Homework 6 Solutions

Homework 6 Solutions - HW 6 SOLUTIONS Distribution Free...

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Unformatted text preview: HW 6 SOLUTIONS Distribution Free Tests (two samples) 1. Fill in the blank. In our class, we apply a non parametric test when we have a small data set that does not come from a normal population. In practice, for certain shaped distributions, we might try a ________________ first, which may allow us to apply a t based inference. Correct: transformation 2. Suppose we have a dependent samples data set. Further suppose that the differences come from a normal distribution. True or False: The dependent samples t test is more powerful than the sign test. TRUE FALSE 3. Consider two samples of size n1 = 9 and n2 = 3. Use Table 6 to bracket the P‐value using a non‐directional alternative. a. Us = 21 Correct: P > 0.146 b. Us = 24 Correct: 0.036 < P < 0.10 c. Us = 25 Correct: P = 0.036 d. Us = 28 Correct: P < 0.0091 4. 7.80. Recall the study described in exercise 7.51 on your previous homework. In this study of hypnosis, breathing patterns were observed in an experimental group of subjects and in a control group. The measurements of total ventilation (liters of air per minute per square meter of body area) are shown in the following table. Experimental Control 5.32 4.50 5.60 4.78 5.74 4.79 6.06 4.86 6.32 5.41 6.34 5.70 6.79 6.08 7.18 6.21 n8 Y 6.169 s 0.621 8 5.291 0.652 a. When you computed a confidence interval using this data on your last homework, you were told to proceed as though the assumptions for validity of a t based inference were checked and deemed acceptable. Use the QQplot to check the normality assumption. Correct: Since we are in the independent samples setting, we must look at both QQplots. The QQplot for the experimental group is not perfect, but shows no systematic departure from linearity. The QQplot for the control group shows sytematic departure from the line. The points are making a shape (an "S" shape?). Also, we only have 8 observations in each group, so we cannot use the CLT. Conclusion: There is almost certainly a violation of the normality assumption here. t distribution inference will not be valid. b. Conduct the appropriate test to see if ventilation tends to be higher in the "to be hypnotized" group. Operate at α = 0.05. Correct: The normality assumption necessary for t based inference was not met, so we will use the non‐parametric test we learned when comparing independent samples ‐ the Wilcoxon‐Mann‐Whitney. (1) α = 0.05 (2) H0: Ventilation is the same for the “to be hypnotized” and the “control” conditions HA: Ventilation tends to be higher for the “to be hypnotized” condition (3 )K1 = 4 + 5 + 6 + 6 + 8 + 8 + 8 + 8 = 53 K2 = 0 + 0 + 0 + 0 +01 + 2 + 4 + 4 = 11 Us = max{53, 11} = 53 (4) The shift in the data is in the direction predicted by HA. With n = n' = 8, ½ (0.021) < P < ½ (0.050) ⇒ 0.0105 < P < 0.025. (5) P < α, reject H0. (6) There is significant evidence to suggest that ventilation rate tends to be higher under the "to be hypnotized" condition than under the "control" condition. 5. 9.53. For an investigation of the mechanism of wound healing, a biologist chose a paired design, using the left and right hindlimbs of the salamander Notophthalmus viridescens. After amputating each limb, she made a small wound in the skin and then kept the limb for 4 hours in either a solution containing benzamil or a control solution. She theorized that the benzamil would impair the healing. The accompanying table shows the amount of healing, expressed as the area (mm2) covered with new skin after 4 hours. Animal 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Control .55 .15 .00 .13 .26 .07 .20 .16 .03 .42 .49 .08 .32 .18 .35 .03 .24 Benzamil .14 .08 .00 .13 .10 .08 .11 .00 .05 .21 .11 .03 .14 .37 .25 .05 .16 a. Use the QQplot to check the assumption of normality. Correct: Note, this is the dependent samples setting, so we need to look at the QQplot of the differences. There is a systematic departure from the line. The points are making a "U" shape, which indicates a right skew ‐ except for that one really low observation, which is not acting like the rest. Anyway, there are only 17 observations, so the CLT may not be appropriate here. There is a probable violation of the normality assumption. b. Use an appropriate hypothesis test to assess the significance of the researcher's hypothesis about the effect of Benzamil. Operate at the α = 0.05 significance level. Correct: Since there is a violation of the normality assumption and we have dependent samples, we'll use the sign test. (1) α = 0.05 (2) H0: Healing is the same with or without Benzamil present HA: Healing tends to be impared when Benzamil is present (3) N+ = 11, N‐ = 4, Bs = 11 (4) Eliminating the two pairs with d = 0, nd = 15. TI‐84 P = Pr{B ≥ 11} = 1 – Pr{B ≤ 10} = 1 – binomcdf(15, 0.5, 10) = 0.0592 (5) P > α, so fail to reject H0 (6) There is not significant evidence to conclude that healing tends to be impaired when Benzamil is present. 