Analyze Phase Hypothesis Testing Normal Data Part 1

Analyze Phase Hypothesis Testing Normal Data Part 1
Now we will continue in the Analyze Phase with “Hypothesis Testing Normal Data Part 1”.

Hypothesis Testing Normal Data Part 1
Testing Means Analyzing Results Sample Size Hypothesis Testing NND P1 Hypothesis Testing ND P1 Intro to Hypothesis Testing Inferential Statistics “X” Sifting Welcome to Analyze Hypothesis Testing ND P2 Wrap Up & Action Items Hypothesis Testing NND P2 The core fundamentals of this phase are Hypothesis Testing, Tests for Central Tendency, Tests for Variance and ANOVA.

Test of Means (t-tests)
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. T-tests are used to compare a Mean against a target, to compare Means from two different samples and to compare paired data. When comparing multiple Means it is inappropriate to use a t-test. Analysis of variance or ANOVA is used when it is necessary to compare more than two Means. They don’t look the same to me!

H0: μsample = μtarget If P-value > 0.05 fail to reject Ho
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 = μtarget If P-value > fail to reject Ho Ha: μsample ≠, <, > μtarget If P-value < reject Ho Target μsample Here we are looking for the region in which we can be 95% certain our true population Mean will lie. This is based on a calculated average, Standard Deviation, number of trials and a given alpha risk of .05. In order for the Mean of the sample to be considered not significantly different than the target, the target must fall within the confidence interval of the sample Mean.

1 Sample t-test Sample Size
Target n = 2 Population n = 30 Cannot tell the difference between the sample and the target. Can tell the difference One common pitfall in statistics is not understanding what the proper sample size should be. When you look at the graphic the question is: Is there a difference between my process Mean and the desired target? If we had population data it would be very easy – no they are not the same but they may be within an acceptable tolerance (or specification window). If we took a sample of 2 can we tell a difference? No, because the spread of the distribution of averages from samples of 2 will create too much uncertainty making it very difficult to statistically say there is a difference. If you remember from earlier 95% of the area under the curve of a Normal Distribution falls within plus or minus 2 Standard Deviations. Confidence intervals are based on your selected alpha level so if you selected an alpha of 5% the confidence interval would be 95%; which is roughly plus or minus 2 Standard Deviations. Using your eye to guesstimate you can see the target value falls within plus or minus 2 Standard Deviations of the sampling distribution of sample size 2. If you used a sample of 30 could you tell if the target was different? Just using your eye it appears the target is outside the 95% confidence interval of the Mean. Luckily, MINITABTM makes this very easy…

Three fields must be filled in and one left blank.
Sample Size Three fields must be filled in and one left blank. Instead of going through the dreadful hand calculations to yield sample size we will use MINITABTM. Three fields must be filled in with one left blank in the sample size window. MINITABTM will solve for the fourth. If you want to know the sample size you must enter the difference which is the shift that must be detected. It is common to state the difference in terms of “generic” Standard Deviations when you do not have an estimate for the Standard Deviation of the process. For example, if you want to detect a shift of 1.5 Standard Deviations enter that in “Differences” and enter 1 for “Standard deviation”. If you knew the Standard Deviation was 0.8 then enter it and 1.2 for the “Differences” (which is a 1.5 Standard Deviation shift in terms of real values). If you are unsure of the desired difference, or in many cases simply get stuck with a sample size you did not have a lot of control over, MINITABTM will tell you how much of a difference can be detected. You as a practitioner must be careful when drawing Practical Conclusions because it is possible to have statistical significance without practical significance. In other words - do a reality check. MINITABTM has made it easy to see an assortment of sample sizes and differences. Try the example shown.

Power and Sample Size 1-Sample t Test
Testing Mean = null (versus not = null) Calculating power for Mean = null + difference Alpha = Assumed Standard Deviation = 1 Sample Size Power Difference The various sample sizes show how much of a difference can be detected assuming a Standard Deviation = 1. Notice as the sample size increases there is not as big an effect on the Difference. If it was only necessary to see a Difference of 0.9 why bother taking any more samples than 15? The Standard Deviation entered has an effect on the Difference calculated. Take a few moments to explore different Standard Deviation sizes in MINITABTM to see their effect on Difference.

