Fundamentals of Hypothesis Testing: One-Sample Tests

Click here to load reader

  • date post

    04-Jan-2016
  • Category

    Documents

  • view

    105
  • download

    7

Embed Size (px)

description

Fundamentals of Hypothesis Testing: One-Sample Tests. What is a Hypothesis?. A hypothesis is a claim (assumption) about a population parameter: population mean population proportion. Example: The mean monthly cell phone bill of this city is μ = $42. - PowerPoint PPT Presentation

Transcript of Fundamentals of Hypothesis Testing: One-Sample Tests

  • Chap 8-*

    Fundamentals of Hypothesis Testing: One-Sample Tests

  • Chap 8-*What is a Hypothesis?A hypothesis is a claim (assumption) about a population parameter:

    population mean

    population proportion

    Example: The mean monthly cell phone bill of this city is = $42Example: The proportion of adults in this city with cell phones is p = .68

  • Chap 8-*The Null Hypothesis, H0States the assumption (numerical) to be testedExample: The average number of TV sets in U.S. Homes is equal to three ( )Is always about a population parameter, not about a sample statistic

  • Chap 8-*The Null Hypothesis, H0Begin with the assumption that the null hypothesis is trueSimilar to the notion of innocent until proven guiltyRefers to the status quoAlways contains = , or signMay or may not be rejected(continued)

  • Chap 8-*The Alternative Hypothesis, H1Is the opposite of the null hypothesise.g., The average number of TV sets in U.S. homes is not equal to 3 ( H1: 3 )Challenges the status quoNever contains the = , or signMay or may not be provenIs generally the hypothesis that the researcher is trying to prove

  • PopulationClaim: thepopulationmean age is 50.(Null Hypothesis:REJECTSupposethe samplemean age is 20: X = 20SampleNull Hypothesis20likely if = 50?=IsHypothesis Testing ProcessIf not likely, Now select a random sample H0: = 50 )X

  • Chap 8-*Sampling Distribution of X = 50If H0 is trueIf it is unlikely that we would get a sample mean of this value ...... then we reject the null hypothesis that = 50.Reason for Rejecting H020... if in fact this were the population meanX

  • Chap 8-*Level of Significance, Defines the unlikely values of the sample statistic if the null hypothesis is trueDefines rejection region of the sampling distributionIs designated by , (level of significance)Typical values are .01, .05, or .10Is selected by the researcher at the beginningProvides the critical value(s) of the test

  • Chap 8-*Level of Significance and the Rejection RegionH0: 3 H1: < 30H0: 3 H1: > 3aa Represents critical valueLower-tail testLevel of significance = a0Upper-tail testTwo-tail testRejection region is shaded /20a /2aH0: = 3 H1: 3

  • Chap 8-*Errors in Making DecisionsType I Error Reject a true null hypothesisConsidered a serious type of error

    The probability of Type I Error is Called level of significance of the testSet by researcher in advance

  • Chap 8-*Errors in Making DecisionsType II ErrorFail to reject a false null hypothesis

    The probability of Type II Error is (continued)

  • Chap 8-*Outcomes and ProbabilitiesActual SituationDecisionDo NotRejectH0No error (1 - )aType II Error ( )RejectH0Type I Error( )aPossible Hypothesis Test Outcomes H0 False H0 TrueKey:Outcome(Probability)No Error ( 1 - )

  • Chap 8-*Type I & II Error Relationship Type I and Type II errors can not happen at the same time Type I error can only occur if H0 is true Type II error can only occur if H0 is false

  • Chap 8-*Factors Affecting Type II ErrorAll else equal, when the difference between hypothesized parameter and its true value

    when when n

  • Chap 8-*Hypothesis Tests for the Mean Known UnknownHypothesis Tests for

  • Chap 8-*Z Test of Hypothesis for the Mean ( Known)Convert sample statistic ( ) to a Z test statistic

    XThe test statistic is: Known UnknownHypothesis Tests for

  • Chap 8-*Critical Value Approach to TestingFor two tailed test for the mean, known:Convert sample statistic ( ) to test statistic (Z statistic )Determine the critical Z values for a specified level of significance from a table or computerDecision Rule: If the test statistic falls in the rejection region, reject H0 ; otherwise do not reject H0

  • Chap 8-*Do not reject H0Reject H0Reject H0There are two cutoff values (critical values), defining the regions of rejection Two-Tail Tests/2-Z0H0: = 3 H1: 3+Z/2Lower critical valueUpper critical value3ZX

  • Chap 8-*Review: 10 Steps in Hypothesis Testing1. State the null hypothesis, H02.State the alternative hypotheses, H13. Choose the level of significance, 4. Choose the sample size, n5. Determine the appropriate statistical technique and the test statistic to use6. Find the critical values and determine the rejection region(s)

  • Chap 8-*Review: 10 Steps in Hypothesis Testing7. Collect data and compute the test statistic from the sample result8.Compare the test statistic to the critical value to determine whether the test statistics falls in the region of rejection9. Make the statistical decision: Reject H0 if the test statistic falls in the rejection region10.Express the decision in the context of the problem

