ( c ) Paired design uses the same aliquot for both measurements, mitigating between-subject variation.

The link between σ bet and σ wit can be illustrated by an experiment to evaluate a weight-loss diet in which a control group eats normally and a treatment group follows the diet. A comparison of the mean weight after a month is confounded by the initial weights of the subjects in each group. If instead we focus on the change in weight, we remove much of the subject variability owing to the initial weight.

If we write the total variance as σ 2 = σ wit 2 + σ bet 2 , then the variance of the observed quantity in Figure 2c is 2 σ 2 for the unpaired design but 2 σ 2 (1 – ρ ) for the paired design, where ρ = σ bet 2 / σ 2 is the correlation coefficient (intraclass correlation). The relative difference is captured by ρ of two measurements on the same aliquot, which must be included because the measurements are no longer independent. If we ignore ρ in our analysis, we will overestimate the variance and obtain overly conservative P values and CIs. In the case where there is no additional variation between aliquots, there is no benefit to using the same aliquot: measurements on the same aliquot are uncorrelated ( ρ = 0) and variance of the paired test is the same as the variance of the unpaired. In contrast, if there is no variation in measurements on the same aliquot except for the treatment effect ( σ wit = 0), we have perfect correlation ( ρ = 1). Now, the difference measurement derived from the same aliquot removes all the noise; in fact, a single pair of aliquots suffices for an exact inference. Practically, both sources of variation are present, and it is their relative size—reflected in ρ —that determines the benefit of using the paired t-test.

We can see the improved sensitivity of the paired design ( Fig. 3a ) in decreased P values for the effects of A and B ( Fig. 3b versus Fig. 1b ). With the between-subject variance mitigated, we now detect an effect for A ( P = 0.013) and an even lower P value for B ( P = 0.0002) ( Fig. 3b ). Testing the difference between ΔA and ΔB requires the two-sample t -test because we are testing different aliquots, and this still does not produce a significant result ( P = 0.18). When reporting paired-test results, sample means ( Fig. 3b ) should never be shown; instead, the mean difference and confidence interval should be shown ( Fig. 3c ). The reason for this comes from our discussion above: the benefit of pairing comes from reduced variance because ρ > 0, something that cannot be gleaned from Figure 3b . We illustrate this in Figure 3c with two different sample simulations with same sample mean and variance but different correlation, achieved by changing the relative amount of σ bet 2 and σ wit 2 . When the component of biological variance is increased, ρ is increased from 0.5 to 0.8, total variance in difference in means drops and the test becomes more sensitive, reflected by the narrower CIs. We are now more certain that A has a real effect and have more reason to believe that the effects of A and B are different, evidenced by the lower P value for ΔB/ΔA from the two-sample t -test (0.06 versus 0.18; Fig. 3c ). As before, P values should be adjusted with multiple-test correction.

( a ) The same n = 17 sample is used to measure the difference between treatment and background (ΔA = A after − A before , ΔB = B after − B before ), analyzed with the paired t -test. Two-sample t -test is used to compare the difference between responses (ΔB versus ΔA). ( b ) Simulated sample means and P values for measurements and comparisons in a . ( c ) Mean difference, CIs and P values for two variance scenarios, σ bet 2 / σ wit 2 of 1 and 4, corresponding to ρ of 0.5 and 0.8. Total variance was fixed: σ bet 2 + σ wit 2 = 1. All error bars show 95% CI.

The paired design is a more efficient experiment. Fewer aliquots are needed: 34 instead of 51, although now 68 fluorescence measurements need to be taken instead of 51. If we assume σ wit = σ bet ( ρ = 0.5; Fig. 3c ), we can expect the paired design to have a power of 97%. This power increase is highly contingent on the value of ρ . If σ wit is appreciably larger than σ bet (i.e., ρ is small), the power of the paired test can be lower than for the two-sample variant. This is because total variance remains relatively unchanged (2 σ 2 (1 – ρ ) ≈ 2 σ 2 ) while the critical value of the test statistic can be markedly larger (particularly for small samples) because the number of degrees of freedom is now n – 1 instead of 2( n – 1). If the ratio of σ bet 2 to σ wit 2 is 1:4 ( ρ = 0.2), the paired test power drops from 97% to 86%.

