Is that there is no association between car color and

  • Westlake High
  • MATH AP
  • Test Prep
  • Syke24
  • 40
  • 97% (129) 125 out of 129 people found this document helpful

This preview shows page 36 - 38 out of 40 pages.

is that there is no association between car color and proportion of cars involved inaccidents.Since theP-value is greater than, we fail to reject the null hypothesis, whichmeans we do not have evidence that there is an association.Fail to rejectH0does notmean the same thing as acceptingH0: the test only provides evidence against the null, notfor it.So (b) is incorrect.Part II11.State: We are testing the hypothesisHo: The proportion of shoppers at the Big Box storefrom each geographic region matches the distribution of the population across those regions,againstHa: The proportion of shoppers at the Big Box store from each geographic region doesnot match the distribution of the population across those regions.We will use a significancelevel of= 0.05.Plan:The procedure is a chi-square goodness-of-fit test.Conditions:Random: the data come from a simple random sample of 250 shoppers.Large sample size:Expected counts are: North: 100; South: 60; East: 55; West: 35all are at least 5 .Independent:Randomly-selected shoppers should be independent; and there are surely at least 10 x 250 =2500 total shoppers.
© 2011 BFW PublishersThe Practice of Statistics, 4/e- Chapter 11539Do:2222212010048606255203513.719100605535,df= 3;P-value = 0.0033Conclude:Since theP-value is much smaller than= 0.05, we can rejectH0.There is goodevidence that the proportion of shoppers at the Big Box store from each geographic region doesnot match the distribution of the population across those regions.12.(a)State: We are testing the hypothesisHo: The distribution of decided/undecided is thesame for both portions of the state, againstHa: The distribution of decided/undecided is differentfor the two portions of the state.We will use a significance level of= 0.05.Plan:Theprocedure is a chi-square test for homogeneity.Conditions:Random: the data come from SRSsof each region.Large sample size: All expected counts are at least 5 (see expected counts tablebelow).Independent: Randomly-selected observations should be independent, and thepopulation of the two regions is doubtless greater than 10 times the sample sizes.Do:Using acalculator,22.930;1;value0.0869.dfP(From Table C,P-value is between 0.05 and0.10).Conclude:Since theP-value is larger than= 0.05, we fail to rejectH0.We do not haveconvincing evidence that the distribution of decided/undecided is different for both portions ofthe state.(b)The responses of these 24 non-responders might change our conclusion.In the most extremecaseif all 24 of them were still undecidedthe chi-square statistic for the test would change to11.408, yielding aP-value of 0.00073, which would provide strong evidence that there is adifference in distribution of decided/undecided between the two regions.Test 11BPart I1.cNull hypothesis should refer to the distribution of the population, not observed values.Inthis setting, the null is that all computer brands have equal popularity, so the proportionpreferred is the same for each brand.2.aStatement III is a condition for a one-sample test of meanscategorical distributions

Upload your study docs or become a

Course Hero member to access this document

Upload your study docs or become a

Course Hero member to access this document

End of preview. Want to read all 40 pages?

Upload your study docs or become a

Course Hero member to access this document

Term
Spring
Professor
N/A
Tags
Statistics, Statistical hypothesis testing, BFW Publishers, 4 e Chapter

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture