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 normally distributed. Observation A B 1 1 3 2 3 4 3 5 7 4 4 4 5 8 9 What is the MD? (one decimal)

The 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…

You 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 4

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 normally distributed. Observation A B 1 1 3 2 3 4 3 5 7 4 4 4 5 8 9 What is the t statistic for the repeated-measures t test to test the null hypothesis H0: µD = 0? (two decimals)

You wish to test the following claim (Ha) at a significance level of α=0.10.      Ho:μ=72.2       Ha:μ≠72.2You 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 Column E 102.6 66.5 52 59.2 91.1 100.3 47.4 77.7 70.8 94.7 83.9 107.6 87.7 65.4 82.3 88.2 87.7 68 60 95.3 78.4 72.9 101.4 58.3 59.2 100.3 71.6 76.1 74.9 74.9 56.2 88.2 77.7 72.9 90.6 81.1 55 86.4 78 53.6 57.3 86 94 50 85.1 105.6 92.2 70.3 73.7 59.2 81.9 78 86.4 75.7 86 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? (Report answer accurate to four decimal places.) p-value = The p-value is... less than (or equal to) α   greater than α

In a test of hypothesis, the null hypothesis is that the population mean is greater than or equal to 37 and the alternative hypothesis is that the population mean is less than 37. The population has a normal distribution with variance of 10.6. The test is to be made at the 5% significance level. In order that the value of power at M₁ = 35. 27 to be 0.9, how large a sample do we need? O d. 20 O e. 38 O c. 6 O a. 31 O b. 63

To determine if their 1.33 centimeter galvanized pipes are properly adjusted, Fogarty Manufacturing has decided to use an X-Chart which uses the range to estimate the variability in the sample. Answer Table Step 4 of 7: Use the following sample data, taken from the next time period, to determine if the process is "In Control" or "Out of Control". Observations: 1.4, 1.42, 1.38, 1.37, 1.4 Sample Mean: 1.394 Period 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Control Chart Copy Table Average Values The average of the selected cell(s) is 1.340. Copy Value obs1 obs2 obs3 obs4 obs5 Sample Mean Sample Range 1.34 1.29 1.35 1.33 1.35 1.3320 1.35 1.35 1.29 1.38 1.28 1.3300 1.34 1.35 1.32 1.33 1.34 1.3360 1.36 1.29 1.37 1.35 1.30 1.3340 1.29 1.31 1.33 1.32 1.38 1.3260 1.31 1.32 1.31 1.31 1.3180 1.34 1.33 1.3300 1.38 1.31 1.29 1.34 1.35 1.33 1.30 1.35 1.32 1.3300 1.34 1.37. 1.28 1.36 1.3440 1.37 1.30 1.3080 1.37 1.29 1.30 1.28 1.35 1.36 1.32 1.31 1.32 1.29 1.36 1.38 1.3400 1.3280 1.30 1.35 1.32 1.32 1.29…