We want to test the samples to determine if their claim is accurate.
1-Sample t Example 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 = β = 0.10 Let’s now try a 1-sample t example. Step 1: Take a moment to review the Practical Problem Step 2: The Statistical Problem is: The null hypothesis is the Mean of the new supplier is equal to 5. The alternative hypothesis is the Mean of the new supplier is not equal to 5. This is considered a 2-tailed test. Step 3: Our selected alpha level is 0.05 and beta is 0.10.

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? Please read the slide.

1-Sample t Example This means we will be able to detect a difference of only 1.24 if the population has a Standard Deviation of 1 unit. Power and Sample Size 1-Sample t Test Testing Mean = null (versus not = null) Calculating power for Mean = null + difference Alpha = Assumed Standard Deviation = 1 Sample Size Power Difference MINITABTM Session Window Follow along in MINITABTM, as you can see, we will be able to detect a difference of 1.24 with the sample of 9. If this was not good enough you would need to request additional samples.

Example: Follow the Road Map
5. State Statistical Solution Stat > Basic Statistics > Normality Test… Are the data in the values column Normal? Now refer to the road map for Hypothesis Testing where you must first check for Normality. In MINITABTM select “Stats>Basic Statistics>Normality Test”. For the “Variable Fields” double-click on “Value” in the left-hand box. Once this is complete select “OK”. Since the P-value is greater than 0.05 we fail to reject the null hypothesis that the data are Normal.

Click “Graphs” Select all 3 Click “Options…” - In CI enter ’95’
1-Sample t Example Click “Graphs” Select all 3 Click “Options…” - In CI enter ’95’ Perform the one sample t-test. In MINITABTM select “Stat>Basic Statistics>1-Sample t”. From the left-hand box double-click on “Values”. In the “Options…” button there is a selection for the alternative hypothesis. We find the default is not equal, which corresponds to our hypothesis. If your alternative hypothesis was a greater than or less than you would have to change the default.

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. Based on the graph we can say there is a statistical difference or reject the null hypothesis for the following reasons; A Histogram is not especially interesting when there are so few data points but it does show the 95% confidence interval of the data along with the hypothesized value of 5 noted as the Ho or null hypothesis.

Box Plot of Values The Box Plot shows a different representation of the data but the conclusion is the same.

Individual Value Plot (Dot Plot)
As you will see the conclusion is the same but the Dot Plot is just another representation of data.

Session Window Ha Ho One-Sample T: Values Test of mu = 5 vs not = 5
Variable N Mean StDev SE Mean % CI T P Values ( , ) Ho Ha 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 and T-Calc = Observed – Expected over SE Mean T-Calc = X-bar – Target over Standard Error T-Calc = – 5 over = Shown here is the MINITABTM Session Window output for the 1-Sample t-test.

Evaluating the Results
Since the P-value of 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. X Ho Please read the slide.

Manual Calculation of 1- Sample t
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 column in the t-table for n -1 (9 - 1) degrees of freedom. Here are the manual calculations of the 1-samle t. Verify MINITABTM is correct.

Manual Calculation of 1- Sample t
α/2=.025 m -2.306 2.306 Critical Regions -2.56 The data supports the alternative hypothesis that the estimate for the Mean of the population is not 5.0. α/2 =.025 If the calculated t-value lies anywhere in the critical regions reject the null hypothesis.

Confidence Intervals for Two-Sided t-test
The formula for a two-sided t-test is: Here is the formula for the confidence interval. Notice we get the same results as MINITABTM. Ho 4.5989 4.9789 4.7889

1-Sample t Exercise 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? Exercise.

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. Please read the slide.

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. Please read the slide.

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. Please read the slide.

In MINITABTM’s Session Window (ctrl – M) you can see the P-value of 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. Please read the slide.

Hypothesis Testing Roadmap
Normal Test of Equal Variance 1 Sample t-test 1 Sample Variance Variance Not Equal Variance Equal 2 Sample T One Way ANOVA Continuous Data Two samples We will now focus on the 2 sample t.