  • Chap 8-*Hypothesis Testing ExampleTest the claim that the true mean # of TV sets in US homes is equal to 3.(Assume = 0.8)1-2. State the appropriate null and alternative hypothesesH0: = 3 H1: 3 (This is a two tailed test)3. Specify the desired level of significanceSuppose that = .05 is chosen for this test4. Choose a sample sizeSuppose a sample of size n = 100 is selected

  • Chap 8-*Hypothesis Testing Example5. Determine the appropriate technique is known so this is a Z test6. Set up the critical valuesFor = .05 the critical Z values are 1.967. Collect the data and compute the test statisticSuppose the sample results are n = 100, X = 2.84 ( = 0.8 is assumed known)So the test statistic is:(continued)

  • Chap 8-*Reject H0Do not reject H08. Is the test statistic in the rejection region? = .05/2-Z= -1.960Reject H0 if Z < -1.96 or Z > 1.96; otherwise do not reject H0Hypothesis Testing Example(continued) = .05/2Reject H0+Z= +1.96Here, Z = -2.0 < -1.96, so the test statistic is in the rejection region

  • Chap 8-* 9-10. Reach a decision and interpret the result-2.0Since Z = -2.0 < -1.96, we reject the null hypothesis and conclude that there is sufficient evidence that the mean number of TVs in US homes is not equal to 3 Hypothesis Testing Example(continued)Reject H0Do not reject H0 = .05/2-Z= -1.960 = .05/2Reject H0+Z= +1.96

  • Chap 8-*p-Value Approach to Testingp-value: Probability of obtaining a test statistic more extreme ( or ) than the observed sample value given H0 is trueAlso called observed level of significanceSmallest value of for which H0 can be rejected

  • Chap 8-*p-Value Approach to TestingConvert Sample Statistic (e.g., ) to Test Statistic (e.g., Z statistic )Obtain the p-value from a table or computerCompare the p-value with If p-value < , reject H0If p-value , do not reject H0 X(continued)

  • Chap 8-*.0228/2 = .025p-Value ExampleExample: How likely is it to see a sample mean of 2.84 (or something further from the mean, in either direction) if the true mean is = 3.0?-1.960-2.0Z1.962.0X = 2.84 is translated to a Z score of Z = -2.0 p-value =.0228 + .0228 = .0456.0228/2 = .025

  • Chap 8-*Compare the p-value with If p-value < , reject H0If p-value , do not reject H0 Here: p-value = .0456 = .05Since .0456 < .05, we reject the null hypothesis(continued)p-Value Example.0228/2 = .025-1.960-2.0Z1.962.0.0228/2 = .025

  • Chap 8-*Example: Upper-Tail Z Test for Mean ( Known) A phone industry manager thinks that customer monthly cell phone bill have increased, and now average over $52 per month. The company wishes to test this claim. (Assume = 10 is known)H0: 52 the average is not over $52 per monthH1: > 52 the average is greater than $52 per month(i.e., sufficient evidence exists to support the managers claim)Form hypothesis test:

  • Chap 8-*Reject H0Do not reject H0Suppose that = .10 is chosen for this test

    Find the rejection region: = .101.280Reject H0Reject H0 if Z > 1.28Example: Find Rejection Region(continued)

  • Chap 8-*Review:One-Tail Critical ValueZ.07.091.1.8790.8810.88301.2.8980.90151.3.9147.9162.9177z01.28.08Standard Normal Distribution Table (Portion)What is Z given a = 0.10?a = .10Critical Value = 1.28.90.8997.10.90

  • Chap 8-*Obtain sample and compute the test statistic

    Suppose a sample is taken with the following results: n = 64, X = 53.1 (=10 was assumed known) Then the test statistic is:Example: Test Statistic(continued)

  • Chap 8-*Reject H0Do not reject H0Example: Decision = .101.280Reject H0Do not reject H0 since Z = 0.88 1.28i.e.: there is not sufficient evidence that the mean bill is over $52Z = .88Reach a decision and interpret the result:(continued)

  • Chap 8-*Reject H0 = .10Do not reject H01.280Reject H0Z = .88Calculate the p-value and compare to (assuming that = 52.0)(continued)p-value = .1894p -Value SolutionDo not reject H0 since p-value = .1894 > = .10

  • Chap 8-*Z Test of Hypothesis for the Mean ( Known)Convert sample statistic ( ) to a t test statistic

    XThe test statistic is: Known UnknownHypothesis Tests for

  • Chap 8-*Example: Two-Tail Test( Unknown) The average cost of a hotel room in New York is said to be $168 per night. A random sample of 25 hotels resulted in X = $172.50 and S = $15.40. Test at the = 0.05 level.(Assume the population distribution is normal)H0: = 168 H1: 168

  • Chap 8-*a = 0.05n = 25 is unknown, so use a t statisticCritical Value: t24 = 2.0639Example Solution: Two-Tail TestDo not reject H0: not sufficient evidence that true mean cost is different than $168Reject H0Reject H0a/2=.025-t n-1,/2Do not reject H00a/2=.025-2.06392.06391.46H0: = 168 H1: 168t n-1,/2