The null hypothesis in words is
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Questions 13 to 19 are based on the following problem:
The following data were obtained from an independent-measures research study comparing three treatment conditions.
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I |
II |
III |
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n = 6 |
n = 4 |
n = 4 |
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M = 2 |
M = 2.5 |
M = 5 |
N = 14 |
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T = 12 |
T = 10 |
T = 20 |
G = 42 |
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SS = 14 |
SS = 9 |
SS = 10 |
ΣX2tot = 182 |
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Use an ANOVA with α = .05 to determine whether there are any significant mean differences among the treatments.
The null hypothesis in words is
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- You wish to test the following claim (HaHa) at a significance level of α=0.01α=0.01. Ho:μ1=μ2Ho:μ1=μ2 Ha:μ1≠μ2Ha:μ1≠μ2You obtain the following two samples of data. Sample #1 Sample #2 65.8 77.7 105.1 78.6 89.2 69.1 96.7 91.3 105.9 87.5 70.5 93.9 83.4 88.3 79.5 66.7 83.4 91.7 106.7 73 75.7 98.7 76.7 80 53.2 92.1 93 78.2 72.4 79.1 82.1 74.1 78.2 85.5 85.9 99.3 84.2 104.3 88.3 107.6 63.8 99.8 97.7 86.3 74.9 72.7 77.5 93.9 76.1 77.1 81.9 74.1 73.6 78.7 81.9 78.9 81.1 88 85.8 75.3 74.3 77.5 69.1 75.5 72.9 80.1 84.3 73.1 67.3 72.2 76.1 67.8 74.9 67.3 74.7 70.1 65.4 76.9 83.8 79.7 72.9 81.7 74.1 68.2 65.4 80.9 69.4 79.5 96 77.5 77.1 75.9 68.7 What is the test statistic for this sample? (Report answer accurate to three decimal places.)test statistic = What is the p-value for this sample? For this calculation, use the degrees of freedom reported from the technology you are using. (Report answer accurate to four…The data below are from a study comparing three different treatment conditions: 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 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. a. The alternative hypothesis isYou wish to test the following claim (HaHa) at a significance level of α=0.05 Ho:μ1=μ2 Ha:μ1>μ2You obtain the following two samples of data. Sample #1 Sample #2 51.6 63.2 60 52.9 54.9 55.6 57.9 61.4 48.8 54.4 49.8 60.6 46 65.7 54.4 62.9 62 52 64.3 50.6 72.3 53.6 65 59.3 55.4 66.6 57.9 54 52.7 46.6 57.1 59.3 56.9 50.9 52.7 50.1 58.9 52 50.6 48.4 53.1 54.4 54.9 58.9 60.9 58.4 66.2 55.7 63.7 57.1 57.9 54 57.9 67.2 52.2 56.7 66.2 50.1 61.1 56.7 25.1 55.1 37 27.3 44 22.4 30 30 46.9 38.5 60.8 50.9 40.5 66.3 32.9 69.5 43.6 30.8 40.5 75.4 31.5 53.8 73.9 53.4 25.1 32.2 69.5 38.5 36.5 60.3 43.1 40.5 15.8 48.9 56 What is the test statistic for this sample? (Report answer accurate to three decimal places.)test statistic = What is the p-value for this sample? For this calculation, use the degrees of freedom reported from the technology you are using. (Report answer accurate to four decimal places.)p-value = The…
- You wish to test the following claim (HaHa) at a significance level of α=0.01α=0.01. Ho:μ1=μ2Ho:μ1=μ2 Ha:μ1<μ2Ha:μ1<μ2You obtain the following two samples of data. Sample #1 Sample #2 52 60.1 61.7 56.9 52 59.7 45 49.9 71.9 31.1 47.1 56.1 54.9 45 85.1 67 66.1 51.2 87.2 50.7 73.1 43.3 65.2 58.5 50.3 71.3 60.5 48 62.2 78.6 50.7 56.5 37.8 45.6 41.5 58.5 54.9 76.8 62.6 29.5 71.3 83.5 71.3 64.3 63 74.5 76.8 50.7 78.6 59.3 31.1 54.9 61.2 50.3 75.1 72.3 68.5 85.7 63.3 50.3 79.1 73.5 65.2 58.4 62.3 57 52.9 72.3 88.2 57.4 75.8 68.5 84.1 45 60 79.7 58.1 56.6 77.1 60.9 68.5 67.3 62 45 66.4 79.7 53.5 52.9 79.7 63.5 67.1 74.8 53.5 66.1 68.7 64.2 72.3 55.2 75.5 What is the test statistic for this sample? (Report answer accurate to three decimal places.)test statistic = What is the p-value for this sample? For this calculation, use the degrees of freedom reported from the technology you are using. (Report answer…You wish to test the following claim (HaHa) at a significance level of α=0.001α=0.001. Ho:μ1=μ2Ho:μ1=μ2 Ha:μ1≠μ2Ha:μ1≠μ2You obtain the following two samples of data. Sample #1 Sample #2 50.8 84.4 72.1 78.4 72.9 48.8 87.4 36.1 78.8 51.7 74.7 56.3 65.2 98.6 72.5 72.1 72.9 54.2 69.8 73.4 51.7 68.5 59.4 86.8 48.8 47.6 48.8 85.6 65.2 43.4 76.1 69.8 43.4 69.4 55.6 74.3 108.9 92.4 77.9 46.4 47.6 93.3 60 72.9 74.7 66.6 72 84.3 66.1 75 86.5 96.6 47.1 79.1 75 74.5 82.2 77.8 79.5 70.9 71.5 104.8 88.8 81.7 99.8 89.3 91.2 87.9 111.1 66.1 68 79.5 82.6 61.4 90.7 115.1 106.9 60.4 50.1 85.7 61.4 100.5 105.8 79.1 88.3 73 68.6 57.1 70.9 67.4 42.4 52.3 80.4 88.8 What is the test statistic for this sample? (Report answer accurate to three decimal places.)test statistic = What is the p-value for this sample? For this calculation, use the degrees of freedom reported from the technology you are using. (Report answer accurate to…Questions 4 to 8 are based on the following problem: The following data summarize the results from an independent measures study comparing three treatment conditions. I II III n = 6 n = 6 n = 6 M = 4 M = 5 M = 6 N = 18 T = 24 T = 30 T = 36 G = 90 SS = 30 SS = 35 SS = 40 ΣX2tot = 567 Use an ANOVA with α = .05 to determine whether there are any significant differences among the three treatment means The null hypothesis in symbols is