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 = μ2 If P-value > fail to reject Ho Ha: μ1 ≠ μ2 If P-value < reject Ho One tailed test: H0: μ1 = μ2 Ha: μ1 > or < μ2 Stat > Basic Statistics > 2-Sample t m1 m2 Notice the difference in the hypothesis for two-tailed vs. one-tailed test. This terminology is only used to know which column to look down in the t-table.

Three fields must be filled in and one left blank.
Sample Size Three fields must be filled in and one left blank. Instead of going through the dreadful hand calculations of sample size we will use MINITABTM. Three fields must be filled in and one left blank in the sample size window. MINITABTM will solve for the third. If you want to know the sample size, you must enter the difference, which is the shift that must be detected. It is common to state the difference in terms of “generic” Standard Deviations when you do not have an estimate for the Standard Deviation of the process. For example, if you want to detect a shift of 1.5 Standard Deviations enter that in “Difference” and enter 1 for “Standard deviation”. If you knew the Standard Deviation and it was 0.8, then enter it for “Standard deviation” and 1.2 for the “Difference” (which is a 1.5 Standard Deviation shift in terms of real values). If you are unsure of the desired difference, or in many cases simply get stuck with a sample size you did not have a lot of control over, MINITABTM will tell you how much of a difference can be detected. You as a practitioner must be careful when drawing Practical Conclusions because it is possible to have statistical significance without practical significance. In other words - do a reality check. MINITABTM has made it easy to see an assortment of sample sizes and differences. Try the example shown. 28 28

Sample Size The various sample sizes show how much of a difference can be detected assuming the Standard Deviation = 1. 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 = Assumed Standard Deviation = 1 Sample Size Power Difference The sample size is for each group. As you can see we used the same command here just as in the 1-sample t. Do you think the results are different? Correct, the results are different.

3. 2-Sample t-test (population Standard Deviations unknown).
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 = β = 0.10 No, not that kind of damper! Over the next several slides we will explore an example for a 2-Sample t-test. Step 1. Read Practical Problem Step 2. The null hypothesis is the Mean of BTU.In for damper 1 is equal to the Mean of BTU.In for damper 2. The alternative hypothesis is the Means are not equal. Step 3. We will use the 2-Sample t-test since the population Standard Deviations are unknown.

4. Sample Size: Open the MINITABTM worksheet: “Furnace.MTW”
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. Now in Step 4. Open the worksheet in MINITABTM called: “Furnace.MTW” How is the data coded? The only way we can work with the data in the BTU.In is by unstacking the data by damper type.

2 Sample t Example Data > Unstack Columns…
We will unstack the data in BTU.In using the subscripts in Damper. Store the unstacked data after the last column in use. Check the “Name the columns containing the unstacked data” box. Then click “OK”.

2 Sample t Example Notice the “unstacked” data for each damper WE NOW HAVE TWO COLUMNS.

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 = Assumed Standard Deviation = 1 Sample Size Power Difference The sample size is for each group. MINITABTM Session Window Now let us perform a 2 Sample t Example. In MINITABTM select “Stat>Power and Sample size>2-Sample t”. For the field “Sample Sizes:” enter ‘40 space 50’ because our data set has unequal sample sizes which is not uncommon. The smallest difference that can be detected is based on the smallest sample size so in this case it is:

Example: Follow the Roadmap…
5. State Statistical Solution Please read the slide.

Normality Test – Is the Data Normal?
The data is considered Normal since the P-value is greater than 0.05.

Normality Test – Is the Data Normal?
This is the Normality Plot for damper 2. Is the data Normal? It is Normal, continuing down the roadmap…

Test of equal Variance (Bartlett’s Test)
In MINITABTM select “Stat>ANOVA>Test for Test for Equal Variances”. This will allow us to perform a Bartlett’s Test.

Sample 1 Sample 2 Test of Equal Variance
The P-value of indicates there is no statistically significant difference in variance. Bartlett’s Test (>2) (f-test 2-samples)

2 Sample t-test Equal Variance
Let’s continue along the roadmap… Perform the 2-Sample t-test; be sure to check the box “Assume equal variances”.

