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LISA Short Course: A Tutorial in t-tests and ANOVA using JMP Anne Ryan Assistant Professor of Practice Department of Statistics, VT [email protected] Laboratory for Interdisciplinary Statistical Analysis Laboratory for Interdisciplinary Statistical Analysis LISA helps VT researchers benefit from the use of Statistics Designing Experiments • Analyzing Data • Interpreting Results Grant Proposals • Using Software (R, SAS, JMP, Minitab...) Walk-In Consulting Collaboration From our website request a meeting for personalized statistical advice Great advice right now: Meet with LISA before collecting your data Available 1-3 PM: Mon—Fri in the GLC Video Conference Room for questions requiring <30 mins See our website for additional times and locations. Short Courses Designed to help graduate students apply statistics in their research All services are FREE for VT researchers. We assist with research—not class projects or homework. www.lisa.stat.vt.edu 3 Defense: Represent the accused (defendant) Prosecution: Hold the “Burden of Proof”—obligation What’s the Assumed Conclusion? to shift the assumed conclusion from an oppositional opinion to one’s own position through evidence ANSWER: The accused is innocent until proven guilty. •Prosecution must convince the judge/jury that the defendant is guilty beyond a reasonable doubt 4 Burden of Proof—Obligation to shift the conclusion using evidence Hypothesis Test Accept the status quo (what is believed before) until the data suggests otherwise Trial Innocent until proven guilty 5 Decision Criteria Hypothesis Test Occurs by chance less than 100α% of the time (ex: 5%) Trial Evidence has to convincing beyond a reasonable 6 … a procedure that allows us to make statements about a general population using the results of a random sample from that population. • Two Types of Inferential Statistics: • Hypothesis Testing • Estimation Point estimates Confidence intervals 7 Hypothesis testing is a detailed protocol for decision-making concerning a population by examining a sample from that population. 8 1. Test 2. Assumptions 3. Hypotheses 4. Mechanics 5. Conclusion 9 Used to test whether the population mean is different from a specified value. 10 In a glaucoma study, the following intraocular pressure (mm Hg) values were recorded from a sample of 21 elderly subjects. Based on this data, can we conclude that the mean intraocular pressure of the population from which the sample was drawn differs from 14 mm Hg?* Intraocular Pressure 14.5 12.9 14 16.1 12 17.5 14.1 12.9 17.9 12 16.4 24.2 12.2 14.4 17 10 18.5 20.8 16.2 14.9 19.6 𝑦 = 15.6238 𝑠 = 3.383 *Wayne, D. Biostatistics: A Foundation for Analysis in the Health Sciences. 5th ed. New York: John Wiley & Sons, 1991. 11 State the name of the testing method to be used It is important to not be off track in the very beginning Example 1: 1. Test: One sample t test for 𝜇 List all the assumptions required for your test to be valid. Example 1: 2. Assumptions • Simple random sample (SRS) was used to collect All tests have assumptions data • The population distribution Even if assumptions from which the sample is are not met you should drawn is normal or still comment on how this affects your approximately normal. results. Claims versus suspicions: The “null hypothesis” is a statement describing a claim about a population constant. - The null hypothesis is denoted as 𝑯𝟎 . The “alternative hypothesis” is a statement describing the researcher’s suspicions about the claim. Also called “research hypothesis”. - The alternative hypothesis is denoted as 𝑯𝒂 . Examples of possible hypotheses: 𝐻0 : 𝜇 = 13 𝑣𝑠 𝐻𝑎 : 𝜇 ≠ 13 For hypothesis testing there are three versions for testing that are determined by the context of the research question. ◦ Left Tailed Hypothesis Test (less than) ◦ Right Tailed Hypothesis Test (greater than) ◦ Two Tailed or Two Sided Hypothesis Test (not equal to) Left Tailed Hypothesis Test: Researchers are only interested in whether the true value is below the hypothesized value. Example— Administrators of a health care center want to know if the mean time spent by patients in the waiting room is less than 20 minutes. 