6. 7.84. Human beta‐endorphin (HBE) is a hormone secreted by the pituitary gland under conditions of stress. An exercise physiologist measured the resting (unstressed) blood concentration of HBE in two groups of men: Group 1 consisted of 11 men who had been jogging regularly for some time, and group 2 consisted of 15 men who had just entered a physical fitness program. (This data was from the same research study on HBE we used in your Inference for the Population Mean HW, exercise 6.16.) The results are given in the following table. Joggers 39 40 32 60 19 52 41 32 13 37 28 Fitness Program Entrants 70 47 54 27 31 42 37 41 9 18 33 23 49 59 41 a. Use the QQplot to check the assumption of normality for a t distribution based inference. Correct: Since we are in the independent samples setting, we need to look at the QQplots for the joggers and the fitness program entrants. The QQplot for the fitness entrants shows an ever so slight hint of being heavy tailed, but is for the most part linear. The QQplot for the joggers, on the other hand, shows a more systematic departure from the line and we only have 12 observations. Conclusion: The normality assumption is may not be met. b. Conduct the appropriate test to compare the two groups. Operate at the α = 0.05 significance level. Correct: The normality assumption necessary for t based inference may not be met, so we will use the non‐parametric test we learned when comparing independent samples ‐ the Wilcoxon‐Mann‐Whitney. (1) α = 0.05 (2) H0: “HBE joggers” = “HBE fitness program entrants” HA: “HBE joggers” ≠ “HBE fitness program entrants” (3 )K1 = 93.5 K2 = 71.5 Us = max{93.5, 71.5} = 93.5 (4) With n = 15 and n' = 11, P > 0.198 (5) P > α, fail to reject H0. (6) There is not significant evidence to conclude that resting blood concentration of HBE for joggers tends to be different from that of fitness program entrants. c. Instead of a non‐directional alternative, suppose the researchers believed the HBE would tend to be higher for the joggers group. What would the P‐value be for this test? Correct: K2 = 71.5 < 93.5 = K1 ‐ the data do not deviate in the direction of HA. P > 0.5 7. 9.33. In an investigation of possible brain damage due to alcoholism, an X‐ray procedure known as a computerized tomography (CT) scan was used to measure brain densities in eleven chronic alcoholics. For each alcoholic, a nonalcoholic control was selected who matched the alcoholic on age, sex, education, and other factors. The brain density measurements on the alcoholics and the matched controls are reported in the accompanying table. Pair Alcoholic Control Difference 1 40.1 41.3 ‐1.2 2 38.5 40.2 ‐1.7 3 36.9 37.4 ‐0.5 4 41.4 46.1 ‐4.7 5 40.6 43.9 ‐3.3 6 42.3 41.9 0.4 7 37.2 39.9 ‐2.7 8 38.6 40.4 ‐1.8 9 38.5 38.6 ‐0.1 10 38.4 38.1 0.3 11 38.1 39.5 ‐1.4 a. Use the QQplot assess whether the normality assumption has been met. Correct: Note, this is the dependent samples setting, so we need to look at the QQplot of the differences. Even though it is slight, there is a systematic departure from the line. The data points are too low at both the low and high ends end of this distribution to come from a normal population. This is essentially, a skewed left distribution. Also, the CLT won't help us here since there are only 11 data points. This is a probable violation of the assumption of normality. a. Use the appropriate test to test the null hypothesis of no difference against the alternative that alcoholism reduces brain density. Let α = 0.05. Correct: (1) α = 0.05 (2) H0: Brain density is not different between alcoholics and healthy individuals HA: Alcoholics tend to have lower brain density (3) N+ = 2, N‐ = 9, Bs = 9 (4) nd = 11 TI‐84 P = Pr{B ≥ 9} = 1 – Pr{B ≤ 8} = 1 – binomcdf(11, 0.5, 8) = 0.0327 (5) P < α, so reject H0 (6) There is significant evidence to conclude that alcoholics tend to have lower brain density. b. Suppose the researcher for this study only wanted to see if there was a significant difference in brain density (i.e. a non‐directional alternative). What would the P‐value be for this test? Correct: P = 2(0.0327) = 0.0654 ...
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This note was uploaded on 12/13/2011 for the course STAT 205 taught by Professor Hendrix during the Fall '09 term at South Carolina.

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