5. State Statistical Conclusions: Fail to reject the null hypothesis.
Box Plot 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. The Box Plots do not show much of a difference between the dampers.

Minitab Session Window
Number of Samples Calculated Average -1.450 0.980 -0.38 Two- Sample T-Test (Variances Equal) Ho: μ1 = μ2 Ha: μ1≠ or < or > μ2 Take a moment to review the MINITABTM Session Window.

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” Exercise. 43 43

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 = 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. Please read the slide. 44 44

“Unstack the data in:” Select ‘Clor.Levl_Post’ “Using subscripts in:” Select ‘Distributor’ Data > Unstack Columns… To unstack the data follow the steps here. This will generate two new columns of data shown on the next page… 45 45

Clor.Lev_Post_1 = Distributor 1 Clor.Lev_Post_2 = Distributor 2 By unstacking the data we how have the Clor.Lev data separated by the distributor it came from. Now let’s move on to trying to determine correct sample size. 46 46

Follow path in MINITABTM: “Stat > Power and Sample Size > 2-Sample t…” 47 47

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. Please read the slide. 48 48

The smallest difference that can be calculated is based on the smallest sample size. In this case: .7339 rounded to.734 Please read the slide. 49 49

Follow the path: “Stat > Basic Statistics > Normality Test…” Follow this trail… 50 50

Check Normality for ‘Clor.Lev_Post_1’ The result shows us a P-value of so our data is Normal. First let’s try Clor.Lev_Post_1. Recall if the P-value is greater than .05 then we will consider our data Normal. 51 51

Check Normality for ‘Clor.Lev_Post_2’ The result shows us a P-value of so our data is also Normal. Now check Normality for Clor.Lev_Post_2. 52 52

Test for Equal Variances MINITABTM Path: “Stat > ANOVA > Test for Equal Variances…” Before calculating a 2 sample t-test we must test for Equal Variances. 53 53

For the “Response:” we select our stacked column ‘Clor.Lev_Post’ For our “Factors:” we select our stacked column ‘Distributor’ Please read the slide. 54 54

Look at the P-value of ~ This tells us there is no statistically significant difference in the variance in these two data sets. What does this mean….We can finally run a 2 sample t–test with Equal Variances? Please read the slide. 55 55

For “Samples:” enter ‘Clor.Lev_Post’ For “Subscripts:” enter ‘Distributors’ Follow the command prompt shown here and enter the data as shown. Remember you must click on graphs and check the Box Plot data option. This way MINITABTM will create a Box Plot. Equal variances can be assumed based on the test for equal variances on the previous page. 56 56

Look at the Box Plot and Session Window. There is NO significant difference between the distributors. Hmm, we’re a lot alike! Two-sample T for Clor.Lev_Post Distributor N Mean StDev SE Mean Difference = mu (1) - mu (2) Estimate for difference: 95% CI for difference: (-1.049, 1.920) T-Test of difference = 0 (vs not =): T-Value = P-Value = DF = 88 Both use Pooled StDev = The Box Plots show VERY little difference between the Distributors. Also note the P-value in the Session Window. There is no difference between the two distributors. 57 57

Normal Test of Equal Variance 1 Sample t-test 1 Sample Variance Variance Not Equal Variance Equal 2 Sample T One Way ANOVA Continuous Data Two samples Now we will now look at an example when variance is not equal.

Unequal Variance Example
Don’t just sit there…. open it! Open the MINITABTM worksheet named “2 SAMPLE UNEQUAL VARIANCE DATA”.

Our data sets are Normally Distributed.
Normality Test Our data sets are Normally Distributed. Run a Normality Test… Let’s compare the data in Sample one and Sample three columns. Check for Normality.

Test for Equal Variance
We use F-Test Statistic because our data is Normally Distributed. P-value is less than 0.05 so our variances are not equal. Stat>ANOVA>Test of Equal Variance Standard Deviation of Samples Medians of Samples This is the output from MINITABTM. Notice even though the names of the columns in MINITABTM were Sample 1 and Sample 3 MINITABTM used Factor levels 1 and 2 to differentiate the outcome. We have to interpret the meaning for factor levels properly. It is simply the difference between the samples labeled one and three in our worksheet.