𝐻0 : 𝜇 ≥ 20 𝑣𝑠. 𝐻𝑎 : 𝜇 < 20 Right Tailed Hypothesis Test: Researchers are only interested in whether the True Value is above the hypothesized value. Example— Administrators of a health care center want to know if the mean time spent by patients in the waiting room is greater than 20 minutes. 𝐻0 : 𝜇 ≤ 20 𝑣𝑠. 𝐻𝑎 : 𝜇 > 20 Two Tailed or Two Sided Hypothesis Test: • The researcher is interested in looking above and below their hypothesized value. • Example— Administrators of a health care center want to know if the mean time spent by patients in the waiting room differs from 20 minutes. 𝐻0 : 𝜇 = 20 𝑣𝑠. 𝐻𝑎 : 𝜇 ≠ 20 ◦ Note: The direction of the alternative hypothesis will be used when determining the p-value at a later step. Example 1: 3. Hypotheses • In a glaucoma study, the following intraocular pressure (mm Hg) values were recorded from a sample of 21 elderly subjects. Based on this data, can we conclude that the mean intraocular pressure of the population from which the sample was drawn differs from 14 mm Hg?* What are the hypotheses for Example 1? 𝑯𝟎 : 𝝁 = 𝟏𝟒 𝒗𝒔. 𝑯𝒂 : 𝝁 ≠ 𝟏𝟒 Where 𝜇 is the true intraocular pressure Computational Part of the Test Parts of the Mechanics Step ◦ ◦ ◦ ◦ Stating the Significance Level Finding the Rejection Rule Computing the Test Statistic Computing the p-value Significance Level: Here we choose a value to use as the significance level, which is the level at which we are willing to start rejecting the null hypothesis. Denoted by α which corresponds to the Type 1 Error for the test. Type 1 Error is error committed when the true null hypothesis is rejected. Ex: You reject 𝑯𝟎 when 𝑯𝟎 is true. * Default value is α=.05, use α=.05 unless otherwise noted! Example 1: 4. Mechanics: Significance Level: 𝛼 = 0.05 *We use 𝛼 = 0.05 here because the significance level was not given in the problem. *Note: The Type I error would be concluding that the true mean intraocular pressure differs from 14 mm Hg, when in fact the pressure is 14 mm Hg. 𝑯𝟎 : 𝝁 = 𝟏𝟒 𝒗𝒔. 𝑯𝒂 : 𝝁 ≠ 𝟏𝟒 Rejection Rule: State our Example 1: criteria for rejecting the null hypothesis 4. Mechanics: Reject the null hypothesis Rejection Rule: (𝑯𝟎 ) if the p-value≤ 𝜶 Reject H0 if 𝑝 − 𝑣𝑎𝑙𝑢𝑒 ≤ 0.05 p-value: The chance of observing your sample results or more extreme results assuming that the null hypothesis is true. If this chance is “small” then you may decide the claim in the null hypothesis is false. Test Statistic: Compute the test statistic, which is usually a standardization of your point estimate. Translates your point estimate, a statistic, to follow a known distribution so that is can be used for a test. A point estimate is a single numerical value used to estimate the corresponding population parameter. • 𝑦 is the point estimate for 𝜇 In many cases, including Example 1, the population standard deviation 𝝈 is unknown because it is a parameter from the population that must be estimated. The best estimate for 𝝈 is 𝒔. • Our standardized value becomes 𝒚 − 𝝁𝟎 𝒕𝒐𝒃𝒔 = 𝒔 ~𝒕𝒏−𝟏 𝒏 𝝁𝟎 : hypothesized mean 𝒚: sample mean 𝑠: sample standard deviation 𝒏: sample size 𝑡𝒐𝒃𝒔 : observed t test statistic Test statistic for a one sample t-test This t observed (𝑡0𝑏𝑠 ) test statistic follows a t distribution with 𝒏 − 𝟏 degrees of freedom. 23 Example 1: 4. Mechanics Test Statistic: *In the example it was given that 𝒚 = 𝟏𝟓. 𝟔𝟐𝟑𝟖 and 𝒔 = 𝟑. 𝟑𝟖𝟑. 𝒕𝒐𝒃𝒔 𝒚 − 𝝁𝟎 𝟏𝟓. 𝟔𝟐𝟑𝟖 − 𝟏𝟒 = = = 𝟐. 𝟐𝟎 𝒔/ 𝒏 𝟑. 𝟑𝟖𝟑/ 𝟐𝟏 p-value: After computing the test statistic, now you can compute the p-value. A p-value is the probability of obtaining a point estimate as “extreme” as the current value where the definition of “extreme” is taken from the alternative hypothesis assuming the null hypothesis is true. The p-value depends on the alternative hypothesis, so there are three ways to compute p-values. p-value: The chance of observing your sample results or more extreme results assuming that the null hypothesis is true. If this chance is “small” then you may decide the claim in the null hypothesis is false. Example 1: 4. Mechanics: P-value (in words): The probability of observing a sample mean of 𝟏𝟓. 𝟔𝟐𝟑𝟖 mm hg or a value more extreme assuming the true mean pressure is 14 mm hg. 1. The p-value is determined based on the sign of the alternative hypothesis. 𝑯𝒂 : 𝝁 ≠ 𝝁𝟎 . If this is the case, then the p-value is the area in both tails of the t distribution. 0.4 0.3 Density 0.2 0.1 1/2 p-value 0.0 1/2 p-value -t_obs 0 t_obs 2. The p-value is determined based on the sign of the alternative hypothesis. 𝑯𝒂 : 𝝁 < 𝝁𝟎 . If this is the case, then the p-value is the area to the left of the observed test statistic. 0.4 p-value 0.3 Density 0.2 0.1 0.0 0 t_obs 3. The p-value is determined based on the sign of the alternative hypothesis. 𝑯𝒂 : 𝝁 > 𝝁𝟎 . If this is the case, then the p-value is the area to the right of the observed test statistic. 0.4 0.3 Density 0.2 0.1 p-value 0.0 0 t_obs Example 1: 4. Mechanics p-value: *In the example the hypotheses are: 𝑯𝟎 : 𝝁 = 𝟏𝟒 𝒗𝒔. 𝑯𝒂 : 𝝁 ≠ 𝟏𝟒 0.4 Density 0.3 0.2 0.1 0.01986 0.0 0.01986 -2.2 0 t 2.2 Example 1: 4. Mechanics p-value: 𝒑 − 𝒗𝒂𝒍𝒖𝒆 = 𝟎. 𝟎𝟏𝟗𝟖𝟔 + 𝟎. 𝟎𝟏𝟗𝟖𝟔 = 𝟎. 𝟎𝟑𝟗𝟕𝟐 0.4 Density 0.3 0.2 0.1 0.01986 0.0 0.01986 -2.2 0 t 2.2 Example 1: 4. Mechanics p-value: 𝒑 − 𝒗𝒂𝒍𝒖𝒆 = 𝟎. 𝟎𝟏𝟗𝟖𝟔 + 𝟎. 𝟎𝟏𝟗𝟖𝟔 = 𝟎. 𝟎𝟑𝟗𝟕𝟐 JMP will give the 3 p-values and you must select the correct p-value based on your alternative hypothesis 𝐻𝑎 : 𝜇 ≠ 14 𝐻𝑎 : 𝜇 > 14 𝐻𝑎 : 𝜇 < 14 Example 1: Conclusion: Last step of the hypothesis test. Conclusions should always include: ◦ Decision: reject or fail to reject (not accept 𝐻0 ). When conducting hypothesis tests, we assume that 𝐻0 is true, therefore the decision cannot be to accept the null hypothesis. ◦ Context: what your decision means in context of the problem. 5. Conclusion: With a pvalue=0.0398, which is less than 0.05, we reject 𝐻0 . There is sufficient sample evidence to conclude that the true mean intraocular pressure differs from 14 mm Hg. Note: The significance level can be thought of as a tolerance for things happening by chance. If we set α=.05 then we are saying that we are willing to say what we observe may be out of the ordinary, but unless it is something that occurs less that 5% of the time we will attribute it to chance. Possible Hypotheses: 2-Tailed Test Right-Tailed Left Tailed Null hypothesis 𝐻0 : 𝜇 = 𝜇0 𝐻0 : 𝜇 ≤ 𝜇0 𝐻0 : 𝜇 ≥ 𝜇0 Alternative hypothesis 𝐻𝑎 : 𝜇 ≠ 𝜇0 𝐻𝑎 : 𝜇 > 𝜇0 𝐻𝑎 : 𝜇 < 𝜇0 Test Statistic: 𝒚−𝝁𝟎 𝒕𝒐𝒃𝒔 = Degrees of Freedom: 𝒏 − 𝟏 𝒔 𝒏 Assumption: The population from which the sample is drawn is normal or approximately normal. 33 Let T be a t random variable with 𝑑. 𝑓. = 𝑛 − 1 and 𝑡𝑜𝑏𝑠 = 𝑦−𝜇0 𝑠/ 𝑛 Left-tailed test 𝐻0 : 𝜇 ≥ 𝜇0 𝑣𝑠. 𝐻𝑎 : 𝜇 < 𝜇0 𝑝 − 𝑣𝑎𝑙𝑢𝑒 = 𝑃 𝑇 ≤ 𝑡 𝜶 *written as Prob<t in jmp Right-tailed test 𝐻0 : 𝜇 ≤ 𝜇0 𝑣𝑠. 𝐻𝑎 : 𝜇 > 𝜇0 𝑝− 𝑣𝑎𝑙𝑢𝑒 = 𝑃 𝑇 ≥ 𝑡 *written as Prob>t in jmp Two-tailed tests 𝐻0 : 𝜇 = 𝜇0 𝑣𝑠. 𝐻𝑎 : 𝜇 ≠ 𝜇0 𝑝− 𝑣𝑎𝑙𝑢𝑒 = 2𝑃 𝑇 ≥ |𝑡| *written as Prob>|t| in jmp 𝜶 𝜶 𝟐 𝜶 𝟐 34 In a glaucoma study, the following intraocular pressure (mm Hg) values were recorded from a sample of 21 elderly subjects. Based on this data, can we conclude that the mean intraocular pressure of the population from which the sample was drawn differs from 14 mm Hg?* 𝑦 = 15.6238 𝑠 = 3.383 T: One sample t-test for 𝜇 A: i) SRS was used ii)The population from which the sample is drawn is normal or approximately normal. H: 𝐻0 : 𝜇 = 14 𝑣𝑠. 𝐻𝑎 : 𝜇 ≠ 14; 𝜇 is the true mean intraocular pressure M: 𝛼 = 0.05 Reject 𝐻0 if p-value≤0.05 𝑦−𝜇0 15.6238−14 𝑡𝑜𝑏𝑠 = = = 2.20 𝑠/ 𝑛 3.383/ 21 p-value=𝑃 𝑇 > 2.20 + 𝑃 𝑇 < 2.20 = 0.0398 (calculated using JMP: Prob>|t|) C: With a p-value less than 0.05, we reject 𝐻0 . There is sufficient sample evidence to conclude that the true mean intraocular pressure differs from 14 mm Hg. 35 • JMP Demonstration • Open Pressure.jmp • AnalyzeDistribution • Complete the dialog box as shown and select OK. • Select the red arrow next to “Pressure” and select Test Mean. • Complete Dialog box as shown and select OK. • Select the red arrow next to “Pressure” and select Confidence Interval->0.95. 36 The normal quantile plot may also be created in JMP to check the normality assumption. The assumption is met if the points fall close to the red line. 37 Two sample t-tests are used to determine whether the population mean of one group is equal to, larger than or smaller than the population mean of another group. 38 The major goal is to determine whether a difference exists between two populations. Examples: ◦ Compare blood pressure for male and females. ◦ Compare the proportion of smokers and nonsmokers with lung cancer. ◦ Compare weight before and after treatment. ◦ Is the mean cholesterol of people taking drug A lower than the mean cholesterol of people taking drug B? 39 The population means of the two groups are not equal. H0: μ1 = μ2 Ha: μ1 ≠ μ2 The population mean of group 1 is greater than the population mean of group 2. H0: μ1 = μ2 Ha: μ1 > μ2 The population mean of group 1 is less than the population mean of group 2. H0: μ1 = μ2 Ha: μ1 < μ2 40 The two samples are random and independent. The populations from which the samples are drawn are either normal or the sample sizes are large. The populations have the same standard deviation. 41 Step 3: Calculate the test statistic 𝑦𝟏 − 𝑦𝟐 𝒕𝒐𝒃𝒔 = 𝟏 𝟏 𝒔𝒑 + 𝒏𝟏 𝒏𝟐 𝒔𝒑 = where 𝒏𝟏 − 𝟏 𝒔𝟐𝟏 + 𝒏𝟐 − 𝟏 𝒔𝟐𝟐 𝒏𝟏 + 𝒏𝟐 − 𝟐 Step 4: Calculate the appropriate p-value. Step 5: Write a Conclusion. 42 Possible Hypotheses: 2-Tailed Test Right-Tailed Left Tailed Null 𝐻0 : 𝜇1 − 𝜇2 = 0 𝐻0 : 𝜇1 − 𝜇2 ≤ 0 𝐻0 : 𝜇1 − 𝜇2 ≥ 0 Alternative 𝐻𝑎 : 𝜇1 − 𝜇2 ≠ 0 𝐻𝑎 : 𝜇1 − 𝜇2 > 0 𝐻𝑎 : 𝜇1 − 𝜇2 < 0 Test Statistic: 𝑦𝟏 − 𝑦𝟐 𝒕𝒐𝒃𝒔 = 𝒔𝒑 𝒔𝒑 = 𝟏 𝟏 + 𝒏𝟏 𝒏𝟐 Degrees of Freedom n1 + n2 − 2 𝒏𝟏 − 𝟏 𝒔𝟐𝟏 + 𝒏𝟐 − 𝟏 𝒔𝟐𝟐 𝒏𝟏 + 𝒏𝟐 − 𝟐 Assumption: The populations from which both samples are drawn are normal or approximately normal. 43 A researcher would like to know whether the mean sepal width of setosa irises is different from the mean sepal width of versicolor irises. The researcher randomly selects 50 setosa irises and 50 versicolor irises and measures their sepal widths. Step 1 Hypotheses: H0: μsetosa = μversicolor Ha: μsetosa ≠ μversicolor http://en.wikipedia.org/ wiki/Iris_flower_data_set http://en.wikipedia.org/ wiki/Iris_versicolor 44 Steps 2-4: JMP Demonstration: Analyze Fit Y By X Y, Response: Sepal Width X, Factor: Species Means/ANOVA/Pooled t Normal Quantile Plot Plot Actual by Quantile 45 -2.33 -1.64 -1.28 -0.67 0.0 setosa 0.67 1.281.64 2.33 0.98 0.9 0.8 0.5 0.2 0.1 0.02 versicolor Normal Quantile Step 5 Conclusion: There is strong evidence (p-value < 0.0001) that the mean sepal widths for the two varieties are different. 46 The paired t-test is used to compare the population means of two groups when the samples are dependent. 47 The objective of paired comparisons is to minimize sources of variation that are not of interest in the study by pairing observations with similar characteristics. Example: A researcher would like to determine if background noise causes people to take longer to complete math problems. The researcher gives 20 subjects two math tests one with complete silence and one with background noise and records the time each subject takes to complete each test. 48 The population mean difference is not equal to zero. H0: μdifference = 0 Ha: μdifference ≠ 0 The population mean difference is greater than zero. H0: μdifference = 0 Ha: μdifference > 0 The population mean difference is less than a zero. H0: μdifference = 0 Ha: μdifference < 0 49 The sample is random. The data is matched pairs. The differences have a normal distribution or the sample size is large. 50 Step 3: Calculate the test Statistic: 𝒚𝒅 𝒕𝒐𝒃𝒔 = 𝒔 𝒅 𝒏 Where 𝑦𝑑 bar is the mean of the differences and sd is the standard deviations of the differences. Step 4: Calculate the p-value. Step 5: Write a conclusion. 51 Possible Hypotheses: 2-Tailed Right Tailed Left Tailed Null 𝐻0 : 𝜇𝑑 = 0 𝐻0 : 𝜇𝑑 ≤ 0 𝐻0 : 𝜇𝑑 ≥ 0 Alternative 𝐻𝑎 : 𝜇𝑑 ≠ 0 𝐻𝑎 : 𝜇𝑑 > 0 𝐻𝑎 : 𝜇𝑑 < 0 Test Statistic: 𝒚𝒅 𝒕𝒐𝒃𝒔 = 𝒔 𝒅 𝒏 Degrees of Freedom: 𝒏 − 𝟏 Assumption: The population of differences is normal or approximately normal. 52 A researcher would like to determine whether a fitness program increases flexibility. The researcher measures the flexibility (in inches) of 12 randomly selected participants before and after the fitness program. Step 1: Formulate a Hypothesis H0: μAfter - Before = 0 Ha: μ After - Before > 0 http://office.microsoft.com/en-us/images 53 Steps 2-4: JMP Analysis: Create a new column of After – Before Analyze Distribution Y, Columns: After – Before Normal Quantile Plot Test Mean Specify Hypothesized Mean: 0 54 Step 5 Conclusion: There is not evidence that the fitness program increases flexibility. 55 ANOVA is used to determine whether three or more populations have different distributions. 56 ANOVA is used to determine whether three or more populations have different distributions. A B C Medical Treatment 57 The first step is to use the ANOVA F test to determine if there are any significant differences among the population means. If the ANOVA F test shows that the population means are not all the same, then follow up tests can be performed to see which pairs of population means differ. 58 yij i ij Where yij is the response of the jth trial on the ith factor level i is the mean of the ith group ij ~ N (0, 2 ) i 1,, r j 1, , ni In other words, for each group the observed value is the group mean plus some random variation. 59 Step 1: We test whether there is a difference in the population means. H 0 : 1 2 r H a : The i are not all equal. 60 The samples are random and independent of each other. The populations are normally distributed. The populations all have the same standard deviations. The ANOVA F test is robust to the assumptions of normality and equal standard deviations. 61 C A B C A B Medical Treatment Compare the variation within the samples to the variation between the samples. 62 F Variation between Groups MSG Variation within Groups MSE Variation within groups small compared with variation between groups → Large F Variation within groups large compared with variation between groups → Small F 63 The mean square for groups, MSG, measures the variability of the sample averages. SSG stands for sums of squares groups. SSG MSG r -1 n1 ( y1 y ) 2 n 2 ( y2 y ) 2 n r ( y1 y ) 2 r -1 64 Mean square error, MSE, measures the variability within the groups. SSE stands for sums of squares error. SSE n-r (n 1 - 1)s12 (n 2 - 1)s 22 (n r - 1)s 2r n-r Where MSE ni si (y j 1 ij yi ) ni 1 65 Step 4: Calculate the p-value. Step 5: Write a conclusion. 66 A researcher would like to determine if three drugs provide the same relief from pain. 60 patients are randomly assigned to a treatment (20 people in each treatment). Step 1: Formulate the Hypotheses H0: μDrug A = μDrug B = μDrug C Ha : The μi are not all equal. http://office.microsoft.com/en-us/images 67 JMP demonstration Analyze Fit Y By X Y, Response: Pain X, Factor: Drug Normal Quantile Plot Plot Actual by Quantile Means/ANOVA 68 -2.33 -1.64 -1.28 -0.67 75 0.0 0.67 1.281.64 2.33 Drug B Drug Drug C A 65 60 0.98 0.9 Drug C 0.8 Drug B Drug 0.5 Drug A 0.2 50 0.1 55 0.02 Pain 70 Normal Quantile Step 5 Conclusion: There is strong evidence that the drugs are not all the same. 69 The p-value of the overall F test indicates that the level of pain is not the same for patients taking drugs A, B and C. We would like to know which pairs of treatments are different. One method is to use Tukey’s HSD (honestly significant differences). 70 Tukey’s test simultaneously tests H 0 : i i ' H a : i i ' for all pairs of factor levels. Tukey’s HSD controls the overall type I error. JMP demonstration Oneway Analysis of Pain By Drug Compare Means All Pairs, Tukey HSD 71 Level Drug C Drug C Drug B - Level Drug A Drug B Drug A Difference 5.850000 3.600000 2.250000 Std Err Dif 1.677665 1.677665 1.677665 Lower CL 1.81283 -0.43717 -1.78717 Upper CL 9.887173 7.637173 6.287173 p-Value 0.0027* 0.0897 0.3786 The JMP output shows that drugs A and C are significantly different. 72 73 We are interested in the effect of two categorical factors on the response. We are interested in whether either of the two factors have an effect on the response and whether there is an interaction effect. ◦ An interaction effect means that the effect on the response of one factor depends on the level of the other factor. 74 No Interaction Interaction Low High Dosage Drug A Drug B Improvement Improvement Drug A Drug B Low High Dosage 75 yijk i j ( ) ij ijk Where yijk is the response of the kth trial on the ith factor A level and the jth factor B level is the overall mean i is the main effect of the ith level of factor A j is the main effect of the jth level of factor B ( ) ij is the interactio n effect of the ith level of factor A and the jth level of factor B ijk ~ N (0, 2 ) i 1, , a j 1, , b k 1,..., nij 76 We would like to determine the effect of two alloys (low, high) and three cooling temperatures (low, medium, high) on the strength of a wire. JMP demonstration Analyze Fit Model Y: Strength Highlight Alloy and Temp and click Macros Factorial to Degree Run Model http://office.microsoft.com/en-us/images 77 Conclusion: There is strong evidence of an interaction between alloy and temperature. 78 The one sample t-test allows us to test whether the population mean of a group is equal to a specified value. The two-sample t-test and paired t-test allow us to determine if the population means of two groups are different. ANOVA allows us to determine whether the population means of several groups are different. 79 For information about using SAS, SPSS and R to do ANOVA: http://www.ats.ucla.edu/stat/sas/topics/anova .htm http://www.ats.ucla.edu/stat/spss/topics/anov a.htm http://www.ats.ucla.edu/stat/r/sk/books_pra. htm 80 Fisher’s Irises Data (used in one sample and two sample t-test examples). Flexibility data (paired t-test example): Michael Sullivan III. Statistics Informed Decisions Using Data. Upper Saddle River, New Jersey: Pearson Education, 2004: 602. 81