Chapter 8 HandoutIntroduction to Hypothesis Testing: One Population Value
Chapter 8 SummaryHypothesis Testing for One Population Value:1. Population Mean (µ) α. σ (population standard deviation) is given (known): Use z/standard normal/bell shaped distribution α. σ (pop std dev) is not given but s (sample std dev) is given Use student’s t distribution2. Population proportion (π) Use z/standard normal/bell shaped distribution3. Population variance (σ2) Use Χ2 (Chi-Square) distribution PS: population standard deviation = σ
The Null Hypothesis, H0• States the Assumption (numerical) to be tested e.g. The average # TV sets in US homes is at least 3 (H0: µ ≥ 3)• Begin with the assumption that the null hypothesis is TRUE. (Similar to the notion of innocent until proven guilty) •Refers to the Status Quo •Always contains the ‘ = ‘ sign•The Null Hypothesis may or may not berejected.
The Alternative Hypothesis, H1or HA • Is the opposite of the null hypothesis e.g. The average # TV sets in US homes is less than 3 (H1: µ < 3) • Challenges the Status Quo • Never contains the ‘=‘ sign • The Alternative Hypothesis may or may not be accepted
Identify the ProblemSteps: State the Null Hypothesis (H : µ ≥ 3) 0 State its opposite, the Alternative Hypothesis (H1: µ < 3) Hypotheses are mutually exclusive & exhaustive Sometimes it is easier to form the alternative hypothesis first.
Hypothesis Testing Process Assume the population mean age is 50. (Null Hypothesis) Population The SampleIs X = 20 ≅ µ = 50? Mean Is 20 No, not likely! REJECT Null Hypothesis Sample
Reason for Rejecting H0Sampling Distribution It is unlikely that we would ... Therefore, we get a sample reject the null mean of this hypothesis that value ... µ = 50. ... if in fact this were the population mean. 20 µ = 50 Sample Mean H0
Level of Significance, α• Defines Unlikely Values of Sample Statistic if Null Hypothesis Is True Called Rejection Region of Sampling Distribution• Designated α (alpha) Typical values are 0.01, 0.05, 0.10• Selected by the Researcher at the Start• Provides the Critical Value(s) of the Test
Level of Significance, α and the Rejection Region αH0: µ ≥ 3 CriticalH1: µ < 3 Value(s) Rejection 0 Regions αH0: µ ≤ 3H1: µ > 3 0 α /2H0: µ = 3H1: µ ≠ 3 0
Errors in Making Decisions• Type I Error Reject True Null Hypothesis Has Serious Consequences Probability of Type I Error Is α Called Level of Significance• Type II Error Do Not Reject False Null Hypothesis Probability of Type II Error Is β (Beta)
Result Possibilities H0: Innocent Jury Trial Hypothesis Test Actual Situation Actual SituationVerdict Innocent Guilty Decision H0 True H0 False Do Not Type IIInnocent Correct Error Reject 1-α Error (β ) H0 Type I PowerGuilty Error Correct Reject Error H0 (1 - β) (α )
α & β Have an Inverse Relationship Reduce probability of one error and the other one goes up. βα
Factors Affecting Type II Error, β• True Value of Population Parameter Increases When Difference Between Hypothesized Parameter & True Value Decreases• Significance Level α β Increases When α Decreases α• Population Standard Deviation σ σ β Increases When σ Increases• Sample Size n β Increases When n Decreases n
3 Methods for Hypotheses TestsRefer to Figure 8-6 (page 299) for a hypothesis test for means (µ) with σ (pop. std. dev.) is given:Method 1: Comparing Xα (X critical) with XMethod 2: Z test, i.e., comparing Zα (Ζ critical) with Z (or Z statistics or Z calculated)Method 3: Comparing α (significance level) with p-valueYou can modify those three methods for other cases. For example, if σ is unknown, you must use student’s t distribution. If you would like to use Method 2, please compare t α (t critical) with t (or t statistics or t calculated). Refer to Figure 8-8 (page 303).
You always get:• Zα (Ζ critical) from Z distribution• tα (t critical) from student’s t distribution• .Χ2α (Χ2critical) from Χ2distributionYou always get:• Z or Z calculated or Z statistics from sample (page 299 and Figure 8-6)• t or t calculated or t statistics from sample (Figure 8-8, page 299)• .Χ2 or Χ2 calculated or Χ2 statistics from sample (Figure 8-19, page 322)
Z-Test Statistics (σ Known)• Convert Sample Statistic (e.g., X ) to Standardized Z Variable X − µX X − µ Z= = Test Statistic σX σ n• Compare to Critical Z Value(s) If Z test Statistic falls in Critical Region, Reject H0; Otherwise Do Not Reject H0
p Value Test• Probability of Obtaining a Test Statistic More Extreme (≤ or ≥) than Actual Sample Value Given H0 Is True• Called Observed Level of Significance Smallest Value of a H0 Can Be Rejected• Used to Make Rejection Decision If p value ≥ α, Do Not Reject H0 If p value < α, Reject H0
Hypothesis Testing: StepsTest the Assumption that the true mean # of TV sets in US homes is at least 3.1. State H0 H0 : µ ≥ 32. State H1 H1 : µ < 33. Choose α α = .054. Choose n n = 1005. Choose Method: Z Test (Method 2)
Hypothesis Testing: Steps (continued) Test the Assumption that the average # of TV sets in US homes is at least 3. 6. Set Up Critical Value(s) Z = -1.645 7. Collect Data 100 households surveyed 8. Compute Test Statistic Computed Test Stat.= -2 9. Make Statistical Decision Reject Null Hypothesis10. Express Decision The true mean # of TV set is less than 3 in the US households.
One-Tail Z Test for Mean (σ Known)• Assumptions Population Is Normally Distributed If Not Normal, use large samples Null Hypothesis Has =, ≤, or ≥ Sign Only• Z Test Statistic: x − µx x −µ z= = σx σ n
Rejection Region H0: µ ≥ 0 H0: µ ≤ 0 H1: µ < 0 H1: µ > 0Reject H0 Reject H 0 α α 0 Z 0 Z Must Be Significantly Small values don’t contradict H0 Below µ = 0 Don’t Reject H0!
Example: One Tail TestDoes an average box ofcereal contain more than368 grams of cereal? Arandom sample of 25 boxes _showed X = 372.5. The 368 gm.company has specified σ tobe 15 grams. Test at the H0:α=0.05 level. µ ≤ 368 H1: µ > 368
Finding Critical Values: One Tail Standardized NormalWhat Is Z Given α = 0.05? Probability Table (Portion) .50 -.05 σZ = 1 Z .04 .05 .06 .45 1.6 .4495 .4505 .4515 α = .05 1.7 .4591 .4599 .4608 0 1.645 Z 1.8 .4671 .4678 .4686 Critical Value 1.9 .4738 .4744 .4750 = 1.645
Example Solution: One TailH0: µ ≤ 368 Test Statistic:H1: µ > 368 X −µα = 0.025 Z= = 1.50n = 25 σ nCritical Value: 1.645 Decision: Reject Do Not Reject Ho at α = .05 .05 Conclusion: No Evidence True Mean 0 1.645 Z Is More than 368
p Value Solution p Value is P(Z ≥ 1.50) = 0.0668Use thealternative p Value 1.0000hypothesis .0668to find the - .9332direction of .0668the test. .9332 0 1.50 Z From Z Table: Z Value of Sample Lookup 1.50 Statistic
p Value Solution(p Value = 0.0668) ≥ (α = 0.05). Do Not Reject. p Value = 0.0668 Reject α = 0.05 0 1.50 Z Test Statistic Is In the Do Not Reject Region
Example: Two Tail TestDoes an average box ofcereal contains 368 grams ofcereal? A random sample of25 boxes showed X = 372.5. The company has specified 368 gm.σ to be 15 grams. Test atthe α=0.05 level. H0: µ = 368 H1: µ ≠ 368
Example Solution: Two TailH0: µ = 386 Test Statistic:H1: µ ≠ 386 X − µ 372.5 − 368α = 0.05 Z= = = 1.50 σ 15n = 25 n 25Critical Value: ±1.96 Decision: Reject Do Not Reject Ho at α = .05 .025 .025 Conclusion: No Evidence that True -1.96 0 1.96 Z Mean Is Not 368
Two tail hypotheses tests = Confidence Intervals _ For X = 372.5oz, σ = 15 and n = 25, The 95% Confidence Interval is: 372.5 - (1.96) 15/ 25 to 372.5 + (1.96) 15/ 25 or 366.62 ≤ µ ≤ 378.38If this interval contains the Hypothesized mean (368), wedo not reject the null hypothesis. It does. Do not reject Ho.
t-Test: σ UnknownAssumptions Population is normally distributed If not normal, only slightly skewed & a large sample takenParametric test proceduret test statistic X −µ t= S n
Example: One Tail t-TestDoes an average box of cerealcontain more than 368 gramsof cereal? A random sample of36 boxes showed X = 372.5,and s= 15. Test at the α=0.01 368 gm.level. σ is not H0: µ ≤ 368 given, H1: µ > 368
Example Solution: One TailH0: µ ≤ 368 Test Statistic:H1: µ > 368 X − µ 372.5 − 368α = 0.01 t= = = 1.80 S 15n = 36, df = 35 n 36Critical Value: 2.4377 Decision: Reject Do Not Reject Ho at α = .01 .01 Conclusion: No Evidence that True 0 2.4377 Z Mean Is More than 368
Proportions• Involves categorical variables• Fraction or % of population in a category• If two categorical outcomes, binomial distribution Either possesses or doesn’t possess the characteristic• Sample proportion (p) p = X = number of successes n samplesize
Example:Z Test for Proportion•Problem: A marketing company claimsthat it receives π = 4% responses from itsMailing.•Approach: To test this claim, a randomsample of n = 500 were surveyed with x =25 responses.•Solution: Test at the α = .05 significancelevel.
Z Test for Proportion: SolutionH0: π = .04 Test Statistic:H1: π ≠ .04α = .05 p-π .05-.04 Z ≅ = = 1.14n = 500, x = 25 π (1 - π ) .04 (1 - .04)p = x/n = 25/500 = 0.05 n 500 Critical Values: ± 1.96 Decision: Reject Reject Do not reject Ho at α = .05 .025 .025 Conclusion: We do not have sufficient evidence to reject the company’s 0 Z claim of 4% response rate.