- QUESTION 2 To test the effect of a new medication on pulse rate, the researcher randomly selected the 60 subjects into two groups of 30. New medicine and placebo were given to Group 1 and Group 2, respectively. Below is the Minitab output. Two-Sample T-Test and CI Group Мean StDev SE Mean 1 30 65.20 7.80 1.4 2 30 70.30 8.40 1.5 Difference = mu (1) - mu (2) Estimate for difference: 95% CI for difference: T-Test of difference = 0 (vs not =): T-Value = -2.44 -5.10 (-9.29, -0.91) P-Value=0.018 DF=57 ii) Using to.025,57 = 2.003 shows that the 95% confidence interval for the difference in mean pulse rate between new medication and placebo is (-9.29, -0.91). Hence, interpret its meaning.Here are the data for an ANOVA experiment, and the APA Source table from Step 3 of that same ANOVA. this information, Calculate the HSD value for a Tukey’s HSD test – including looking up the value of q from the table. Use α = .05. GROUP 1 GROUP 2 GROUP 3 GROUP 4 10 8 6 4 11 9 7 4 12 10 5 7 13 10 6 5 SOURCE SS df MS Fobt Between 107.19 3 35.73 27.22 Within 15.75 12 1.31 Total 122.94 15 Group of answer choices q = 4.20, HSD = 2.40 q = 6.20, HSD = 4.13 q = 4.53, HSD = 3.02 q = 5.53, HSD = 3.15Refer to the below table. Using an alpha = 0.05, test the claim that IQ scores are the same for children in three different blood lead level groups: low lead level, medium lead level, and high lead level). One-Way Analysis of Variance Summary Table for IQ Measurements for Children among Three Blood Lead Level Groups: Low Lead Level, Medium Lead Level, and High Lead Level. Source df SS MS F p Between-group (treatment) 2 469.1827 2677.864 2.30 0.104 Within-group (error) 118 203.6918 11745.05 Total 120
- The authors of the paper "Statistical Methods for Assessing Agreement Between Two Methods of Clinical Measurement"† compared two different instruments for measuring a person's ability to breathe out air. (This measurement is helpful in diagnosing various lung disorders.) The two instruments considered were a Wright peak flow meter and a mini-Wright peak flow meter. Seventeen people participated in the study, and for each person air flow was measured once using the Wright meter and once using the mini-Wright meter. Subject Mini-WrightMeter WrightMeter Subject Mini-WrightMeter WrightMeter 1 512 494 10 445 433 2 430 395 11 432 417 3 520 516 12 626 656 4 428 434 13 260 267 5 500 476 14 477 478 6 600 557 15 259 178 7 364 413 16 350 423 8 380 442 17 451 427 9 658 650 (a) Suppose that the Wright meter is considered to provide a better measure of air flow, but the mini-Wright meter is easier to transport and to use. If the two types of meters produce…You wish to test the following daim (Ha) at a significance level of a = 0.005. H.:P1 > P2 Ha:P1 < P2 You obtain a sample from the first population with 303 successes and 238 failures. You obtain a sample from the second population with 155 successes and 79 failures. critical value = [three decimal accuracy] test statistic = [three decimal accuracy] The test statistic is... O in the critical region O not in the critical region This test statistic leads to a decision to... reject the null hypothesis fail to reject the null hypothesis As such, the final condusion is that... O There is sufficient evidence to support that the first population proportion is less than the second population proportion. O There is not sufficient evidence to support that the first population proportion is less than the second population proportion.Use the following scenario for questions 1-4: Dr. Wiese wants to determine whether duration of pregnancy and type of opiate use among pregnant women predict developmental delays in their children. In a sample of 30 new mothers, Dr. Wiese measured the duration of their pregnancies (premature, mature, postmature) and the type of opiate that they used during pregnancy (oxycodone, heroin). All participants' babies were then rated on how closely they have met developmental milestones on a scale ranging from 1-10. 1. Identify one of the independent variables. How many levels does this variable have? 2. Identify the other independent variable. How many levels does this variable have? 3. This study is a .............x ...................factorial design.