To analyze experimental designs that have more than two levels, or additional factors, a method called analysis of variance is used. This generalizes the t -test for comparing three or more levels while maintaining better power than comparing all sets of two levels. Experiments with two or more levels will be our next topic.

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Martin Krzywinski is a staff scientist at Canada's Michael Smith Genome Sciences Centre.,

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Naomi Altman is a Professor of Statistics at The Pennsylvania State University.,

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Krzywinski, M., Altman, N. Designing comparative experiments. Nat Methods 11 , 597–598 (2014). https://doi.org/10.1038/nmeth.2974

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Published : 29 May 2014

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Simple Comparative Experiments

Many of the most useful designs are extremely simple.

Sir Ronald Fisher

We now look at a method for making simple comparisons of two or more “treatments,” such as varying brands of toothpaste, and testing their efficacy for preventing cavities. Called the F-test , it is named after Sir Ronald Fisher, a geneticist who developed the technique for application to agricultural experiments. The F-test compares the variance among the treatment means versus the variance of individuals within the specific treatments. High values of F indicate that one (or more) of the means differs from another. This can be very valuable information when, for example, you must select from several suppliers or materials or levels ...

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What Are Comparative Experiments?

Definitions of Control, Constant, Independent and Dependent Variables ...

Many students of science understand the basic idea of the comparative experiment because the name "comparative experiment" mostly explains itself. Students would be correct in defining a comparative experiment as one that compares the effects of two treatments. However, like most anything in science, the comparative experiment has advantages and disadvantages. Students must understand these aspects at a deep level before fully understanding the comparative experiment itself.

Asking the Right Question

According to Penn State, a comparative experiment starts with a question or hypothesis that asks how two or more treatments affect some response. When a scientist wants to know the difference between the effects of treatment A and treatment B on dependent variable C, he will run an experiment in which all of the conditions are the same except for one: the treatment -- A or B -- given to the subject. After receiving the results of the experiment, the scientist can then compare the difference in the dependent variable C for each treatment, concluding either that one treatment is more effective than the other or that both treatments have about the same effectiveness.

The keys to a comparative treatment are control and randomization. Control refers to holding constant all of the other variables that could affect the outcome. For example, a comparative experiment comparing the effects of two diets of different nutritional value on the growth of mice should ensure that the mice eat at the same time, regardless of which diet they are assigned to eat. Randomization refers to randomly assigning the experiment’s subjects, such as mice, to the two or more treatment groups. This randomization allows for valid conclusions and statistical analysis across treatments.

The Advantage

To many students of science, the comparative experiment is a time-saver. Standard, non-comparative experiments use a “control,” which refers to a group of subjects that receive no treatment or a placebo. Scientists engaging in non-comparative experiments in their research would need to run the experiment twice, once with each treatment. For many experiments, however, running just one experiment can be a remarkable expense in both time and money. Thus, a comparative experiment can save a scientist the trouble of having to allocate resources to a second run with a different treatment.

Comparative treatments do not need to include a control, which can be a problem if both treatments yield similar results. For example, if two different injections lead to a similar amount of increased activity in mice, a scientist might be tempted to conclude that both of the injected drugs are effective at inciting activity. The truth is that without a control, the scientist cannot make such a conclusion, as other factors might be influencing the enhanced activity of the mice, such as anxiety from the injection or being handled by the scientists. A comparative experiment is generally limited to conclude the relative effectiveness of one treatment compared to the other.

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• Experiment Design and Statistical Methods for Behavioural and Social Research; David Boniface
• Encyclopedia of Research Design; Neil Salkind
• East Tennessee State University: Producing Data: Experiments

About the Author

Having obtained a Master of Science in psychology in East Asia, Damon Verial has been applying his knowledge to related topics since 2010. Having written professionally since 2001, he has been featured in financial publications such as SafeHaven and the McMillian Portfolio. He also runs a financial newsletter at Stock Barometer.

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Chapter Questions

The breaking strength of a fiber is required to be at least 150 psi. Past experience has indicated that the standard deviation of breaking strength is $\sigma=3$ psi. A random sample of four specimens is tested, and the results are $y_{1}=145, y_{2}=153, y_{3}=150$, and $y_{4}=$ 147. (a) State the hypotheses that you think should be tested in this experiment. (b) Test these hypotheses using $\alpha=0.05 .$ What are your conclusions? (c) Find the $P$-value for the test in part (b). (d) Construct a 95 percent confidence interval on the mean breaking strength.

The viscosity of a liquid detergent is supposed to average 800 centistokes at $25^{\circ} \mathrm{C} . \mathrm{A}$ random sample of 16 batches of detergent is collected, and the average viscosity is $812 .$ Suppose we know that the standard deviation of viscosity is $\sigma=25$ centistokes. (a) State the hypotheses that should be tested. (b) Test these hypotheses using $\alpha=0.05 .$ What are your conclusions? (c) What is the $P$-value for the test? (d) Find a 95 percent confidence interval on the mean.

The diameters of steel shafts produced by a certain manufacturing process should have a mean diameter of $0.255$ inches. The diameter is known to have a standard deviation of $\sigma=0.0001$ inch. A random sample of 10 shafts has an average diameter of $0.2545$ inch. (a) Set up appropriate hypotheses on the mean $\mu$. (b) Test these hypotheses using $\alpha=0.05 .$ What are your conclusions? (c) Find the $P$-value for this test. (d) Construct a 95 percent confidence interval on the mean shaft diameter.

A normally distributed random variable has an unknown mean $\mu$ and a known variance $\sigma^{2}=9$. Find the sample size required to construct a 95 percent confidence interval on the mean that has total width of $10 .$

The sheif life of a carbonated beverage is of interest. Ten bottles are randomly selected and tested, and the following results are obtained: $$ \begin{array}{lr} \hline \multicolumn{2}{c}{\text { Days }} \\ \hline 108 & 138 \\ 124 & 163 \\ 124 & 159 \\ 106 & 134 \\ 115 & 139 \\ \hline \end{array} $$ We would like to demonstrate that the mean shelf life exceeds 120 days. Set up appropriate hypotheses for investigating this claim. Test these hypotheses using $\alpha=0.01$. What are your conclusions? (c) Find the $P$-value for the test in part (b). (d) Construct a 99 percent confidence interval on the mean shelf life.

Consider the shelf life data in Problem 2-5. Can shelf life be described or modeled adequately by a normal distribution? What effect would violation of this assumption have on the test procedure you used in solving Problem 2-5?

The time to repair an electronic instrument is a normally distributed random variable measured in hours. The repair times for 16 such instruments chosen at random are as follows: $$ \begin{array}{lccc} \hline \multicolumn{4}{c}{\text { Hours }} \\ \hline 159 & 280 & 101 & 212 \\ 224 & 379 & 179 & 264 \\ 222 & 362 & 168 & 250 \\ 149 & 260 & 485 & 170 \\ \hline \end{array} $$ (a) You wish to know if the mean repair time exceeds 225 hours. Set up appropriate hypotheses for investigating this issue. (b) Test the hypotheses you formulated in part (a). What are your conclusions? Use $\alpha=0.05$ (c) Find the $P$-value for the test. (d) Construct a 95 percent confidence interval on mean repair time.

Reconsider the repair time data in Problem 2-7. Can repair time, in your opinion, be adequately modeled by a normal distribution?

Two machines are used for filling plastic bottles with a net volume of $16.0$ ounces. The filling processes can be assumed to be normal, with standard deviations of $\sigma_{1}=0.015$ and $\sigma_{2}=0.018$. The quality engineering department suspects that both machines fill to the same net volume, whether or not this volume is $16.0$ ounces. An experiment is performed by taking a random sample from the output of each machine. $$ \begin{array}{llll} \hline \multicolumn{2}{c}{\text { Machine 1 }} & \multicolumn{2}{c}{\text { Machine 2 }} \\ \hline 16.03 & 16.01 & 16.02 & 16.03 \\ 16.04 & 15.96 & 15.97 & 16.04 \\ 16.05 & 15.98 & 15.96 & 16.02 \\ 16.05 & 16.02 & 16.01 & 16.01 \\ 16.02 & 15.99 & 15.99 & 16.00 \\ \hline \end{array} $$ (a) State the hypotheses that should be tested in this experiment. (b) Test these hypotheses using $\alpha=0.05$. What are your conclusions? (c) Find the $P$-value for this test. (d) Find a 95 percent confidence interval on the difference in mean fill volume for the two machines.

Two types of plastic are suitable for use by an electronic calculator manufacturer. The breaking strength of this plastic is important. It is known that $\sigma_{1}=\sigma_{2}=1.0$ psi. From random samples of $n_{1}=10$ and $n_{y}=12$ we obtain $\bar{y}_{1}=162.5$ and $\bar{y}_{2}=155.0$. The company will not adopt plastic 1 unless its breaking strength exceeds that of plastic 2 by at least 10 psi. Based on the sample information, should they use plastic $1 ?$ In answering this question, set up and test appropriate hypotheses using $\alpha=0.01$. Construct a 99 percent confidence interval on the true mean difference in breaking strength.

The following are the buming times of chemical flares of two different formulations. The design engineers are interested in both the mean and variance of the burning times. \begin{array}{lllr} \hline \multicolumn{2}{l}{\text { Type 1 }} & \multicolumn{2}{c}{\text { Type 2 }} \\ \hline 65 & 82 & 64 & 56 \\ (a) Test the hypothesis that the two variances are equal. Use $\alpha=0.05$. (b) Using the results of (a), test the hypothesis that the mean burning times are equal. Use $\alpha=0.05$. What is the $P$-value for this test? (c) Discuss the role of the normality assumption in this problem. Check the assumption of normality for both types of flares. 81 & 67 & 71 & 69 \\ 57 & 59 & 83 & 74 \\ 66 & 75 & 59 & 82 \\ 82 & 70 & 65 & 79 \\ \hline \end{array}

An article in Solid State Technology, "Orthogonal Design for Process Optimization and Its Application to Plasma Etching" by G. Z. Yin and D. W. Jillie (May, 1987 ) describes an experiment to determine the effect of the $\mathrm{C}_{2} \mathrm{~F}_{6}$ flow rate on the uniformity of the etch on a silicon wafer used in integrated circuit manufacturing. Data for two flow rates are as follows: \begin{array}{ccccccc} \hline \multirow{3}{\mathrm{C}_{2}\mathrm{F}_{6}\text{Flow}}{{\text { (SCCM) }}} & \multicolumn{5}{c}{\text { Uniformity Observation }} \\ \cline { 1 - 7 } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 125 & 2.7 & 4.6 & 2.6 & 3.0 & 3.2 & 3.8 \\ 200 & 4.6 & 3.4 & 2.9 & 3.5 & 4.1 & 5.1 \\ \hline \end{array} (a) Does the $\mathrm{C}_{2} \mathrm{~F}_{6}$ flow rate affect average etch uniformity? Use $\alpha=0.05$. (b) What is the $P$-value for the test in part (a)? (c) Does the $\mathrm{C}_{2} \mathrm{~F}_{6}$ flow rate affect the wafer-to-wafer variability in etch uniformity? Use $\alpha=0.05$. (d) Draw box plots to assist in the interpretation of the data from this experiment.