2-Sample t-test Unequal Variance
UNCHECK “Assume equal variances” box. You can see there is very little difference in the 2-Sample t-tests.

Indicates Sample Means The Box Plot shows no difference between the Means. The overall box is smaller for sample on the left which is an indication for the difference in variance.

Indicates Sample Means By looking at this Individual Value Plot you can notice a big spread or variance of the data.

Two-Sample T-Test (Variances Not Equal) Ho: μ1 = μ2 (P-value > 0.05) Ha: μ1 ≠ or < or > μ2 (P-value < 0.05) Stat>Basic Stats> 2 sample T (Deselect Assume Equal Variance) What does the P-value of mean? After conducting a 2-sample t-test there is no significant difference between the Means.

Normal Test of Equal Variance 1 Sample t-test 1 Sample Variance Variance Not Equal Variance Equal 2 Sample T One Way ANOVA Continuous Data Two samples Now we will cover paired t-tests.

Paired t-test A Paired t-test is used to compare the Means of two measurements from the same samples generally used as a before and after test. MINITABTM performs a paired t-test. This is appropriate for testing the difference between two Means when the data are paired and the paired differences follow a Normal Distribution. Use the Paired t command to compute a confidence interval and perform a Hypothesis Test of the difference between population Means when observations are paired. A paired t-procedure matches responses that are dependent or related in a pair-wise manner. This matching allows you to account for variability between the pairs usually resulting in a smaller error term, thus increasing the sensitivity of the Hypothesis Test or confidence interval. Ho: μδ = μo Ha: μδ ≠ μo Where μδ is the population Mean of the differences and μ0 is the hypothesized Mean of the differences, typically zero. mbefore delta (d) mafter Stat > Basic Statistics > Paired t Please read the slide.

3. Paired t-test (comparing data that must remain paired).
Example Practical Problem: We are interested in changing the sole material for a popular brand of shoes for children. In order to account for variation in activity of children wearing the shoes each child will wear one shoe of each type of sole material. The sole material will be randomly assigned to either the left or right shoe. 2. Statistical Problem: Ho: μδ = 0 Ha: μδ ≠ 0 3. Paired t-test (comparing data that must remain paired). α = β = 0.10 Just checking your souls, er…soles! Please read the slide.

Example 4. Sample Size: How much of a difference can be detected with 10 samples? Open Minitab Worksheet “EXH_STAT DELTA.MTW” Now let’s open EXH_STAT Delta.MTW for analysis. Use columns labeled Mat-A and Mat-B.

Now that’s a tee test! Paired t-test Example
Power and Sample Size 1-Sample t Test Testing Mean = null (versus not = null) Calculating power for Mean = null + difference Alpha = Assumed Standard Deviation = 1 Sample Size Power Difference MINITABTM Session Window This means we will be able to detect a difference of only 1.15 if the Standard Deviation is equal to 1. Now that’s a tee test! In MINITABTM open “Stat>Power and Sample size>1-Sample t”. Enter in the appropriate Sample Size, Power Value and Standard Deviation. Given the sample size of 10 we will be able to detect a Difference of If this was your process you would need to decide if this was good enough. In this case is a difference of 1.15 enough to practically want to change the material used for the soles of the children’s shoes?

5. State Statistical Solution
Paired t-test Example Check this box so MinitabTM will recalculate as new data is entered. 5. State Statistical Solution We need to calculate the difference between the two distributions. We are concerned with the delta; is the Ho outside the t-calc (confidence interval)? Calc > Calculator For the next test we must first calculate the difference between the two columns. In MINITABTM open “Calc > Calculator”. We placed Mat-B first in the equation shown because it was generally higher than the values for Mat-A.

Analyzing the Delta Following the Hypothesis Test roadmap we first test the AB-Delta distribution for Normality. Please read the slide.

1-Sample t Stat > Basic Statistics > 1-Sample t-test… Since there is only one column, AB Delta, we do not test for Equal Variance per the Hypothesis Testing roadmap. Check this data for statistical significance in its departure from our expected value of zero. Please read the slide.

