EGR 252 Spring 2004 TEST 2
Dr. Joan Burtner Fall 2020 Hypothesis Testing Examples Minitab 17
One-way ANOVA with Tukey Comparisons
Testing for Normality
Experimental Design
(One-way ANOVA)
One factor hypothesis test for three or more samples
Hypothesis test based on three or more levels of a single factor
Use Minitab 17 to conduct a One-way ANOVA and Tukey Analysis
Enter data in worksheet. (See page 4 of this handout for data formatting guidelines.)
Select Stat/ANOVA/One-way/ from the pull-down menu
Select response data format. For “Response data are in one column for all factor levels:
Enter Column for Response:
Enter Column for Factor:
Select Comparisons
Tukey
Interval Plot
Tests
Select Graphs
Four in one
Hypothesis about the distribution of the data (Normality Test)
Use Minitab 17 to conduct a hypothesis test to determine if a data set is normally distributed.
Method:
Enter data in worksheet.
Select Stat/Basic Statistics/Normality Test from the pull-down menu.
Choose any one (or all) of the three tests:
Anderson-Darling Ryan-Joiner Kolmogorov-Smirnov
It is often helpful to compare the outputs of the three tests.
Single Factor Hypothesis Testing Template with Definitions
Problem Statement:
Response: (What is being measured?) ___________________________
Factor and Levels (What are the groups or categories that are being compared?)
Hypotheses:
H0:
H1:
Justification of correct experimental design and test statistic (factors, levels, sample size, etc):
Computer Output (Include calculated test statistic, p-value and ANOVA Table if applicable)
Graphic: (Place an arrow at the approximate location of the p-value.)
0 0.05 0.10 0.15 1 (p-value)
Decision: ________________H0
Conclusion: Use complete sentences. (Refer to problem statement and managerial decision based on p-values)
______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Dr. Burtner Fall 2020 Single Factor Hypothesis Testing with Tukey Analysis Example for Minitab 17
Problem Statement:
A quality researcher is interested in comparing the assembly times (in seconds) of four workers. All four workers received the same training at an automobile manufacturer in Alabama. Each of the workers has been employed at the plant for at least two years. Does this data suggest that there is a significant difference in the assembly times of the four workers?
Steve
244
245
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245
244
Betty
240
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246
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241
242
241
Karen
256
243
249
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253
Jose
246
243
245
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247
Minitab 17 Method: Select Stat/ANOVA/One-way/ from the pull-down menu
Select response data format.
Here is how you should enter the data based on your selection.
Dr. Burtner Fall 2020 Single Factor Hypothesis Testing Template Example
Problem Statement:
A quality researcher is interested in comparing the assembly times (in seconds) of four workers. All four workers received the same training at an automobile manufacturer in Alabama. Each of the workers has been employed at the plant for at least two years. Does this data suggest that there is a significant difference in the assembly times of the four workers?
Response: (What is being measured?) assembly time (in seconds)
Factor and Levels (What are the groups or categories that are being compared?)
Factor: Worker Levels: Steve, Betty Karen Jose
Hypotheses:
H0: ( Steve = ( Betty = ( Karen = ( Jose
H1: At least one population mean assembly time is different.
Justification of correct experimental design and test statistic:
One factor, four levels, normally-distributed data: Use F statistic
Computer Output (Include calculated test statistic, p-value and ANOVA Table if applicable)
Minitab 17 Output:
One-way ANOVA: time versus worker
Method
Null hypothesis All mean assembly times are equal.
Alternative hypothesis At least one mean assembly time is different.
Significance level α = 0.05
Equal variances were assumed for the analysis.
Factor Information
Factor Levels Values
worker 4 Betty, Jose, Karen, Steve
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
worker 3 396.7 132.235 17.12 0.000
Error 25 193.1 7.724
Total 28 589.8
Model Summary
S R-sq R-sq(adj) R-sq(pred)
2.77913 67.26% 63.33% 56.96%
Means
worker N Mean StDev 95% CI
Betty 8 241.750 1.832 (239.726, 243.774)
Jose 7 244.429 1.813 (242.265, 246.592)
Karen 8 251.380 4.57 ( 249.35, 253.40)
Steve 6 245.000 0.894 (242.663, 247.337)
Pooled StDev = 2.77913
Tukey Pairwise Comparisons
Grouping Information Using the Tukey Method and 95% Confidence
worker N Mean Grouping
Karen 8 251.380 A
Steve 6 245.000 B
Jose 7 244.429 B
Betty 8 241.750 B
Means that do not share a letter are significantly different.
Graphic:
0 0.05 0.10 0.15 1 p-value
Decision: Reject H0
Conclusion: Based on a p-value of 0.000, the data suggest that there is a statistically significant difference in the mean assembly time for at least one worker.
Based on the Tukey 95% Simultaneous Confidence Intervals, we conclude that the mean assembly time for Karen is significantly different from the mean assembly time of the other three workers. Karen’s mean assembly time is significantly longer.
****
You can also use Minitab to generate a graphic representation of the Tukey intervals.
[pic]
Based on this graphic, Karen’s time is significantly different from the other three workers.
*******************
Consider this Tukey graphic for a different data set.
[pic]
Based on this graphic, Kate’s time is significantly different from Belle’s time and Steven’s time; however Kate’s time is not significantly different from Juan’s time.
Testing for Normality Using Minitab 17
Normality tests are goodness-of-fit hypothesis tests.
H0: The data are normally distributed.
H1: The data are not normally distributed.
or
H0: The data follow a normal distribution.
H1: The data do not follow a normal distribution.
Consider the following data set. Are the data normally distributed?
Sodium_content
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Method: Select Stat/Basic Statistics/Normality Test
Choose Anderson-Darling
or
Choose Ryan-Joiner
or
Choose Kolmogorov-Smirnov
It is often best to do all three tests and compare results.
Minitab shows the results of the normality test in the form of a graph as well as relevant statistics.
[pic]
KS 0.116 P-Value > 0.150
Decision: Fail to reject the null hypothesis.
Conclusion: The sodium content data follow a normal distribution.
[pic]
AD 0.293 P-Value 0.574
Decision: Fail to reject the null hypothesis.
Conclusion: The sodium content data are normally distributed.
-----------------------
Partial Minitab input for “Response data are in one column…” (All responses are in one column. All factor levels are in a different column.)
times people
244 Steve
245 Steve.6ABDRi~€?‚”•¤¥íÞ̺¨ºÌí–„r„¨`UJ?J4Jh±*¶5?>*[pic]CJaJhuz
5?>*[pic]CJaJh&JÌ5?>*[pic]CJaJh?C5?>*[pic]CJaJ#hs¼h&JÌ5?>*[pic]B*CJaJphS?5#hs¼h±*¶5?>*[pic]B*CJaJphS?5#hs¼hÀ˜5?>*[pic]B*CJaJphS?5#hs¼huz
5?>*[pic]B*CJaJphS?5#
246 Steve
246 Steve
245 Steve
244 Steve
240 Betty
241 Betty
246 Betty
242 Betty
241 Betty
… …
… …
Partial Minitab input for “Response data are in a separate column…”
Steve Betty Karen Jose
244 240 256 246
245 241 243 243
246 246 249 245
… … … …
Note that the Karen-Betty interval does not contain 0.
[pic] Note that the Steven-Belle interval contains 0.
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