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Analyze Phase Hypothesis Testing Normal Data Part 1

Welcome to Analyze “X” phase, focusing on hypothesis testing with t-tests on normal data. Learn to compare means, interpret results, calculate sample sizes, and draw practical conclusions using inferential statistics.

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Analyze Phase Hypothesis Testing Normal Data Part 1

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  1. Analyze PhaseHypothesis Testing Normal Data Part 1

  2. Hypothesis Testing Normal Data Part 1 Welcome to Analyze “X” Sifting Inferential Statistics Intro to Hypothesis Testing Sample Size Hypothesis Testing ND P1 Testing Means Analyzing Results Hypothesis Testing ND P2 Hypothesis Testing NND P1 Hypothesis Testing NND P2 Wrap Up & Action Items

  3. t-tests are used: To compare a Mean against a target. i.e.; The team made improvements and wants to compare the Mean against a target to see if they met the target. To compare Means from two different samples. i.e.; Machine one to machine two. i.e.; Supplier one quality to supplier two quality. To compare paired data. Comparing the same part before and after a given process. Test of Means (t-tests) They don’t look the same to me!

  4. 1 Sample t • A 1-sample t-test is used to compare an expected population Mean to a target. • MINITABTM performs a one sample t-test or t-confidence interval for the Mean. • Use 1-sample t to compute a confidence interval and perform a Hypothesis Test of the Mean when the population Standard Deviation, σ, is unknown. For a one or two-tailed 1-sample t: • H0: μsample = μtargetIf P-value > 0.05 fail to rejectHo • Ha: μsample≠, <, > μtargetIf P-value < 0.05 rejectHo Target μsample

  5. Target 1 Sample t-test Sample Size Population n = 2 Cannot tell the difference between the sample and the target. n = 30 Can tell the difference between the sample and the target.

  6. Sample Size Three fields must be filled in and one left blank.

  7. Sample Size Power and Sample Size 1-Sample t Test Testing Mean = null (versus not = null) Calculating power for Mean = null + difference Alpha = 0.05 Assumed Standard Deviation = 1 Sample Size Power Difference 10 0.9 1.15456 15 0.9 0.90087 20 0.9 0.76446 25 0.9 0.67590 30 0.9 0.61245 35 0.9 0.56408 40 0.9 0.52564 The various sample sizes show how much of a difference can be detected assuming a Standard Deviation = 1.

  8. 1. Practical Problem: We are considering changing suppliers for a part we currently purchase from a supplier that charges us a premium for the hardening process. The proposed new supplier has provided us with a sample of their product. They have stated they can maintain a given characteristic of 5 on their product. We want to test the samples to determine if their claim is accurate. 2. Statistical Problem: Ho: μN.S. = 5 Ha: μN.S.≠ 5 3. 1-sample t-test (population Standard Deviation unknown, comparing to target). α = 0.05 β = 0.10 1-Sample t Example

  9. Example • 4. Sample Size: • Open the MINITABTM worksheet: “Exh_Stat.MTW”. • Use the C1 column: Values • In this case, the new supplier sent 9 samples for evaluation. • How much of a difference can be detected with this sample?

  10. 1-Sample t Example MINITABTM Session Window Power and Sample Size 1-Sample t Test Testing Mean = null (versus not = null) Calculating power for Mean = null + difference Alpha = 0.05 Assumed Standard Deviation = 1 Sample Size Power Difference 9 0.9 1.23748 This means we will be able to detect a difference of only 1.24 if the population has a Standard Deviation of 1 unit.

  11. Example: Follow the Road Map 5. State Statistical Solution Stat > Basic Statistics > Normality Test… Are the data in the values column Normal?

  12. 1-Sample t Example • Click “Graphs” • Select all 3 • Click “Options…” • - In CI enter ’95’

  13. Histogram of Values Note our target Mean (represented by red Ho) is outside our population confidence boundaries which tells us there is a significant difference between population and target Mean.

  14. Box Plot of Values

  15. Individual Value Plot (Dot Plot)

  16. Session Window Ha Ho One-Sample T: Values Test of mu = 5 vs not = 5 Variable N Mean StDev SE Mean 95% CI T P Values 9 4.78889 0.24721 0.08240 (4.59887, 4.97891) -2.56 0.034 T-Calc = Observed – Expected over SE Mean T-Calc = X-bar – Target over Standard Error T-Calc = 4.7889 – 5 over .0824 = - 2.56 N – sample size Mean – calculate mathematic average StDev – calculated individual Standard Deviation (classical method) SE Mean – calculated Standard Deviation of the distribution of the Means Confidence Interval that our population average will fall between 4.5989 and 4.9789

  17. Ho X Evaluating the Results Since the P-value of 0.034 is less than 0.05 reject the null hypothesis. Based on the samples given there is a difference between the average of the sample and the desired target. 6. State Practical Conclusions The new supplier’s claim they can meet the target of 5 for the hardness is not correct.

  18. Let’s compare the manual calculations to what the computer calculates. Calculate t-statistic from data: Determine critical t-value from t-table in reference section. When the alternative hypothesis has a not equal sign it is a two-sided test. Split the α in half and read from the 0.975 column in the t-table for n -1 (9 - 1) degrees of freedom. Manual Calculation of 1- Sample t

  19. The data supports the alternative hypothesis that the estimate for the Mean of the population is not 5.0. Manual Calculation of 1- Sample t m -2.56 -2.306 2.306 α/2=.025 α/2=.025 0 Critical Regions

  20. The formula for a two-sided t-test is: Confidence Intervals for Two-Sided t-test Ho 4.5989 4.9789 4.7889

  21. Exercise objective: Utilize what you have learned to conduct and analyze a one sample t-test using MINITABTM. The last engineering estimation said we would achieve a product with average results of 32 parts per million (ppm). We want to test if we are achieving this performance level, we want to know if we are on target with 95% confidence in our answer. Use worksheet HYPOTTESTSTUD with data in column “ppm VOC” Are we on Target? 1-Sample t Exercise

  22. 1-Sample t Exercise: Solution • Since we do not know the population Standard Deviation we will use the 1 sample T test to determine if we are at Target.

  23. 1-Sample t Exercise: Solution • After selecting column C1 and setting “Test mean” to 32.0, click “Graphs…” and select “Histogram of data” to get a good visualization of the analysis. • Depending on the test you are running you may need to select “Options…” to set your desired Confidence Interval and hypothesis. In this case the MINITABTM defaults are what we want.

  24. 1-Sample t Exercise: Solution • Because we used the option of “Graphs…” we get a nice visualization of the data in a histogram AND a plot of the null hypothesis relative to the confidence level of the population Mean. • Because the null hypothesis is within the confidence level you know we will “fail to reject” the null hypothesis and accept the equipment is running at the target of 32.0.

  25. In MINITABTM’s Session Window (ctrl – M) you can see the P-value of 0.201. Because it is above 0.05 we “fail to reject” the null hypothesis so we accept the equipment is giving product at a target of 32.0 ppm VOC. 1-Sample t Exercise: Solution

  26. Hypothesis Testing Roadmap Normal Continuous Data Test of Equal Variance 1 Sample Variance 1 Sample t-test Variance Equal Variance Not Equal Two samples Two samples 2 Sample T One Way ANOVA 2 Sample T One Way ANOVA

  27. 2 Sample t-test • A 2-sample t-test is used to compare two Means. • MINITABTM performs an independent two-sample t-test to generate a confidence interval. • Use 2-Sample t to perform a Hypothesis Test and compute a confidence interval of the difference between two population Means when the population Standard Deviations, σ’s, are unknown. • Two tailed test: • H0: μ1 = μ2If P-value > 0.05 fail to rejectHo • Ha: μ1≠ μ2If P-value < 0.05 rejectHo • One tailed test: • H0: μ1 = μ2 • Ha: μ1> or <μ2 Stat > Basic Statistics > 2-Sample t m1 m2

  28. Sample Size Three fields must be filled in and one left blank. 28

  29. Sample Size Power and Sample Size 2-Sample t Test Testing Mean 1 = Mean 2 (versus not equal) Calculating power for Mean 1 = Mean 2 + difference Alpha = 0.05 Assumed Standard Deviation = 1 Sample Size Power Difference 10 0.9 1.53369 15 0.9 1.22644 20 0.9 1.05199 25 0.9 0.93576 30 0.9 0.85117 35 0.9 0.78605 40 0.9 0.73392 The sample size is for each group. The various sample sizes show how much of a difference can be detected assuming the Standard Deviation = 1.