A new filtering device is installed in a chemical unit. Before its installation, a random sample yielded the following information about the percentage of impurity: $\bar{y}_{1}=12.5$, $S_{1}^{2}=101.17$, and $n_{1}=8$. Afler installation, a random sample yielded $\bar{y}_{2}=10.2$. $S_{2}^{2}=94.73, n_{2}=9$ (a) Can you conclude that the two variances are equal? Use $\alpha=0.05$. (b) Has the filtering device reduced the percentage of impurity significantly? Use $\alpha=0.05$

Twenty observations on etch uniformity on silicon wafers are taken during a qualification experiment for a plasma etcher. The data are as follows: $$ \begin{array}{lllll} 5.34 & 6.65 & 4.76 & 5.98 & 7.25 \\ 6.00 & 7.55 & 5.54 & 5.62 & 6.21 \\ 5.97 & 7.35 & 5.44 & 4.39 & 4.98 \\ 5.25 & 6.35 & 4.61 & 6.00 & 5.32 \end{array} $$ (a) Construct a 95 percent confidence interval estimate of $\sigma^{2}$. (b) Test the hypothesis that $\sigma^{2}=1.0$. Use $\alpha=0.05$. What are your conclusions? (c) Discuss the normality assumption and its role in this problem. (d) Check nomality by constructing a normal probability plot. What are your conclusions?

The diameter of a ball bearing was measured by 12 inspectors, each using two different kinds of calipers. The results were \begin{array}{ccc} \hline \text { Inspector } & \text { Caliper 1 } & \text { Caliper 2 } \\ \hline 1 & 0.265 & 0.264 \\ 2 & 0.265 & 0.265 \\ 3 & 0.266 & 0.264 \\ 4 & 0.267 & 0.266 \\ 5 & 0.267 & 0.267 \\ 6 & 0.265 & 0.268 \\ 7 & 0.267 & 0.264 \\ 8 & 0.267 & 0.265 \\ 9 & 0.265 & 0.265 \\ 10 & 0.268 & 0.267 \\ 11 & 0.268 & 0.268 \\ 12 & 0.265 & 0.269 \\ \hline \end{array} (a) Is there a significant difference between the means of the population of measurements from which the two samples were selected? Use $\alpha=0.05$. (b) Find the $P$-value for the test in part (a). (c) Construct a 95 percent confidence interval on the difference in mean diameter measurements for the two types of calipers.

An article in the Journal of Strain Analysis (vol. 18, no. 2,1983$)$ compares several procedures for predicting the shear strength for steel plate girders. Data for nine girders in the form of the ratio of predicted to observed load for two of these procedures, the Karlsruhe and Lehigh methods, are as follows: \begin{array}{lcc} \hline \text { Girder } & \text { Karlsruhe Method } & \text { Lehigh Method } \\ \hline \mathrm{S} 1 / 1 & 1.186 & 1.061 \\ \mathrm{~S} 2 / 1 & 1.151 & 0.992 \\ \mathrm{~S} 3 / 1 & 1.322 & 1.063 \\ S 4 / 1 & 1.339 & 1.062 \\ \mathrm{~S} 5 / 1 & 1.200 & 1.065 \\ \mathrm{~S} 2 / 1 & 1.402 & 1.178 \\ \mathrm{~S} 2 / 2 & 1.365 & 1.037 \\ \$ 2 / 3 & 1.537 & 1.086 \\ \$ 2 / 4 & 1.559 & 1.052 \\ \hline \end{array} (a) Is there any evidence to support a claim that there is a difference in mean performance between the two methods? Use $\alpha=0.05$. (b) What is the $P$-value for the test in part (a)? (c) Construct a 95 percent confidence interval for the difference in mean predicted to observed load. (d) Investigate the normality assumption for both samples. (e) Investigate the normality assumption for the difference in ratios for the two methods. (f) Discuss the role of the normality assumption in the paired t-test.

The deflection temperature under load for two different formulations of ABS plastic pipe is being studied. Two samples of 12 observations each are prepared using each formulation and the deflection temperatures (in ${ }^{\circ} \mathrm{F}$ ) are reported below: \begin{array}{lcllcr} \hline \multicolumn{3}{c}{\text { Formulation 1 }} & \multicolumn{3}{c}{\text { Formulation 2 }} \\ \hline 206 & 193 & 192 & 177 & 176 & 198 \\ 188 & 207 & 210 & 197 & 185 & 188 \\ 205 & 185 & 194 & 206 & 200 & 189 \\ 187 & 189 & 178 & 201 & 197 & 203 \\ \hline \end{array} (a) Construct normal probability plots for both samples. Do these plots support assumptions of normality and equal variance for both samples? (b) Does the data support the claim that the mean deflection temperature under load for formulation 1 exceeds that of formulation $2 ?$ Use $\alpha=0.05$. (c) What is the $P$-value for the test in part (a)?

