Statistical Hypothesis Testing - Part 3
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EGR 252 S10 JMB Ch.10 Part 3 Slide 1 Statistical Hypothesis Testing - Part 3 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: H 0 : null hypothesis H 1 : alternate hypothesis New test statistic of interest: n i i i i E E O 1 2 2 ) (
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Statistical Hypothesis Testing - Part 3. 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: H 0 : null hypothesis H 1 : alternate hypothesis - PowerPoint PPT Presentation
Transcript of Statistical Hypothesis Testing - Part 3
EGR 252 S10 JMB Ch.10 Part 3 Slide 1
Statistical Hypothesis Testing - Part 3
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 New test statistic of interest:
n
i i
ii
E
EO
1
22 )(
EGR 252 S10 JMB Ch.10 Part 3 Slide 2
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.
Example:
H0: The data come from a normal distribution.
H1: The data do not come from a normal distribution.
EGR 252 S10 JMB Ch.10 Part 3 Slide 3
Goodness of Fit Tests (cont.)Test statistic is χ2
Draw the picture Determine the critical value for goodness of fit test
χ2 with parameters α, ν = k – 1
Calculate χ2 from the sample
Compare χ2calc to χ2
crit
Make a decision about H0
State your conclusion.Discussion: Look at Table 10.4 in text.
n
i i
ii
E
EOcalc
1
22 )(
EGR 252 S10 JMB Ch.10 Part 3 Slide 4
Tests of Independence (without computer)
Example: Worker type and Choice of pension plan Hypotheses
H0: Pension Plan Choice and Worker Type are independent
H1: Pension Plan Choice and Worker Type are not independent
1. Develop a Contingency Table of Observed Values
Worker Type
Pension Plan
Total#1 #2 #3
Salaried 160 140 40 340
Hourly 40 60 60 160
Total 200 200 100 500
EGR 252 S10 JMB Ch.10 Part 3 Slide 5
Worker vs. Pension Plan Example
2. Calculate expected probabilities. Multiply by total observations to determine expected values for each cell.
P(#1 ∩ S) = P(#1)*P(S) = (200/500)*(340/500)=0.272 0.272*500 = 136
P(#1 ∩ H) = P(#1)*P(H) = (200/500)*(160/500)=0.128 0.128*500 = 64
Worker Type
Pension Plan
Total#1 #2 #3
Salaried 160 140 40 340
Hourly 40 60 60 160
Total 200 200 100 500
#1 #2 #3
S (exp.) 136 136
H (exp.) 64 64
EGR 252 S10 JMB Ch.10 Part 3 Slide 6
Hypotheses
Recall the general format of the hypotheses
H0: the categories (worker & plan) are independent
H1: the categories are not independent
3. Calculate the sample-based statistic
(160-136)^2/136 + (140-136)^2/136 + (40-68)^2/68 + (40-64)^2/64 + (60-64)^2/64 + (60-32)^2/32
= 49.63
n
i i
ii
E
EO
1
22 )(
EGR 252 S10 JMB Ch.10 Part 3 Slide 7
The Chi-Squared Test of Independence4. Compare to the critical statistic for a test of independence, χ2
α, r
where r = (a – 1)(b – 1) a = # of columnsb = # of rows
For our example, let’s use α = 0.01 _
χ20.01,2 = 9.210 (from Table A.5, pp 756)
Comparison: χ2 calc > χ2 crit
Decision: Reject the null hypothesis
Conclusion: Worker and plan are not independent.
****Note that we are stating that there is an association between the two categories. We are not claiming a cause and effect relationship.
Chi-Square Test Using Minitab
plan1 plan2 plan3
salaried 160 140 40
hourly 40 60 60
EGR 252 S10 JMB Ch.10 Part 3 Slide 8
Minitab Output:Chi-Square Test: plan1, plan2, plan3 Expected counts are printed below observed countsChi-Square contributions are printed below expected counts
plan1 plan2 plan3 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
Minitab Input:
