Lecture 8 Hypothesis Testing


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1 Lecture 8 Hypothesis Testing Fall 2013 Prof. Yao Xie, H. Milton Stewart School of Industrial Systems & Engineering Georgia Tech
2 Midterm 1 Score 46 students Highest score: 98 Lowest score: 34 Mean: 70.2 Median: 72
3 Outline General concept Type of errors and power z test for the mean, variance known t test for the mean, variance unknown test for sample proporton Confidence interval and hypothesis test P value
4 Hypotheses Testing The hypothesis test has two possible but contradictory truths, written in terms of Hypotheses: H 0 Null Hypothesis The prior belief with the data. H 1 Alternative Hypothesis The alternative belief we have with the data Varies from problem to problem Goal of hypothesis testing is to use data to tell which hypothesis is true.
5 Mathematical formulation For example, comparing the mean and variance of samples from normal distributon Null Hypothesis AlternaTve Hypothesis " H 0 : µ = µ 0 # $ H 1 : µ µ 0 " H 0 :σ 2 2 $ = σ 0 # H 1 :σ 2 2 %$ σ 0 population N( µ =?, σ 2 =?) sample X, X 2,, 1! X n Conclusion about null hypothesis Null Hypothesis AlternaTve Hypothesis
6 Example Example: measure the diameter of a batch of screws
7 Examples of Hypotheses Test (1) Is the coin fair? Prior assumption: coin is fair. Let p = P(heads), H 0 : p = 1 2 H 1 : p 1 2 (our prior belief) (the opposite of our prior belief) (2) Apple produces a batch of iphones, with battery life (X) with mean µ, variance = σ 2 Is the variability of battery under control? Here the variability is under control when it is smaller than a known value H 0 : σ 2 2 σ 0 H 1 : σ 2 2 > σ 0 (3) Does this batch of iphones have sufficient battery life? H 0 : µ > µ 0 H 1 : µ µ 0 σ 0 2 7
8 Class Activity (1) Is the coin fair? P=P(heads) A. H 0 : P = ½ B. H A : P = ½ C. H 0 : P ½ (2) A machine produces product (X) with mean µ, variance σ 2 Is the variability under control? A. H A : σ 2 σ 0 2 B. H 0 : σ 2 > σ 0 2 C. H 0 : σ 2 σ 0 2 Do we support the hypothesis that the machine produce an item of a size larger than a known µ 0? A. H A : µ µ 0 B. H 0 : µ > µ 0 C. H 0 : µ µ 0 8
9 Twosided versus onesided Two sided AlternaTve Hypothesis X: innerdiameter of screws null hypothesis alternatve hypothesis One sided AlternaTve Hypotheses H H 0 1 : µ = 10 : µ < 10 X: customers waiting time in a bank H H 0 1 : µ = 100 : µ > 100 X: time to failure of machines
10 Simple versus composite Simple Hypothesis Testing two possible values of the parameter H 0 : µ = 12 H 1 : µ = 24 null hypothesis alternatve hypothesis Composite Hypotheses : µ = 10 Testing a range 0 of values H : µ < 10 Remaining quantity in a vending machine, 1 dozen or 2 dozens H 1 X: customers waiting time in a bank Average diameter of screw
11 Outline General concept Type of errors and power z test for the mean, variance known t test for the mean, variance unknown test for sample proporton Confidence interval and hypothesis test P value
12 Errors in Hypothesis Testing α = P(type I error) = P(Reject H 0, when H 0 is true) This error rate can be set low enough to ensure the test is safe. β = P(type II error) = P(Accept H 0, when H 1 is true) truth H 0 is True H 0 is False decision Accept H 0 Reject H 0 Correct Decision Type I Error Type II Error Correct Decision 12
13 Court room Suppose you are the prosecutor in a courtroom trial. The defendant is either guilty or not. The jury will either convict or not. Not Guilty truth Guilty decision Freeofguilt Right decision Wrong decision Convict Wrong decision Right decision
14 Significant level The classic hypothesis test fixes α (type I error) to be some small (tolerable) value accept the corresponding β (type II error) that results from this. The level fixed for α is called significance level. Typical value for α: 0.1, 0.05, 0.001
15 Statistical power Power = 1 β = P(reject H 0 when H 1 is true) Example: power functon for the test H 0 : µ = 0 H 1 : µ 0 At µ = 0, H 0 is true, so this point is equal to α, the type I error μ Yaxis shows power, Xaxis is the value of µ. 15
16 Class Activity 2 In 1990, a study on the weight of students at GT provided an average weight of µ = 160 lbs. We would like to test our belief that the GT student weight average did not decrease in the past 15 years. 1. What is the alternative hypothesis? A. H 1 : µ = 160 B. H 1 : µ > 160 C. H 1 : µ < Test H 0 : µ = 160 vs. H 1 : µ < 160. What is P(Reject H 0 µ = 160)? A. Type I error B. Type II error C. Power 3. Test H 0 : µ = 160 vs. H 1 : µ < 160. What is P(Reject H 0 µ < 160)? A. Type I error B. Type II error C. Power 16
17 Outline General concept Type of errors and power z test for the mean, variance known t test for the mean, variance unknown test for sample proporton P value
18 Onesided ztest For the GT student average example, we want to test H 0 : µ =160 H 1 : µ <160 The conventional way to test this hypothesis is to find the test for which the typei error is fixed at a particular value (e.g., α = 0.01, 0.05, 0.10). Sample mean x x as the test statistic. is a good estimator for µ, so here we use We should reject the null hypothesis when x is small 18
19 Determine threshold from significance level The test is to reject H 0 if x < b. To determine the threshold b, α = P(X<b µ =160) = P(Z< b160 σ n ) This determines b because b160 σ n Φ 1 α ( ) b = ( σ n )Φ 1 α ( )
20 Numerical values b = ( σ n )Φ 1 α If α = 0.05, σ = 10, n = 100, Φ 1 ( 0.05) = ( ) +160 We reject H 0 when 20 x< ( σ n )z α +160 = 10 / = Note that even when x<160 we stll accept H 0 => we take into account data variability.
21 Power Suppose μ = 150 (so H 1 is true) Power when μ = 150 P{ x < } = P x / 100 < / 100 = Φ( 8.355) 1 EnTre Power FuncTon = Φ( μ)
22 Twosided ztest For the GT student average example, suppose now we want to test that the average weight has not changed in the past 15 years Reject H 0 when k H 0 : µ = 160 H 1 : µ 160 x [160 k,160 + k] k α = P( x160 > k µ = 160) = P( Z > σ n ) σ n = z α /2 α = 0.05, z = 1.96 σ = 10, n = 100, k = z α /2 σ n = 1.96 Reject H 0 when x [158.04,161.96]
23 Outline General concept Type of errors and power z test for the mean, variance known t test for the mean, variance unknown test for sample proporton Confidence interval and hypothesis test P value
24 ttest, twosided For the GT student average example, we want to test H 0 : µ = 160 H 1 : µ 160 We do not know the true variance, but just the sample variance s. Note that before we test using standardized test statistics Known variance Unknown variance ztest ttest x 160 σ / n > k x 160 s / n > k 24
25 Determine threshold for ttest The test is to reject H 0 if To determine the threshold b, x 160 s / n > k α = P( x 160 s / n > k µ = 160) = P( T > k) k = t α /2,n 1 Φ(x) CDF of t random variable We reject H 0 when x 160 s / n > t α /2,n 1 In comparison, when variance is known, we reject H 0 when x 160 σ / n > z α /2
26
27 Example: coke machine twoside ttest A softdrink machine at a steak house is regulated so that the amount of drink dispensed is approximately normally distributed with a mean of 200 ml. The machine is checked periodically by taking a sample of 9 drinks and computing the average content. If for one batch of samples variance s = 30 ml, and mean is x = 215 ml, do we believe the machine is operating running OK, for a significant level α = 0.05? 27
28 Formulation: coke machine H 0 : µ = 200 H 1 : µ 200 x 200 Test statstcs s/ n Threshold k = t 0.025,8 = x 200 Reject H0 when > s/3 x = = 1 < With our data, 30 / 3 s/ n So machine is running OK.