  30. 2-Sample t Example • 1. Practical Problem: • We have conducted a study in order to determine the effectiveness of a new heating system. We have installed two different types of dampers in home ( Damper = 1 and Damper = 2). • We want to compare the BTU.In data from the two types of dampers to determine if there is any difference between the two products. • 2. Statistical Problem: • Ho: μ1 = μ2 • Ha: μ1≠μ2 • 3. 2-Sample t-test (population Standard Deviations unknown). • α = 0.05 β = 0.10 No, not that kind of damper!

  31. 2 Sample t Example • 4. Sample Size: • Open the MINITABTM worksheet: “Furnace.MTW” • Scroll through the data to see how the data is coded. • In order to work with the data in the BTU.In column we will need to unstack the data by damper type.

  32. 2 Sample t Example Data > Unstack Columns…

  33. 2 Sample t Example

  34. 2 Sample t Example MINITABTM Session Window Power and Sample Size 2-Sample t Test Testing Mean 1 = Mean 2 (versus not =) Calculating power for Mean 1 = Mean 2 + difference Alpha = 0.05 Assumed Standard Deviation = 1 Sample Size Power Difference 40 0.9 0.733919 50 0.9 0.654752 The sample size is for each group.

  35. Example: Follow the Roadmap… 5. State Statistical Solution

  36. Normality Test – Is the Data Normal?

  37. Normality Test – Is the Data Normal?

  38. Test of equal Variance (Bartlett’s Test)

  39. Test of Equal Variance Sample 1 Sample 2

  40. 2 Sample t-test Equal Variance

  41. 5. State Statistical Conclusions: Fail to reject the null hypothesis. 6. State Practical Conclusions: There is no difference between the dampers for BTU’s in. Box Plot

  42. Minitab Session Window Calculated Average Two- Sample T-Test (Variances Equal) -1.450 0.980 Number of Samples -0.38 Ho: μ1 = μ2 Ha: μ1≠ or < or > μ2

  43. Exercise Exercise objective: Utilize what you have learned to conduct and analyze a 2 sample t-test using MINITABTM. Billy Bob’s Pool Care has conducted a study on the effectiveness of two chlorination distributors in a swimming pool. (Distributor 1 & Distributor 2). The up and coming Billy Bob Jr., looking to prove himself, wants a comparison done on the Clor.Lev_Post data from the two types of distributors in order to determine if there is any difference between the two products. With 95% confidence is there a significant difference between the two distributors? Use data within MINITABTM Worksheet “Billy Bobs Pool.mtw” 43

  44. 2 Sample T-Test: Solution • 1.What do we want to know: With 95% confidence is there a significant difference between the two distributors? • 2.Statistical Problem: • Ho: μ1 = μ2 • Ha: μ1≠μ2 3.2-Sample t-test (population Standard Deviations unknown). α = 0.05 β = 0.10 4.Now we need to look at the data to determine the Sample Size but let’s see how the data is formatted first. 44

  45. 2 Sample T-Test: Solution Data > Unstack Columns… • “Unstack the data in:” Select ‘Clor.Levl_Post’ • “Using subscripts in:” Select ‘Distributor’ 45

  46. 2 Sample T-Test: Solution • Clor.Lev_Post_1 = Distributor 1 • Clor.Lev_Post_2 = Distributor 2 46

  47. 2 Sample T-Test: Solution 47

  48. 2 Sample T-Test: Solution • We want to determine what is the smallest difference that can be detected based on our data. • Fill in the three areas and leave “Differences:” blank so that MINITABTM will tell us the differences we need. 48

  49. 2 Sample T-Test: Solution • The smallest difference that can be calculated is based on the smallest sample size. • In this case: • .7339 rounded to.734 49

  50. 2 Sample T-Test: Solution Follow the path: “Stat > Basic Statistics > Normality Test…” 50

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