- One sample T-tests (Student's T-Test) are utilized to resolve hypothesis testing by comparing process means.
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• When to Use 1 Sample T-test:
Pre-project
- Verify the Mean of the process is significantly different from the Target validating the need for an improvement project.
Mid-project
- Verify the improvement in Mean from the preproject standard, throughout the course of making improvements.
1 Sample T-Test
Pre-requisite for 1 Sample T-Test
- Data must be continuous and normal
- Samples must be independentto each other
- Samples size less than 30
- It must not have any outliers
1 Sample T-Test method
Hypothesis of 1 Sample T-Test
1 Sample T-Test Example -1
- The average time to resolve a IT ticket is 8 hours in an organization. The automation team implemented Robotic solution to reduce the routine work. Leadership want to understand that at 95% confidence level is there a significance reduction in current ticket resolution time compared to earlier 8 hours. Analytics team collected last 30 resolved tickets.
- Let us conduct one sample T test to find out if the difference is significant enough
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Step 1.a: Conduct Normality test in Minitab
Note: You can also evaluate the normality test by selecting
- Minitab -> Stats -> Basic Statistics -> Normality Tests
(or)
- Minitab -> Graph -> Probability Plot -
Step 1.b:NormalityCheck and Interpret
Interpret:
- As P-value is greater than 0.05, we can conclude that the data are normal. Data also does not have any any outliers. -
Step 2: Hypothesis
- Null Hypothesis Ho:
- The average ticket resolving time is 8 hours i.e. μ = 8 hours
- Alternate Hypothesis Ha:
- The average ticket resolving time is less than 8 hours i.e. μ < 8 hours
Note: 8 hours was the original resolution time prior automation and μ is the current population mean of resolution time post automation -
Step 3: Conduct 1-Sample T-Test
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Step 4: Interpretation
Sample Mean = 7.233 hours
- At 95% confidence level, the upper bound for mean is 7.685 hours i.e. in the given scenario the maximum mean would be 7.685hours
- The upper bound mean (7.685 hours) is less than the target (8 hours) leads to p-value (0.004) less than alpha (0.05), so,we can rejectthe null hypothesis.
- Therefore, the new system has ticket resolving time less than 8 hours.
- P value 0.004 means that there is 0.4% chance that the populationmean will be equal to 8 hours
1 Sample T-Test Example -2
- A large real estate company wants to confirm whether the average cost per square foot for office rental space in the XYZ tech park is $25.50. The firm randomly conducts a telephone survey of 30 offices in that tech park and asks the office managers how much they pay in rent per square foot. The firm conducts a hypothesis test at 95% confidence level to determine whether the cost per square foot reported by the real estate company is valid
- Let us conduct one sample T test to find out if the difference is significant enough
-
Step 1.a: Conduct Normality test in Minitab
Note: You can also evaluate the normality test by selecting
- Minitab -> Stats -> Basic Statistics -> Normality Tests
(or)
- Minitab -> Graph -> Probability Plot -
Step 1.b:NormalityCheck and Interpret
Interpret:
- As P-value is greater than 0.05, we can conclude that the data are normal and doesn’thave any outliers. -
Step 2: Hypothesis
- Null HypothesisHo:
- The average cost per square foot for office rental space in the XYZ tech park is $25.50, i.e. μ = $25.50
- Alternate Hypothesis Ha:
- The average cost per square foot for office rental space in the XYZ tech park is not equal to $25.50, i.e. μ ≠ $25.50
Note: $25.50 hours was the original cost per square foot and μ is the current population mean of office rental space cost per square foot -
Step 3: Conduct 1-Sample T-Test
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Step 4: Interpretation
Sample Mean = $24.70
- At 95% confidence level, the confidence interval is $24.142-$25.258
- The target (μa = $25.50) does not fall in the confidence interval. The probability of $25.50 falling within the confidence interval is 0.7% and the same is indicated by p-value (0.007) which is less than alpha (0.05), so, we can reject the null hypothesis.
- Inference: The average cost per square foot for office rental space in the XYZ tech park is not equal to $25.50.
- P value 0.007 means that there is 0.7% chance that the population mean will be equal to $25.50