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Using the data from Problem #1 above, we want to use the Bonferonni method to test the following hyptheses at the 3% significance level:
| H0 : μ1 =
μ2 H1 : μ1
≠ μ2 H0 : μ1 = μ3 H1 : μ1 ≠ μ3 H0 : μ2 = μ3 H1 : μ2 ≠ μ3 |
| (a) | Find the value of the test statistic for each of the above three hypotheses (in the above order). (3 test statistics - correct to 2 decimal places) |
| (b) | Which pairs of means are significantly different (using the
Bonferonni method at the 3% significance level)? Which pairs? |
(A) 1 and 2 only (B) 1 and 3, 2 and 3 only (C) none of them (D) 2 and 3 only (E) 1 and 2, 2 and 3 only (F) 1 and 3 only (G) all of them (H) 1 and 2, 1 and 3 only
- Strategies for treating hypertensive patients by nonpharmacologic methods are compared by establishing three groups of hypertensive patients who receive the following types of nonpharmacologic therapy: Group 1: Patients receive counseling for weight reduction Group 2: Patients receive counseling for meditation Group 3: Patients receive no counseling at all month period and are given in the The reduction in diastolic blood pressure is noted in these patients after a table below. 4.2 Group 1 Group 2 Group 3 4.5 1.2...
1.Wait-Times: There are three registers at the local grocery store. I suspect the mean wait-times for the registers are different. The sample data is depicted below. The second table displays results from an ANOVA test on this data with software. Wait-Times in Minutes x Register 1 2.0 2.0 1.1 2.0 1.0 2.0 1.0 1.3 1.55 Register 2 1.8 2.0 2.2 2.6 1.8 2.1 2.2 1.7 2.05 Register 3 2.1 2.1 1.8 1.5 1.4 1.4 2.0 1.7 1.75 ANOVA Results...
There are three registers at the local grocery store. I suspect the mean wait-times for the registers are different. The sample data is depicted below. The second table displays results from an ANOVA test on this data with software. Wait-Times in Minutes x Register 1 2.0 2.0 1.1 2.0 1.0 2.0 1.0 1.3 1.55 Register 2 1.8 2.0 2.2 1.9 1.8 2.1 2.2 1.7 1.96 Register 3 2.1 2.1 1.8 1.5 1.4 1.4 2.0 1.7 1.75 ANOVA Results F...
PLEASE SOLVE USING ALL MANUAL CALCULATIONS, Z-SCORE
TABLES, T-SCORE TABLES, ETC.
Problem #2: Using the data from Problem #1 above, we want to use the Bonferonni method to test the following hyptheses at the 3% significance level (a) Find the value of the test statistic for each of the above three hypotheses (in the above order) (b) which pairs of means are significantly different (using the Bonferonni method at the 3% significance level)? 3 test statistics (correct to 2 decimals)...
Strategies for treating hypertensive patients by nonpharmacologic methods are compared by establishing three groups of hypertensive patients who receive the following types of nonpharmacologic therapy: Group 1 Patients receive counseling for weight reduction Group 2: Patients receive counseling for meditation Group 3: Patients receive no counseling at all The reduction in diastolic blood pressure is noted in these patients after a 1-month period and are given in the table below Group 1 Group 2 Group 3 4.2 5.7 4.5 1.2...
Consider the data set that is summarized in the Minitab Output
below.
(a)
Find a 93% Bonferonni confidence interval for
μ2 − μ1.
(b)
Which pairs of means are significantly different (using the
Bonferonni method at the 7% significance level)?
Problem #4(a):
(A) 2 and 3 only (B) 1 and 3
only (C) 1 and 2, 2 and 3 only
(D) all of them (E) none of them
(F) 1 and 3, 2 and 3 only (G) 1
and 2,...
1. Test the claim that the mean GPA of night students is significantly different than 2.4 at the 0.2 significance level. The null and alternative hypothesis would be: a) H0:μ=2.4 H1:μ>2.4 b) H0:μ=2.4 H1:μ<2.4 c) H0:p=0.6 H1:p<0.6 d) H0:p=0.6 H1:p>0.6 e) H0:p=0.6 H1:p≠0.6 f) H0:μ=2.4 H1:μ≠2.4 2. The test is: a) left-tailed b) right-tailed c) two-tailed 3. Based on a sample of 35 people, the sample mean GPA was 2.44 with a standard deviation of 0.04 The test statistic is:...
Treatment 1 Treatment 2 Treatment 3 0 1 6 1 4 5 0 1 6 3 2 3 T = 4 T = 8 T = 20 SS = 6 SS = 6 SS = 6 N = 12 G = 32 ƩX2= 138 1a. Conduct a single-factor independent-measures ANOVA to test the hypothesis that there are significant differences in the mean scores among the three treatment conditions. Use α = .01. The alternative hypothesis is Group of answer choices...
Consider the data set that is summarized in the Minitab Output below. -------------------------------------------------------------------------------------------------- Source DF Adj SS Adj MS F-Value P-Value C2 2 67.46 33.73 7.36 0.002 Error 34 155.71 4.58 Total 36 223.17 S = 2.140 R-Sq = 30.23% R-Sq(adj) = 26.12% Level N Mean StDev 1 12 8.386 2.071 2 12 10.638 2.269 3 13 11.608 2.080 Pooled StDev = 2.140 Fisher Individual Tests for Differences of Means Difference of Levels Difference of Means 97.6667% CI T-Value Adjusted...
The following data was reported on total Fe for four types of iron formation (1 = carbonate, 2 = silicate, 3 = magnetite, 4 = hematite). : 1: 21.0 28.1 27.8 27.0 28.0 25.2 25.3 27.1 20.5 31.2 2: 26.9 24.0 26.2 20.2 23.7 34.0 17.1 26.8 23.7 25.2 3: 29.8 34.0 27.5 29.4 28.5 26.2 29.9 29.5 30.0 36.4 4: 37.2 44.2 34.1 30.3 31.8 33.1 34.1 32.9 36.3 25.7 1.Carry out an analysis of variance F test at...