29 Outline General concept Type of errors and power z test for the mean, variance known t test for the mean, variance unknown test for sample proporton Confidence interval and hypothesis test P value
30 Test on population proportion Sample proportion, X ~ BIN(n, p) Based on CLT approximation, n > 30 H 0 : p = p 0 H 1 : p p 0 Under H 0, X approximately normal random variable ( ( ) ) X ~ Ν np 0, np 0 1 p 0 standardized test sta+s+cs X np 0 ( ) np ( 0 1 p ) ~ Ν 0, 1 0 Test with significance level: α Reject H 0 when ˆp p 0 p ( 0 1 p ) 0 / n > z, α /2 ˆp = X / n
31 Example: obesity The Associated Press (October 9, 2002) reported that 1276 individuals in a sample of 4115 adults were found to be obese. A 1998 survey based on people's own assessment revealed that 20% of adult Americans considered themselves obese. Does the recent data suggest that the true proportion of adults who are obese is consistent with the selfassessment survey, with significance level 0.05?
32 Example continued H 0 : p = 0.2 H 1 : p 0.2 Data from 2002 study: 1276 of 4115 were reported with obesity ˆp = 1276 / 4115 = 0.31 α = 0.05 z = 1.96 ˆp p 0 ( ) / n = p 0 1 p 0 ( ) / 4115 = 17.74>1.96 Reject the null hypothesis: the proporton is not 0.2 (very likely to be greater than 0.2)
33 Summary of Tests
34 Summary: test for mean Null Hypothesis H 0 : µ = µ 0 Test StaTsTc x Significance level: α Alterna+ve Known Variance Unknown Variance Hypothesis H0 is rejected if H0 is rejected if H 1 : µ µ 0 x µ 0 > z α 2 σ / n x µ 0 > t α 2,n 1 s / n H 1 : µ > µ 0 x > µ 0 + z α σ / n x > µ 0 + t α,n 1 s / n H 1 : µ < µ 0 x < µ 0 z α σ / n x < µ 0 t α,n 1 s / n
35 Summary: test for sample proportion Null Hypothesis H 0 : p = p 0 Significance level: α Test StaTsTc ˆp p 0 p ( 0 1 p ) 0 / n Alterna+ve Hypothesis H0 is rejected if H 1 : p p 0 ˆp p 0 ( ) / n > z α /2 p 0 1 p 0
36 Outline General concept Type of errors and power z test for the mean, variance known t test for the mean, variance unknown test for sample proporton p value Confidence interval and hypothesis test
37 Motivation for pvalue Think about the GT student average weight example. Assume average weight is 160lb. Based on the test, we reject H 0 if sample average is less than , and this gives significance level 0.05 (i.e. chance of making type I error = 0.05) Consider case1: sample average = 150lb case2: sample average = 155lb In both cases we reject H 0 However, if sample sample average = 150lb, we should reject H 0 with more confidence than if sample average = 155lb. 37 In other words, when sample average = 150 lb, the chance for H 0 to happen is smaller.
38 pvalue p value = probability of observing some data even more extreme than the given data It is a measure of the null hypothesis plausibility based on the samples The smaller the p value is, the less likely H 0 is true Reject H 0 when p value is small Test can also be performed as rejectng H 0 when p value is less than prescribed significant level
39 Convention: decisions based on Pvalue If Pvalue < 0.01 H 0 is not plausible/ H 1 is supported If Pvalue > 0.1 H 0 is plausible If 0.01 < Pvalue < 0.1 we have some evidence that H 0 is not plausible but we need further investigation
40 1. pvalue for onesided Test H 0 :µ = µ 0 H 1 :µ > µ 0 Test statistic: Z = Xµ 0 σ / n observed value of test statistic: x pvalue: P(X> x)=p(z> xµ 0 σ / n ) = 1 Φ xµ 0 σ / n
41 Example: pvalue for onesided Test Claim: mean baoery life of a cellphone is 9 hours. Observed mean baoery life: 8.5 hours, from 44 observatons, σ = 1 hour Test: the mean baoery life of a cellphone exceeds 10 hours H 0 :µ = 9 H 1 :µ < 9 Test statistic: Z = Xµ 0 σ / n observed value of test statistic: x=8.5 pvalue: P(X< x)=p(z< xµ 0 σ / n ) = Φ =
42 Example: pvalue nicotine content A cigarette manufacturer claims that the average nicotine content of a brand of cigarettes is at most What is the alternative hypothesis? A. H A : m 1.5 B. H A : m < 1.5 C. H A : m > What is the pvalue? Denote A. P(Z z) B. P(Z z) C. P( Z z ) Z = n( X µ 0) / S and z = 100( ) /1.3
43 Calculation for nicotine example We observe the nicotne content for 100 cigareoes with a sample mean 1.7 and sample standard error 1.3. Observing data more extreme: P(X > 100( ) /1.3) = P( n(x µ 0 ) / S> 100( ) /1.3) = P(t n 1 > ) = Not a very small p value
44 2. pvalue for twosided Test H 0 :µ = µ 0 H 1 :µ µ 0 Test statistic: Z = Xµ 0 σ / n observed value of test statistic: x pvalue: P( Xµ 0 σ / n > xµ 0 σ / n ) = P( Z > xµ 0 σ / n ) = 1 2Φ xµ 0 σ / n
45 Example: compute pvalue n = 13, x = 2.879,σ = H 0 : µ = 3 H 1 : µ 3 pvalue = P(getting x even further away from 3 than x = 2.879) = P( X 3 > ) = P( Z > = P( Z > 1.34) ) = 2Φ(1.34) =
46 Example: pvalue for twosided test Test to see if the mean is significantly far from 90. The sample mean is 87.9 with known standard deviation of 5.9 for a sample of size What is the alternatve hypothesis? A. H A : m 90 B. H A : m < 90 C. H A : m > At a significance level α = 0.1, what is your decision? A. strongly suggest H A B. H 0 is plausible C. strongly support H 0 3. What is the p value? Denote Z = n( X µ 0) / S and z = 44( ) /5.9 A. P(Z z) B. P(Z z) C. P( Z z )
47 Comparison of significance level and p values Two ways of reportng results of a hypothesis test 1: report the results of a hypothesis test is to state that the null hypothesis was or was not rejected at a specified level of significance. This is called fixed significance level testng. 2: The p value is the probability that the test statstc will take on a value that is at least as extreme as the observed value of the statstc when the null hypothesis H 0 is true.
48 Relate pvalue and significance level: example H 0 : µ = 90 H 1 : µ 90 reject if z > z α 2 where z = x n Note: if we have an α = 0.10 level test, then Z.05 = and we would reject H 0 when Z > That means the actual pvalue <
49 Outline General concept Type of errors and power z test for the mean, variance known t test for the mean, variance unknown test for sample proporton P value Confidence interval and hypothesis test
50 Connection between confidence Intervals and Hypothesis Test CI : X  k X X + k We believe the true parameter µ 0 [X k, X + k] with probability 1 α, X [µ 0 k, µ 0 + k] Test : If we wish to test H 0 : µ = µ o H 1 : µ µ o [µ 0 k, µ 0 + k] If X not in, reject H o Under H 0, making error with probability 1 α
51 More Examples
52 Two types of approaches to hypothesis testing Fixed significance level Decide decision statstcs Reject H0, when Decision statstc >< threshold p value Calculate p value: what is the probability to observe a statstc more extreme than data The smaller p value is, the more likely H1 is true à reject H0
53 Procedure of hypothesis test (sec ) 1. Set the significance level (.01,.05,.1) 2. Set null and alternatve hypothesis 3. Determine other parameters 4. Decide type of the test  test for mean with known variance (z test)  test for mean with unknown variance (t test)  test for sample proporton parameter 6. Use data available:  perform test to reach a decision  and report p value
54 Example 1: change of average height? ztest The average height of females in the freshman class at GT has been cm with a a standard deviation of 6.9 cm. Is there reason to believe that there has been a change in the average height if a random sample of 50 females in the present freshman class has an average height of 165.2cm? Assume the standard deviation remains the same, and significance level
55 Solution to example 1 1. Significance level: 2. Set null and the alternatve H 0 : µ = H 1 : µ Determine other parameters: 4. Decide type of test: known variance, test for mean, use z test Reject H 0 if Plug in values: α = 0.05 Reject H 0 if 5. Use data to perform test: n = 50,σ = 6.9 x µ 0 > z α 2 σ / n z = 1.96 x > / 50 = x = 165.2, x = 2.7 > Reject H 0
56 6. Report p value: pvalue = P( Z > /(6.9 / 50)) = P( Z > ) = 2(1 Φ(2.7669)) =
57 Example 2: effective hours of drug ttest In a study for the effective hours of certain drug, the sample average time for n = 9 individual was 6.32 hours and the sample standard deviation was 1.65 hours. It has previously been assumed that the average adaptation time was at least 7 hours. Assuming effective hours to be normally distributed, does the data contradict the prior belief? 57
58 Solution to example 2 1. Significance level: 2. Set null and the alternatve H 0 : µ = 7 H 1 : µ 7 3. Determine other parameters: 4. Decide type of test: unknown variance, test for mean, use t test Reject H 0 if Plug in values: Reject H 0 if 5. Use data to perform test: 6. Report p value: α = 0.05 n = 9,s = 1.65 x µ 0 > t α 2,n 1 s / n t 0.025,8 = x 7 > / 9 = x = 6.32, x 7 = 0.68 < Accept H 0
59 6. Report p value: pvalue = P( T 8 > /(1.65 / 9)) = P( T 8 > ) = =
60 Example 3: sample proportion In a survey from 2000, it has been found that 33% of the adults favored paying traffic ticket rather than attending traffic school. Now we did a online survey and found out from a sample of 85 adults, 26 favor paying traffic ticket. Do we have reason to believe that the proportion of adults favoring paying traffic tickets has decreased today with a significant level 0.05? 60
61 Summary: test for sample proportion Null Hypothesis H 0 : p = p 0 Significance level: α Test StaTsTc ˆp p 0 p ( 0 1 p ) 0 / n Alterna+ve Hypothesis H0 is rejected if H 1 : p p 0 ˆp p 0 ( ) / n > z α /2 p 0 1 p 0
62 Solution to example 3 1. Significance level: 2. Set null and the alternatve H 0 : p = 0.33 H 1 : p Determine other parameters: 4. Decide type of test: sample proporton test α = 0.05 n = 85 Reject H 0 if ˆp p 0 Plug in values: Reject H 0 if p ( 0 1 p ) 0 / n > z α /2 z = 1.96 ˆp ( ) / 85 > 1.96
63 Solution to example 3 5. Use data to perform test: ˆp = 26 / 85 = Report p value: ( ) / 85 = < 1.96 Accept H 0 pvalue = P( Z > = P( Z > ) = 2(1 Φ(0.4725)) = ( ) / 85 )
64 Summary for performing hypothesis test Follow the procedure Examples: Test for normal mean, variance known, use ztest Test for normal mean, variance unknown, use ttest Test for sample proportion, using normal approximation and ztest
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