EGR252 F11 Ch 10 9th edition rev2 Slide 1

Statistical Hypothesis Testing Review A 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 : μ = 17H1 : μ ≠ 17

We sometimes refer to the null hypothesis as the “equals” hypothesis.

EGR252 F11 Ch 10 9th edition rev2 Slide 2

Potential errors in decision-making

α Probability of committing a

Type I error Probability of rejecting the

null hypothesis given that the null hypothesis is true

P (reject H0 | H0 is true)

β Probability of committing a

Type II error Power of the test = 1 - β

(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

EGR252 F11 Ch 10 9th edition rev2 Slide 3

Hypothesis Testing – Approach 1 Approach 1 - Fixed probability of Type 1 error.

1. State the null and alternative hypotheses.2. Choose a fixed significance level α.3. Specify the appropriate test statistic and establish

the critical region based on α. Draw a graphic representation.

4. Calculate the value of the test statistic based on the sample data.

5. Make a decision to reject H0 or fail to reject H0, based on the location of the test statistic.

6. Make an engineering or scientific conclusion.

EGR252 F11 Ch 10 9th edition rev2 Slide 4

Hypothesis Testing – Approach 2 Approach 2 - Significance testing based on the calculated P-

value

1. State the null and alternative hypotheses.2. Choose an appropriate test statistic.3. Calculate value of test statistic and determine P-

value. Draw a graphic representation.4. Make a decision to reject H0 or fail to reject H0,

based on the P-value. 5. Make an engineering or scientific conclusion.

P-value 0 1.000.25 0.50 0.75

p = 0.05 ↓

P-value

EGR252 F11 Ch 10 9th edition rev2 Slide 5

Example: Single Sample Test of the Mean P-value Approach

A 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. μ < 35.

Step 1: State the hypotheses.H0: μ = 35

H1: μ < 35

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

EGR252 F11 Ch 10 9th edition rev2 Slide 6

Single Sample Example (cont.)Approach 2:

= -1.11842

Find probability from chart or use Excel’s 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.

nSXT

/

EGR252 F11 Ch 10 9th edition rev2 Slide 7

Example (concl.)

Approach 1: Predetermined significance level (alpha)Step 1: Use same hypotheses.Step 2: Let’s 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 = 1.729

Since H1 specifies “< ” we declare tcrit = -1.729

Step 4: Using the equation, we calculate tcalc = -1.11842

Step 5: Decision Fail to reject H0

Step 6: Conclusion: The mean is not significantly less than 35 mpg.

EGR252 F11 Ch 10 9th edition rev2 Slide 8

Your turn … same data, different hypotheses

A 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. μ ≠ 35 at an α level of 0.05. Be sure to draw the picture.Step 1Step 2Step 3Step 4Step 5Step 6 (Conclusion will be different.)

EGR252 F11 Ch 10 9th edition rev2 Slide 9

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

EGR252 F11 Ch 10 9th edition rev2 Slide 10

Two-Sample Hypothesis Testing Define the difference in the two means as:

μ1 - μ2 = d0

where d0 is the actual value of the hypothesized difference

What are the Hypotheses?

H0: _______________

H1: _______________

orH1: _______________

orH1: _______________

EGR252 F11 Ch 10 9th edition rev2 Slide 11

Our Example Using ExcelReading: n1 = 4 mean x1 = 2.075 s1

2 = 1.363

No reading: n2 = 5 mean x2 = 2.860 s22 = 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.

t-Test: Two-Sample Assuming Equal Variances

  Read DoNotRead

Mean 2.075 2.860

Variance 1.3625 3.883

Observations 4 5

Pooled Variance 2.8027857

Hypothesized Mean Difference 0

df 7

t Stat -0.698986

P(T<=t) one-tail 0.2535567

t Critical one-tail 1.8945775

P(T<=t) two-tail 0.5071134

t Critical two-tail 2.3646226  

EGR252 F11 Ch 10 9th edition rev2 Slide 12

Your turn … Lower-tail test (μ1 - μ2 < 0)

“Fixed α” approach (“Approach 1”) at α = 0.05 level. “p-value” approach (“Approach 2”)

Upper-tail test (μ2 – μ1 > 0) “Fixed α” approach at α = 0.05 level. “p-value” approach

Two-tailed test (μ1 - μ2 ≠ 0) “Fixed α” approach at α = 0.05 level. “p-value” approach

Recall 21

021

/1/1)(

nnsdxxt

pcalc

EGR252 F11 Ch 10 9th edition rev2 Slide 13

Our Example – Hand CalculationReading:

n1 = 4 mean x1 = 2.075 s12 = 1.363

No reading:n2 = 5 mean x2 = 2.860 s2

2 = 3.883

To conduct the test by hand, we must calculate sp2 .

= 2.803 s = 1.674

and = ???? 2

)1()1(

21

222

2112

nn

snsnsp

21

021

/1/1)(

nnsdxxt

pcalc

EGR252 F11 Ch 10 9th edition rev2 Slide 14

Lower-tail test (μ1 - μ2 < 0) Why?

Draw the picture:Approach 1: df = 7, t0.5,7 = 1.895 tcrit = -1.895

Calculation: tcalc = ((2.075-2.860)-0)/(1.674*sqrt(1/4 + 1/5)) =

-0.70Graphic:

Decision:

Conclusion:

EGR252 F11 Ch 10 9th edition rev2 Slide 15

Upper-tail test (μ2 – μ1 > 0)Conclusions

The data do not support the hypothesis that the mean time to complete homework is less for students who read the textbook.

or There 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.

or The data do not support the hypothesis that the

mean completion time is less for readers than for non-readers.

EGR252 F11 Ch 10 9th edition rev2 Slide 16

Our Example Using ExcelReading: n1 = 4 mean x1 = 2.075 s1

2 = 1.363

No reading: n2 = 5 mean x2 = 2.860 s22 = 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.

t-Test: Two-Sample Assuming Equal Variances

  Read DoNotRead

Mean 2.075 2.860

Variance 1.3625 3.883

Observations 4 5

Pooled Variance 2.8027857

Hypothesized Mean Difference 0

df 7

t Stat -0.698986

P(T<=t) one-tail 0.2535567

t Critical one-tail 1.8945775

P(T<=t) two-tail 0.5071134

t Critical two-tail 2.3646226  

t-Test: Two-Sample Assuming Unequal Variances

  Read DoNotRead

Mean 2.075 2.86

Variance 1.3625 3.883

Observations 4 5

Hypothesized Mean Difference 0

df 7

t Stat -0.7426759

P(T<=t) one-tail 0.2409258

t Critical one-tail 1.8945775

P(T<=t) two-tail 0.4818516

t Critical two-tail 2.3646226  

EGR252 F11 Ch 10 9th edition rev2 Slide 17

Another Example: Low Carb Meals

Suppose 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 = 15 x1 = 27.2 s12 = 11

n2 = 10 x2 = 23.9 s22 = 23

tcalc = ______________________

ν = ______________ (using equation in table 10.3)

EGR252 F11 Ch 10 9th edition rev2 Slide 18

Example (cont.) What are our options for hypotheses?

H0: μ1 - μ2 = 0 or H0: μ1 - μ2 = 0H1: μ1 - μ2 > 0 H1: μ1 - μ2 ≠ 0

At an α level of 0.05,One-tailed test, t0.05, 15 = 1.753Two-tailed test, t0.025, 15 = 2.131

How are our conclusions affected? Our data don’t 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 ?)

EGR252 F11 Ch 10 9th edition rev2 Slide 19

Special Case: Paired Sample T-TestWhich designs are paired-sample?

A. Car Radial Belted 1 ** ** Radial, Belted tires 2 ** ** placed on each car. 3 ** ** 4 ** **

B. Person Pre Post 1 ** ** Pre- and post-test 2 ** ** administered to each 3 ** ** person. 4 ** **

C. Student Test1 Test2 1 ** ** 4 scores from test 1, 2 ** ** 4 scores from test 2. 3 ** ** 4 ** **

EGR252 F11 Ch 10 9th edition rev2 Slide 20

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.

EGR252 F11 Ch 10 9th edition rev2 Slide 21

Sheer Strength Example Data

GirderKarlsruhe Method

LehighMethod

Difference (d)

1 1.186 1.061 0.1252 1.151 0.992 0.1593 1.322 1.063 0.2594 1.339 1.062 0.2775 1.200 1.065 0.1356 1.402 1.178 0.2247 1.365 1.037 0.3288 1.537 1.086 0.4519 1.559 1.052 0.507

EGR252 F11 Ch 10 9th edition rev2 Slide 22

Sheer Strength Example CalculationsHypotheses:

H0: μD = 0H1: μD ≠ 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)

EGR252 F11 Ch 10 9th edition rev2 Slide 23

What does this mean?Draw the picture:

Decision:

Conclusion:

EGR252 F11 Ch 10 9th edition rev2 Slide 24

Goodness-of-Fit TestsProcedures for confirming or refuting

hypotheses about the distributions of random variables.

Hypotheses:

H0: The population follows a particular distribution.H1: The population does not follow the distribution.

Examples:H0: The data come from a normal distribution.H1: The data do not come from a normal distribution.

EGR252 F11 Ch 10 9th edition rev2 Slide 25

Goodness of Fit Tests: Basic MethodTest statistic is χ2

Draw the pictureDetermine the critical value

χ2 with parameters α, ν = k – 1Calculate χ2 from the sample

Compare χ2calc to χ2

crit Make a decision about H0

State your conclusion

k

i i

ii

EEOcalc

1

22 )(

EGR252 F11 Ch 10 9th edition rev2 Slide 26

Tests of Independence Example: 500 employees were surveyed with respect to

pension plan preferences. HypothesesH0: Worker Type and Pension Plan are independent.

H1: Worker Type and Pension Plan are not independent.

Develop a Contingency Table showing the observed values for the 500 people surveyed.

Worker TypePension Plan

Total#1 #2 #3

Salaried 160 140 40 340Hourly 40 60 60 160

Total 200 200 100 500

EGR252 F11 Ch 10 9th edition rev2 Slide 27

Calculation of Expected Values

2. Calculate expected probabilities

P(#1 ∩ S) = P(#1)*P(S) = (200/500)*(340/500)=0.272 E(#1 ∩ S) = 0.272 * 500 = 136

Worker TypePension Plan

Total#1 #2 #3

Salaried 160 140 40 340Hourly 40 60 60 160

Total 200 200 100 500

#1 #2 #3S (exp.) 136 ? 68

H (exp.) 64 ? 32

EGR252 F11 Ch 10 9th edition rev2 Slide 28

Calculate the Sample-based Statistic

Calculation of the sample-based statistic

= (160-136)^2/(136) + (140-136)^2/(136) + … (60-32)^2/(32)

= 49.63

k

i i

ii

EEOcalc

1

22 )(

EGR252 F11 Ch 10 9th edition rev2 Slide 29

The Chi-Squared Test of Independence

5. Compare to the critical statistic, χ2α, r

where r = (a – 1)(b – 1)

For our example, suppose α = 0.01

χ2 0.01,2 = ___________

χ2 calc = ___________

Decision:Conclusion:

EGR252 F11 Ch 10 9th edition rev2 Slide 30

The Chi-Squared Test in Minitab 15

Chi-Square Test: pp1, pp2, pp3 Expected counts are printed below observed countsChi-Square contributions are printed below expected counts pp1 pp2 pp3 Total 1 160 140 40 340 136.00 136.00 68.00 4.235 0.118 11.529

2 40 60 60 160 64.00 64.00 32.00 9.000 0.250 24.500

Total 200 200 100 500 Chi-Sq = 49.632, DF = 2, P-Value = 0.000