As p value 01 we do not reject h cannot accept h 1

This preview shows page 13 - 18 out of 18 pages.

As p-value > 0.1, we do not reject H0, cannot accept H1. Therefore, we cannot agree that there is a difference in the mean mounting times.t-Test: Two-Sample Assuming Equal VariancesWelles(minutes)Atkins(minutes)Mean45Variance8.54.4Observations56Pooled Variance6.222222222Hypothesized Mean Difference0df9t Stat-0.662051122P(T<=t) one-tail0.262263176t Critical one-tail1.383028738P(T<=t) two-tail0.524526351t Critical two-tail1.833112933
8/26/20171411-27Activity 2.4 – Paired t-TestNickel Savings and Loan wishes to compare the two companies it uses to appraise the value of residential homes. Nickel Savings selected a sample of 10 residential properties and scheduled both firms for an appraisal. The results, reported in $000, are shown on the table (right). At the 0.05 significance level, can we conclude there is a difference in the mean appraised values of the homes?10-28Tasks: In groups of 3 or 4 persons, 1.Open the Activity2.4_Data.xls file2.Follow the 5-steps procedure to conduct an appropriate hypothesis testing.3.Discuss whether there was a difference in the mean appraised values of the homes.You may refer to “Excel Self Practice Prior Seminar 2” Practice 2.4..Activity 2.4 Cont’d
8/26/201715Activity 2.4 AnswerTwo-Sample Paired t-Test for MeanStep 1: State the null and alternate hypotheses. H0: d= 0H1: d≠ 0where µdis the mean difference between the appraised value by Schadek and BowyerStep 2:State the level of significance: α = 0.05.Step 3: Find the appropriate test statistic.We will use the paired t-test.11-30Step 4: State the decision rule.Reject H0if p-value < 0.05Activity 2.4 Answer Cont’dStep 5:Determine the p-value and make a decisionSince we are performing a two-tailed test, the corresponding p-value is 0.00916. As p-value < 0.05, we reject H0and accept H1. Therefore, we conclude that there is a difference in the appraised value by Schadek and Bowyer.t-Test: Paired Two Sample for MeansSchadekBowyerMean226.8222.2Variance208.844204.1778Observations1010Pearson Correlation0.95314Hypothesized Mean Difference0df9t Stat3.3045P(T<=t) one-tail0.00458t Critical one-tail1.83311P(T<=t) two-tail0.00916t Critical two-tail2.26216
8/26/201716Summary1-31Point estimation and Confidence intervalsoPoint estimateoConfidence intervals𝑋± 𝑧or 𝑋± 𝑡ଶ,௡ିଵMeasureParameterEstimateMean𝑋Standard deviationsProportion p = Summary1-32Hypothesis tests: One-sample𝐻: = 6.125𝐻: ≠ 6.125𝐻: ≥ 6.125𝐻: < 6.125𝐻: ≤ 6.125𝐻: > 6.125Hypothesis tests: Two-sample𝐻: = 0𝐻:≠ 0𝐻:≥ 0𝐻:< 0𝐻:≤ 0𝐻:> 0Analysis of variance (ANOVA)𝐻: ==𝐻: not all means are equal
8/26/201717AssumptionsTwo-sample tests: independent samplesoBoth populations are normaloBoth samples are independentoKnown standard deviationsANOVAoThe populations are normalo

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture