Statistical Hypothesis Testing Review

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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
• Example
• H0 µ 17
• H1 µ ? 17
• We sometimes refer to the null hypothesis as the
  equals hypothesis.

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Potential errors in decision-making

• a
• 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)

• ß
• Probability of committing a Type II error
• Power of the test 1 - ß
• (probability of rejecting the null hypothesis
  given that the alternate is true.)
• Power P (reject H0 H1 is true)

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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 a.
3. Specify the appropriate test statistic and
   establish the critical region based on a. 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.

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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 0.05 ?
P-value 0
1.00
P-value
0.75
0.25
0.50
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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 lt 35.
• Step 1 State the hypotheses.
• H0 µ 35
• H1 µ lt 35
• Step 2 Determine the appropriate test statistic.
  
• s unknown, n 20 Therefore, use t
  distribution

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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.

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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 1.729
• Since H1 specifies lt 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.

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Your turn same data, different hypotheses

• A sample of 20 cars driven under varying highway
  conditions achieved fuel efficiencies as follows
• Sample mean 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 a level of 0.05.
  Be sure to draw the picture.
• Step 1
• Step 2
• Step 3
• Step 4
• Step 5
• Step 6 (Conclusion will be different.)

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Two-Sample Hypothesis Testing

• A 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

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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 _______________
• or
• H1 _______________
• or
• H1 _______________

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Our Example Using Excel

• Reading n1 4 mean x1 2.075 s12 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 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(Tltt) one-tail 0.2535567
t Critical one-tail 1.8945775
P(Tltt) two-tail 0.5071134
t Critical two-tail 2.3646226  
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Your turn

• Lower-tail test (µ1 - µ2 lt 0)
• Fixed a approach (Approach 1) at a 0.05
  level.
• p-value approach (Approach 2)
• Upper-tail test (µ2 µ1 gt 0)
• Fixed a approach at a 0.05 level.
• p-value approach
• Two-tailed test (µ1 - µ2 ? 0)
• Fixed a approach at a 0.05 level.
• p-value approach
• Recall ?

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Our Example Hand Calculation

• Reading
• n1 4 mean x1 2.075 s12 1.363
• No reading
• n2 5 mean x2 2.860 s22 3.883
• To conduct the test by hand, we must calculate
  sp2 .
• 2.803 sp 1.674
• and ????

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Lower-tail test (µ1 - µ2 lt 0) Why?

• Draw the picture
• Approach 1 df 7, t0.05,7 1.895 ? tcrit
  -1.895
• Calculation
• tcalc ((2.075-2.860)-0)/(1.674sqrt(1/4 1/5))
  
• -0.70
• Graphic
• Decision
• Conclusion

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Upper-tail test (µ2 µ1 gt 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.

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Our Example Using Excel

• Reading n1 4 mean x1 2.075 s12 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 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(Tltt) one-tail 0.2535567
t Critical one-tail 1.8945775
P(Tltt) two-tail 0.5071134
t Critical two-tail 2.3646226  
t-Test Two-Sample Assuming Unequal Variances 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(Tltt) one-tail 0.2409258
t Critical one-tail 1.8945775
P(Tltt) two-tail 0.4818516
t Critical two-tail 2.3646226  
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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 ______________________
• ? ______________
• (using equation in table 10.3)

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Example (cont.)

• What are our options for hypotheses?
• H0 µ1 - µ2 0 or H0 µ1 - µ2 0
• H1 µ1 - µ2 gt 0 H1 µ1 - µ2 ? 0
• At an a level of 0.05,
• One-tailed test, t0.05, 15 1.753
• Two-tailed test, t0.025, 15 2.131
• How are our conclusions affected?
• Our data dont support a conclusion that the mean
  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 average carbs than Meal 2 at an alpha level
  of .05. (What is H1 ?)

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Special Case Paired Sample T-Test

• Which designs are paired-sample?
• Car Radial Belted
• 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 4 scores from test 1,
• 2 4 scores from test 2.
• 3
• 4

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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.

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Sheer Strength Example Data
Girder Karlsruhe Method Lehigh Method Difference (d)
1 1.186 1.061 0.125
2 1.151 0.992 0.159
3 1.322 1.063 0.259
4 1.339 1.062 0.277
5 1.200 1.065 0.135
6 1.402 1.178 0.224
7 1.365 1.037 0.328
8 1.537 1.086 0.451
9 1.559 1.052 0.507
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Sheer Strength Example Calculations

• Hypotheses
• H0 µD 0
• H1 µD ? 0 t0.025,8 2.306 Why 8?
• Calculation of difference scores (d), mean and
  standard deviation, and tcalc
• d 0.2739
• sd 0.1351
• tcalc ( d d0 ) (0.2739 - 0)
  6.082
• sd / sqrt(n) (1.1351 / 3)

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What does this mean?

• Draw the graphic
• Decision
• Conclusion

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Goodness-of-Fit Tests

• Procedures 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.

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Goodness of Fit Tests Basic Method

• Test statistic is ?2
• Draw the picture
• Determine the critical value
• ?2 with parameters a, ? k 1
• Calculate ?2 from the sample
• Compare ?2calc to ?2crit
• Make a decision about H0
• State your conclusion

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Tests of Independence

• Example 500 employees were surveyed with respect
  to pension plan preferences.
• Hypotheses
• H0 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 Type Pension Plan Pension Plan Pension Plan Total
Worker Type 1 2 3 Total
Salaried 160 140 40 340
Hourly 40 60 60 160
Total 200 200 100 500
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Calculation of Expected Values
Worker Type Pension Plan Pension Plan Pension Plan Total
Worker Type 1 2 3 Total
Salaried 160 140 40 340
Hourly 40 60 60 160
Total 200 200 100 500

• 2. Calculate expected probabilities
• P(1 n S) P(1)P(S) (200/500)(340/500)0.272
  
• E(1 n S) 0.272 500 136

1 2 3
S (exp.) 136 ? 68
H (exp.) 64 ? 32
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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

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The Chi-Squared Test of Independence

• 5. Compare to the critical statistic, ?2a, v
• where v (r 1)(c 1) Note v is the
  symbol for degrees of freedom
• For our example, suppose a 0.01
• ?2 0.01,2 ___________
• ?2 calc ___________
• Decision
• Conclusion

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The Chi-Squared Test in Minitab 15

• Chi-Square Test pp1, pp2, pp3
• Expected counts are printed below observed counts
• Chi-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
• Test statistic Chi-Sq calc 49.632, DF 2,
  P-Value 0.000 Reject Ho.
• Conclude that worker and plan are not
  independent.