Since we are performing a left tailed test the

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Since we are performing a left- tailed test, the corresponding p - value is 0.0071. As p -value < 0.01, we reject H 0 and accept H 1 . Therefore, we conclude that there was an improvement in the cost cutting measures. Step 5: Make a decision and interpret the result. Excel Output : t-Test for Mean Cost Mean 57.2866667 Variance 14.0640952 Observations 15 Pearson Correlation #DIV/0! Hypothesized Mean Difference 60 df 14 t Stat -2.8021597 P(T<=t) one-tail 0.00706009 t Critical one-tail 2.62449407 P(T<=t) two-tail 0.01412017 t Critical two-tail 2.97684273
Activity 3.5 A random sample of 400 families who planned to buy a vacation residence revealed that 228 families want to buy a condominium in Florida. At the 0.01 significance level, test the statement that 55% of those families who plan to purchase a vacation residence want a condominium in Florida. To decide if the sample data support the 55%, state your decision in terms of the null hypothesis. Use a 0.01 level of significance.
Activity 3.5 - Answer Step 1: State the null hypothesis and the alternate hypothesis. Claim: 55% of those families who plan to purchase a vacation residence want a condominium in Florida (i.e. 𝜋 =0.55) H 0 : 𝜋 = 0.55 H 1 : 𝜋 ≠ 0.55 Step 2: Select the level of significance. α = 0.01 as stated in the problem Step 3: Select the test statistic. Use z -distribution for the proportion
Activity 3.5 - Answer Step 4: Formulate the decision rule. Reject the null if the Z-statistic is outside the critical region. The Z-statistic is The critical z-statistic for a two- tailed hypothesis test with α = 0.01 is ± 2.576 Step 5: Make a decision and interpret the result. Since the test statistic (+0.804) is not greater than +2.576, we fail to reject the null hypothesis.
BUS105 Statistics Seminar 4
Topics Unit 2 1 Point Estimation and Confidence Intervals 2.1 Hypothesis Tests: One-Sample 2.2 Hypothesis Tests: Two-Sample 3 Analysis of Variance (ANOVA)
Review: Hypothesis Tests: Two-Sample Two-sample tests of hypothesis About mean difference Independent samples Known/Unknown standard deviation Paired About proportion difference
Recall: Hypothesis Testing Steps
Activity 4.1 Paired t -Test Nickel 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?
Activity 4.1 Cont’d Tasks: In groups of 4 persons, 1.Open the Activity4.1_Data.xls file 2.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 3” Practice 2.4..
Activity 4.1 Answer Two-Sample Paired t-Test for Mean Step 1: State the null and alternate hypotheses.

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