Consider the following data from a repeated-measures design. You want to use a repeated-measures t test to test the null hypothesis H0: µD = 0 (the null hypothesis states that the mean difference for the general population is zero). The data consist of five observations, each with two measurements, A and B, taken before and after a treatment. Assume the population of the differences in these measurements are
Observation | A | B |
1 | 1 | 3 |
2 | 3 | 4 |
3 | 5 | 7 |
4 | 4 | 4 |
5 | 8 | 9 |
For a repeated-measures t test, you need to calculate the t statistic, which requires you to calculate s and SMD· What is the estimated standard error of the mean difference scores? (Note: For best results, retain at least six decimal places from your calculation of s.) (round your answer to this question to four decimals)
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- Using the following table: 1. Determine a hypothesis to test the relationship between students enrolled 12 months or less and sleep of students, use descriptive statistics to describe the variables both numerically and graphically, identify the appropriate type of statistical test using alpha = 0.05, perform the statistical test (please show steps). ID Months enrolled Sleep 1 12 360 2 6 400 3 8 420 4 10 440 5 48 540 6 48 480 7 36 320 8 32 440 9 6 300 10 22 420 11 15 500 12 20 400 13 11 440 14 18 300 15 29 360 16 13 480 17 49 410 18 16 430 19 24 330 20 8 480arrow_forwardYou wish to test the following claim (HaHa) at a significance level of α=0.05α=0.05. Ho:μ=63.7Ho:μ=63.7 Ha:μ<63.7Ha:μ<63.7You believe the population is normally distributed, but you do not know the standard deviation. You obtain the following sample of data: Column A Column B Column C Column D 66.7 56.1 82.1 74.8 60.4 57.5 55.2 52.4 74.8 56.1 83.6 98 49.5 33.7 83.6 44.9 27.3 29.8 39 92.9 42.2 29.8 25.8 42.7 64.5 69.7 58 38.4 76.6 50.9 57.5 69.7 47 43.9 65 67.8 What is the critical value for this test? (Report answer accurate to three decimal places.)critical value = What is the test statistic for this sample? (Report answer accurate to three decimal places.)test statistic =arrow_forwardThe following data represent the results from an independent-measures study comparing two treatment conditions. Treatment I II 10 7 N = 16 8 4 G = 120 7 9 ∑X² = 1036 9 3 13 7 7 6 6 10 12 2 M = 9.000 M = 6.000 T = 72 T = 48 SS = 44 SS = 56 Use an independent-measures t test with α = .05 to determine whether there is a significant mean difference between the two treatments. (Use three decimal places; subtract MIIII from MII.) t-critical = ± t = t Distribution Degrees of Freedom = 21 -3.0-2.0-1.00.01.02.03.0t Conclusion: Fail to reject the null hypothesis; there are significant differences between the two treatments. Reject the null hypothesis; there are significant differences between the two treatments. Fail to reject the null hypothesis; there are no significant differences between the two treatments. Reject the null hypothesis; there are…arrow_forward
- You may need to use the appropriate technology to answer this question. In a completely randomized design, 15 experimental units were used for the first treatment, 17 for the second treatment, and 20 for the third treatment. Complete the following analysis of variance. (Round your values for MSE and F to two decimal places, and your p-value to four decimal places.) Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments 1,100 Error Total 1,800 At a 0.05 level of significance, is there a significant difference between the treatments? State the null and alternative hypotheses. H0: ?1 = ?2 = ?3Ha: ?1 ≠ ?2 ≠ ?3H0: ?1 ≠ ?2 ≠ ?3Ha: ?1 = ?2 = ?3 H0: At least two of the population means are equal.Ha: At least two of the population means are different.H0: Not all the population means are equal.Ha: ?1 = ?2 = ?3H0: ?1 = ?2 = ?3Ha: Not all the population means are equal. Find the value of the test statistic. (Round your…arrow_forwardPart A of question one asks for the variance (sigma squared), not the equation of regression line. Is the variance 2.79 as I listed on my answer sheet? Maybe my image didn't go through, but I already had the answer to subparts B and C to the question, which you list as #2 and #3. But I need help with parts D, E, and F. I have the following answers for D. d. Y hat (fitted value = 426.46), with residual of -1.62. e. r squared value = 1.00 g. 99% confidence interval = (-11.63, -1.05) Thanks thanksarrow_forwardYou are conducting a test of homogeneity for the claim that two different populations have the same proportions of the following two characteristics. Here is the sample data. Population Population Category #1 #2 A 40 102 28 84 The expected observations for this table would be Population Population Category #1 #2 A 38.02 103.98 29.98 82.02 The resulting Pearson residuals are: Population Population Category #1 #2 0.103 0.038 0.131 0.048 Chp -> Ce & 5 7. 00 16 %24 4arrow_forward
- The following data represent the results from an independent-measures experiment comparing three treatment conditions. Conduct an analysis of variance with α=0.05 to determine whether these data are sufficient to conclude that there are significant differences between the treatments. Treatment A Treatment B Treatment C 8 10 16 10 9 12 9 10 14 9 8 15 9 13 13 F-ratio = p-value = Conclusion: There is a significant difference between treatments These data do not provide evidence of a difference between the treatments The results obtained above were primarily due to the mean for the third treatment being noticeably different from the other two sample means. For the following data, the scores are the same as above except that the difference between treatments was reduced by moving the third treatment closer to the other two samples. In particular, 3 points have been subtracted from each score in the third sample.Before you begin the calculation, predict how the…arrow_forwardRefer 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 120arrow_forward
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