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### Transcript of EGR 252 S10 Ch.10 8th edition Slide 1 Statistical Hypothesis Testing Review  A statistical...

• EGR 252 S10 Ch.10 8th edition Slide *Statistical Hypothesis Testing ReviewA statistical hypothesis is an assertion concerning one or more populations.In statistics, a hypothesis test is conducted on a set of two mutually exclusive statements:H0 : null hypothesis H1 : alternate hypothesis ExampleH0 : = 17H1 : 17We sometimes refer to the null hypothesis as the equals hypothesis.

• EGR 252 S10 Ch.10 8th edition Slide *Potential errors in decision-making Probability of committing a Type I errorProbability of rejecting the null hypothesis given that the null hypothesis is trueP (reject H0 | H0 is true)Probability of committing a Type II errorPower of the test = 1 - (probability of rejecting the null hypothesis given that the alternate is true.)Power = P (reject H0 | H1 is true)

H0 True

H0 False

Do not reject H0

Correct

Decision

Type II

error

Reject H0

Type I

error

Correct

Decision

• EGR 252 S10 Ch.10 8th edition Slide *Hypothesis Testing Approach 1Approach 1 - Fixed probability of Type 1 error.State the null and alternative hypotheses.Choose a fixed significance level .Specify the appropriate test statistic and establish the critical region based on . Draw a graphic representation.Calculate the value of the test statistic based on the sample data.Make a decision to reject H0 or fail to reject H0, based on the location of the test statistic. Make an engineering or scientific conclusion.

• EGR 252 S10 Ch.10 8th edition Slide *Hypothesis Testing Approach 2Approach 2 - Significance testing based on the calculated P-valueState the null and alternative hypotheses.Choose an appropriate test statistic.Calculate value of test statistic and determine P-value. Draw a graphic representation.Make a decision to reject H0 or fail to reject H0, based on the P-value. Make an engineering or scientific conclusion. P-value 01.000.250.500.75p = 0.05 P-value

• EGR 252 S10 Ch.10 8th edition Slide *Example: Single Sample Test of the Mean P-value ApproachA sample of 20 cars driven under varying highway conditions achieved fuel efficiencies as follows:Sample mean x = 34.271 mpg Sample std dev s = 2.915 mpg

Test the hypothesis that the population mean equals 35.0 mpg vs. < 35.

Step 1: State the hypotheses.H0: = 35 H1: < 35

Step 2: Determine the appropriate test statistic. unknown, n = 20 Therefore, use t distribution

• EGR 252 S10 Ch.10 8th edition Slide *Single Sample Example (cont.)Approach 2: = -1.11842

Find probability from chart or use Excels tdist function.P(x -1.118) = TDIST (1.118, 19, 1) = 0.139665

p = 0.14

0______________1 Decision: Fail to reject null hypothesis

Conclusion: The mean is not significantly less than 35 mpg.

• EGR 252 S10 Ch.10 8th edition Slide *Example (concl.)Approach 1: Predetermined significance level (alpha)Step 1: Use same hypotheses.Step 2: Lets set alpha at 0.05.Step 3: Determine the critical value of t that separates the reject H0 region from the do not reject H0 region. t, n-1 = t0.05,19 = 1.729Since H1 specifies < we declare tcrit = -1.729Step 4: Using the equation, we calculate tcalc = -1.11842

Step 5: Decision Fail to reject H0Step 6: Conclusion: The mean is not significantly less than 35 mpg.

• EGR 252 S10 Ch.10 8th edition Slide *Your turn same data, different hypothesesA sample of 20 cars driven under varying highway conditions achieved fuel efficiencies as follows:Sample mean x = 34.271 mpg Sample std dev s = 2.915 mpgTest the hypothesis that the population mean equals 35.0 mpg vs. 35 at an level of 0.05. Be sure to draw the picture.Step 1Step 2Step 3Step 4Step 5Step 6 (Conclusion will be different.)

• EGR 252 S10 Ch.10 8th edition Slide *Two-Sample Hypothesis TestingA professor has designed an experiment to test the effect of reading the textbook before attempting to complete a homework assignment. Four students who read the textbook before attempting the homework recorded the following times (in hours) to complete the assignment:

3.1, 2.8, 0.5, 1.9 hours

Five students who did not read the textbook before attempting the homework recorded the following times to complete the assignment:

0.9, 1.4, 2.1, 5.3, 4.6 hours

• EGR 252 S10 Ch.10 8th edition Slide *Two-Sample Hypothesis TestingDefine the difference in the two means as:1 - 2 = d0 where d0 is the actual value of the hypothesized difference

What are the Hypotheses?

H0: _______________

H1: _______________ orH1: _______________ orH1: _______________

• EGR 252 S10 Ch.10 8th edition Slide *Our Example Using ExcelReading:n1 = 4mean x1 = 2.075s12 = 1.363No reading:n2 = 5mean x2 = 2.860s22 = 3.883

If we have reason to believe the population variances are equal, we can conduct a t- test assuming equal variances in Minitab or Excel.

• EGR 252 S10 Ch.10 8th edition Slide *Your turn Lower-tail test (1 - 2 < 0)Fixed approach (Approach 1) at = 0.05 level.p-value approach (Approach 2)Upper-tail test (2 1 > 0)Fixed approach at = 0.05 level.p-value approach Two-tailed test (1 - 2 0)Fixed approach at = 0.05 level.p-value approach

Recall

• EGR 252 S10 Ch.10 8th edition Slide *Our Example Hand CalculationReading:n1 = 4 mean x1 = 2.075s12 = 1.363No reading:n2 = 5 mean x2 = 2.860s22 = 3.883

To conduct the test by hand, we must calculate sp2 .= 2.803 s = 1.674

and = ????

• EGR 252 S10 Ch.10 8th edition Slide *Lower-tail test (1 - 2 < 0) Why?Draw the picture:Approach 1: df = 7, t0.5,7 = 1.895 tcrit = -1.895Calculation:tcalc = ((2.075-2.860)-0)/(1.674*sqrt(1/4 1/5)) = -0.70Graphic:

Decision:

Conclusion:

• EGR 252 S10 Ch.10 8th edition Slide *Upper-tail test (2 1 > 0)Conclusions

The data do not support the hypothesis that the mean time to complete homework is less for students who read the textbook.orThere is no statistically significant difference in the time required to complete the homework for the people who read the text ahead of time vs those who did not.orThe data do not support the hypothesis that the mean completion time is less for readers than for non-readers.

• EGR 252 S10 Ch.10 8th edition Slide *Our Example Using ExcelReading:n1 = 4mean x1 = 2.075s12 = 1.363No reading:n2 = 5mean x2 = 2.860s22 = 3.883

What if we do not have reason to believe the population variances are equal? We can conduct a t- test assuming unequal variances in Minitab or Excel.

• EGR 252 S10 Ch.10 8th edition Slide *Another Example: Low Carb MealsSuppose we want to test the difference in carbohydrate content between two low-carb meals. Random samples of the two meals are tested in the lab and the carbohydrate content per serving (in grams) is recorded, with the following results:

n1 = 15x1 = 27.2s12 = 11n2 = 10x2 = 23.9s22 = 23

tcalc = ______________________ = ______________ (using equation in table 10.2 Round up df)

• EGR 252 S10 Ch.10 8th edition Slide *Example (cont.)What are our options for hypotheses?H0: 1 - 2 = 0 orH0: 1 - 2 = 0H1: 1 - 2 > 0 H1: 1 - 2 0At an level of 0.05,One-tailed test, t0.05, 15 = 1.753Two-tailed test, t0.025, 15 = 2.131How are our conclusions affected?Our data dont support a conclusion that the carb content of the two meals are different at an alpha level of .05 (What is H1 ?)Our data do support a conclusion that meal 1 has more carbs than meal 2 at an alpha level of .05 (What is H1 ?)

• EGR 252 S10 Ch.10 8th edition Slide *Special Case: Paired Sample T-TestWhich designs are paired-sample?CarRadialBelted 1 ** **Radial, Belted tires 2 ** ** placed on each car. 3 ** ** 4 ** **Person Pre Post 1 ** **Pre- and post-test 2 ** **administered to each 3 ** **person. 4 ** **Student Test1 Test2 1 ** **5 scores from test 1, 2 ** **5 scores from test 2. 3 ** ** 4 ** **

• EGR 252 S10 Ch.10 8th edition Slide *Sheer Strength Example*An article in the Journal of Strain Analysis compares several methods for predicting the shear strength of steel plate girders. Data for two of these methods, when applied to nine specific girders, are shown in the table on the next slide. We would like to determine if there is any difference, on average, between the two methods. Procedure: We will conduct a paired-sample t-test at the 0.05 significance level to determine if there is a difference between the two methods.* adapted from Montgomery & Runger, Applied Statistics and Probability for Engineers.

• EGR 252 S10 Ch.10 8th edition Slide *Sheer Strength Example Data

GirderKarlsruhe MethodLehighMethodDifference (d)11.1861.0610.12521.1510.9920.15931.3221.0630.25941.3391.0620.27751.2001.0650.13561.4021.1780.22471.3651.0370.32881.5371.0860.45191.5591.0520.507

• EGR 252 S10 Ch.10 8th edition Slide *Sheer Strength Example CalculationsHypotheses:H0: D = 0H1: D 0 t0.025,8 = 2.306 Why 8?

Calculation of difference scores (d), mean and standard deviation, and tcalc

d = 0.2739sd = 0.1351

tcalc = ( d d0 ) = (0.2739 - 0) = 6.082sd / sqrt(n) 1.1351 / 3

• EGR 252 S10 Ch.10 8th edition