Refer to the data in Problem 2-17. Do the data support a claim that the mean deflection temperature under load for formulation 1 exceeds that of formulation 2 by at least $3^{\circ} \mathrm{F}$ ?

In semiconductor manufacturing wet chemical etching is often used to remove silicon from the backs of wafers prior to metalization. The etch rate is an important characteristic of this process. Two different etching solutions are being evaluated. Eight randomly selected wafers have been etched in each solution and the observed etch rates (in mils/min) are shown below. \begin{array}{rrrr} \hline \multicolumn{2}{c}{\text { Solution 1 }} & \multicolumn{2}{c}{\text { Solution 2 }} \\ \hline 9.9 & 10.6 & 10.2 & 10.6 \\ 9.4 & 10.3 & 10.0 & 10.2 \\ 10.0 & 9.3 & 10.7 & 10.4 \\ 10.3 & 9.8 & 10.5 & 10.3 \\ \hline \end{array} (a) Do the data indicate that the claim that both solutions have the same mean etch rate is valid? Use $\alpha=0.05$ and assume equal variances. (b) Find a 95 percent confidence interval on the difference in mean etch rates. (c) Use normal probability plots to investigate the adequacy of the assumptions of normality and eaual variances.

Two popular pain medications are being compared on the basis of the speed of absorption by the body. Specifically, tablet 1 is claimed to be absorbed twice as fast as tablet 2 . Assume that $\sigma_{1}^{2}$ and $\sigma_{2}^{2}$ are known. Develop a test statistic for $$ \begin{aligned} &H_{0}: 2 \mu_{1}=\mu_{2} \\ &H_{1}: 2 \mu_{1} \neq \mu_{2} \end{aligned} $$

Suppose we are testing $$ \begin{aligned} &H_{0}: \mu_{1}=\mu_{2} \\ &H_{1}: \mu_{1} \neq \mu_{2} \end{aligned} $$ where $\sigma_{1}^{2}$ and $\sigma_{2}^{2}$ are known. Our sampling resources are constrained such that $n_{1}+n_{2}=N .$ How should we allocate the $N$ observations between the two populations to obtain the most powerful test?

Develop Equation $2-46$ for a $100(1-\alpha)$ percent confidence interval for the variance of a normal distribution.

Develop Equation $2-50$ for a $100(1-\alpha)$ percent confidence interval for the ratio $\sigma_{7}^{2} / \sigma_{2}^{2}$ where $\sigma_{1}^{2}$ and $\sigma_{2}^{2}$ are the variances of two normal distributions.

Develop an equation for finding a $100(1-\alpha)$ percent confidence interval on the difference in the means of two normal distributions where $\sigma_{1}^{2} \neq \sigma_{2}^{2}$. Apply your equation to the portland cement experiment data, and find a 95 pecent confidence interval.

Construct a data set for which the paired $t$-test statistic is very large, but for which the usual two-sample or pooled $t$-test statistic is small. In general, describe how you created the data. Does this give you any insight regarding how the paired $t$-test works?

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6 6 Simple comparative experiments: comparing drug treatments for chronic asthmatics

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This chapter describes a clinical trial to be used to compare two anti-congestant drugs, Formoterol and Salbutamol. It considers the ways in which the trial might have been designed (parallel group design and paired design), and in each case how the data would have been analysed.

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Chapter 2 Simple Comparative Experiments

Solutions 2-2 The viscosity of a liquid detergent is supposed to average 800 centistokes at 25°C. A random sample of 16 batches of detergent is collected, and the average viscosity is 812. Suppose we know that the standard deviation of viscosity is σ = 25 centistokes. (a) State the hypotheses that should be tested. H 0 : µ = 800 H 1 : µ ≠ 800 (b) Test these hypotheses using α = 0.05. What are your conclusions? 812 800 12 1.92 25 25 4 16 o o y z n µ σ − − = = = = Since z α/2 = z 0.025 = 1.96, do not reject. (c) What is the P-value for the test? P = = 2 0 0274 0 0549 (.). (d) Find a 95 percent confidence interval on the mean.