5. State Statistical Conclusions: Reject the null hypothesis
Box Plot 5. State Statistical Conclusions: Reject the null hypothesis 6. State Practical Conclusions: We are 95% confident there is a difference in wear rates between the two materials. Box Plot of AB Delta One-Sample T: AB Delta Test of mu = 0 vs not = 0 Variable N Mean StDev SE Mean AB Delta 95% CI T P ( , ) MINITABTM Session Window Analyzing the Box Plot we see the null hypothesis falls outside the confidence interval so we reject the null hypothesis. The P-value is also less than Given this we are 95% confident there is a difference in the wear rates between the two materials used for the soles of children’s shoes.

Another way to analyze this data is to use the paired t-test command.
Stat>Basic Statistics>Paired T-test Click on “Graphs…” and select the graphs you would like to generate. An additional way to analyze this data is to use the paired t-test command.

Paired T-Test Paired T-Test and CI: Mat-A, Mat-B Paired T for Mat-A - Mat-B N Mean StDev SE Mean Mat-A Mat-B Difference 95% CI for Mean difference: ( , ) T-Test of Mean difference = 0 (vs not = 0): T-Value = P-Value = 0.009 The P-value from this Paired T-Test tells us the difference in materials is statistically significant. As you will see the conclusions are the same but just presented differently.

The wrong way to analyze this data is to use a 2-sample t-test:
Paired T-Test The wrong way to analyze this data is to use a 2-sample t-test: MINITABTM Session Window Two-sample T for Mat-A vs Mat-B N Mean StDev SE Mean Mat-A Mat-B Difference = mu (Mat-A) - mu (Mat-B) Estimate for difference: 95% CI for difference: ( , ) T-Test of difference = 0 (vs not =): T-Value = P-Value = DF = 18 Both use Pooled StDev = If you analyze this as a 2-sample t–test it simply compares the Means of Material A to Material B. The power of the paired test is that it increases the sensitivity of the test without having to look at a series of other factors.

Paired t-test Exercise
Exercise objective: Utilize what you have learned to conduct an analysis a paired t-test using MINITABTM. 1. A corrugated packaging company produces material that uses creases to make boxes easier to fold. It is a Critical to Quality characteristic to have a predictable Relative Crease Strength. The quality manager is having her lab test some samples labeled Then those same samples are being sent to her colleague at another facility who will report their measurements on those same 1-11 samples. 2. The US quality manager wants to know with 95% confidence what the average difference is between the lab located in Texas and the lab located in Mexico when measuring Relative Crease Strength. 3. Use the data in columns “Texas” & “Mexico” in “HypoTestStud.mtw” to determine the answer to the quality manager’s question. Exercise.

Paired t-test Exercise: Solution
Because the two labs agreed to exactly report measurement results for the same parts and the results were put in the correct corresponding row we are able to do a paired t-test. The first thing we must do is create a new column with the difference between the two test results. Calc > Calculator… Please read the slide.

We must confirm the differences (now in a new calculated column) are from a Normal Distribution. This was confirmed with the Anderson-Darling Normality Test by doing a graphical summary under Basic Statistics. Please read the slide.

As we have seen before this 1 Sample T analysis is found with: Stat>Basic Stat>1-sample T Please read the slide.

Even though the Mean difference is 0.23 we have a 95% confidence interval that includes zero so we know the 1-sample t-test’s null hypothesis was “failed to be rejected”. We cannot conclude the two labs have a difference in lab results. The P-value is greater than 0.05 so we do not have the 95% confidence we wanted to confirm a difference in the lab Means. This confidence interval could be reduced with more samples taken next time and analyzed by both labs. Please read the slide.

Normal Test of Equal Variance 1 Sample t-test 1 Sample Variance Variance Not Equal Variance Equal 2 Sample T One Way ANOVA Continuous Data Two samples Please read the slide.

At this point you should be able to:
Summary At this point you should be able to: Determine appropriate sample sizes for testing Means Conduct various Hypothesis Tests for Means Properly analyze results Please read the slide.

Analyze Phase Hypothesis Testing Normal Data Part 1

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