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Juan Valencia

Uriel Vazquez

Vinh San Nguyen

1. There are two requirements for using the methods of this section, and each of them is violated. (1) The samples should be two sample random samples that are independent. These samples are convenience samples, not simple random samples. These samples are likely not independent. Since she surveyed her friends, she may well have males and females that are dating each other (or least that associate with each other) – and people tend to associate with those that have similar behaviors. (2) The number of successes for each sample should be at least 5, and the number of failures for each sample should be at least 5. This is not true for the males, for which x=4. NOTE: This is the same requirement from previous chapters for using the normal distribution to approximate the binomial that required np ≥ 5 and nq ≥ 5. Usingˆp =x/n to estimate p andˆq = 1-x/n = (n-x)/n to estimate q, nˆp ≥ 5 n ˆ q ≥ 5 n[x/n] ≥ 5 n[(n-x)/n] ≥ 5 x≥ 5 (n-x) ≥ 5 These inequalities state that the number of successes must be greater than 5, and the number of failures must be greater than 5. 2. We have 95% confidence that the limits of-0.0518 and 0.0194 contain the true difference between the population proportions of subjects who experience headaches. Repeating the trials many times would result in confidence limits that would include the true difference between the population proportions 95% of the time. Since the interval includes the value 0, there is no significant difference between the two populations proportions. 3. In this context, 1 ˆ p = 15/1583 = 0.00948 2 ˆ p = 8/157 = 0.05096 p = (15+8)/(1583+157) = 23/1740 = 0.01322 p 1 denotes the rue proportion of all Zocor users who experience headaches p 2 denotes the true proportion of all placebo users who experience headaches 4. No. The P-value method and the traditional method will always agree, but it is possible for the confidence interval approach to lead to a different conclusion. The P-value and traditional methods used a standard deviation for the sampling distribution of 1

Nirian Martín , Ayanendranath Basu

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Welcome to stat 503, about this course.

Welcome to the course notes for STAT 503: Design of Experiments . These notes are designed and developed by Penn State's Department of Statistics and offered as open educational resources. These notes are free to use under Creative Commons license CC BY-NC 4.0 .

This course is part of the Online Master of Applied Statistics program offered by Penn State's World Campus.

Currently enrolled?

If you are a current student in this course, please see Canvas for your syllabus, assignments, lesson videos and communication from your instructor.

How to enroll?

If you would like to enroll and experience the entire course for credit please see ' How to enroll in a course ' on the World Campus website.

Course Overview Section  

Statistics is often taught as though the design of the data collection and the data cleaning have already been done in advance. However, as most practicing statisticians quickly learn, typically problems that arise at the analysis stage, could have been avoided if the experimenter had consulted a statistician before the experiment was done and the data were conducted. This course is created to provide an understanding of how experiments should be designed so that when the data are collected, these shortcomings are avoided.

This course covers the following topics:

• Understanding basic design principles
• Working in simple comparative experimental contexts
• Working with single factors or one-way ANOVA in completely randomized experimental design contexts
• Implementing randomized blocks, Latin square designs and extensions of these
• Understanding factorial design contexts
• Working with two level, \(2^k\), designs
• Implementing confounding and blocking in \(2^k\) designs
• Working with 2-level fractional factorial designs
• Working with 3-level and mixed-level factorials and fractional factorial designs
• Simple linear regression models
• Understanding and implementing response surface methodologies
• Understanding robust parameter designs
• Working with random and mixed effects models
• Understanding and implementing nested and split-plot and strip-plot designs
• Using repeated measures designs, unbalanced AOV and ANCOVA

Summer holiday science: turn your home into a lab with these three easy experiments

Associate Professor in Biology, University of Limerick

Disclosure statement

Audrey O'Grady receives funding from Science Foundation Ireland. She is affiliated with Department of Biological Sciences, University of Limerick.

University of Limerick provides funding as a member of The Conversation UK.

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Many people think science is difficult and needs special equipment, but that’s not true.

Science can be explored at home using everyday materials. Everyone, especially children, naturally ask questions about the world around them, and science offers a structured way to find answers.

Misconceptions about the difficulty of science often stem from a lack of exposure to its fun and engaging side. Science can be as simple as observing nature, mixing ingredients or exploring the properties of objects. It’s not just for experts in white coats, but for everyone.

Don’t take my word for it. Below are three experiments that can be done at home with children who are primary school age and older.

Extract DNA from bananas

DNA is all the genetic information inside cells. Every living thing has DNA, including bananas.

Did you know you can extract DNA from banana cells?

What you need: ¼ ripe banana, Ziploc bag, salt, water, washing-up liquid, rubbing alcohol (from a pharmacy), coffee filter paper, stirrer.

What you do:

Place a pinch of salt into about 20ml of water in a cup.

Add the salty water to the Ziploc bag with a quarter of a banana and mash the banana up with the salty water inside the bag, using your hands. Mashing the banana separates out the banana cells. The salty water helps clump the DNA together.

Once the banana is mashed up well, pour the banana and salty water into a coffee filter (you can lay the filter in the cup you used to make the salty water). Filtering removes the big clumps of banana cells.

Once a few ml have filtered out, add a drop of washing-up liquid and swirl gently. Washing-up liquid breaks down the fats in the cell membranes which makes the DNA separate from the other parts of the cell.

Slowly add some rubbing alcohol (about 10ml) to the filtered solution. DNA is insoluble in alcohol, therefore the DNA will clump together away from the alcohol and float, making it easy to see.

DNA will start to precipitate out looking slightly cloudy and stringy. What you’re seeing is thousands of DNA strands – the strands are too small to be seen even with a normal microscope. Scientists use powerful equipment to see individual strands.

Learn how plants ‘drink’ water

What you need: celery stalks (with their leaves), glass or clear cup, water, food dye, camera.

• Fill the glass ¾ full with water and add 10 drops of food dye.
• Place a celery stalk into the glass of coloured water. Take a photograph of the celery.
• For two to three days, photograph the celery at the same time every day. Make sure you take a photograph at the very start of the experiment.

What happens and why?

All plants, such as celery, have vertical tubes that act like a transport system. These narrow tubes draw up water using a phenomenon known as capillarity.

Imagine you have a thin straw and you dip it into a glass of water. Have you ever noticed how the water climbs up the straw a little bit, even though you didn’t suck on it? This is because of capillarity.

In plants, capillarity helps move water from the roots to the leaves. Plants have tiny tubes inside them, like thin straws, called capillaries. The water sticks to the sides of these tubes and climbs up. In your experiment, you will see the food dye in the water make its way to the leaves.

Build a balloon-powered racecar

What you need: tape, scissors, two skewers, cardboard, four bottle caps, one straw, one balloon.

• Cut the cardboard to about 10cm long and 5cm wide. This will form the base of your car.
• Make holes in the centre of four bottle caps. These are your wheels.
• To make the axles insert the wooden skewers through the holes in the cap. You will need to cut the skewers to fit the width of the cardboard base, but leave room for the wheels.
• Secure the wheels to the skewers with tape.
• Attach the axles to the underside of the car base with tape, ensuring the wheels can spin freely.
• Insert a straw into the opening of a balloon and secure it with tape, ensuring there are no air leaks.
• Attach the other end of the straw to the top of the car base, positioning it so the balloon can inflate and deflate towards the back of the car. Secure the straw with tape.
• Inflate the balloon through the straw, pinch the straw to hold the air, place the car on a flat surface, then release the straw.

The inflated balloon stores potential energy when blown up. When the air is released, Newton’s third law of motion kicks into gear: for every action, there is an equal and opposite reaction.

As the air rushes out of the balloon (action), it pushes the car in the opposite direction (reaction). The escaping air propels the car forward, making it move across the surface.

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