Question

HELP

Anorexia Nervosa (AN) is a psychiatric condition leading to substantial weight loss among women who are fearful of becoming fat. An article used whole-body magnetic resonance imagery to determine various tissue characteristics for both an AN sample of individuals who had undergone acute weight restoration and maintained their weight for a year and a comparable (at the outset of the study) control sample. Here is summary data on intermuscular adipose tissue (IAT; in kg). Condition Sample Size Sample Mean Sample SD AN 16 0.53 0.27 Control 8 0.36 0.17 Assume that both samples were selected from normal distributions. in USE SALT (a) Calculate an estimate for true average IAT under the described AN protocol, and do so in a way that conveys information about the reliability and precision of the estimation. Do this by calculating a 99% CI for the true average IAT under the described AN protocol. (Round your answers to two decimal places.) kg (b) Calculate an estimate for the difference between the true average AN IAT and true average control IAT, and do so in a way that conveys information about the reliability and precision of the estimation. (Use a 99% confidence interval for Man - M control. Round your answers to two decimal places.) kg What does your estimate suggest about true average AN IAT relative to true average control IAT? O Since this confidence interval includes zero, it is plausible that the true average AN IAT and true average control IAT are equal. O Since this confidence interval includes only positive numbers, it is plausible that the true average AN IAT is greater than the true average control IAT are equal. O Since this confidence interval includes only negative numbers, it is plausible that the true average AN IAT is less than the true average control IAT. Since this confidence interval includes only negative numbers, it is plausible that the true average AN IAT is greater than the true average control IAT are equal. O Since this confidence interval includes zero, it is plausible that the true average AN IAT is greater than the true average control IAT are equal. You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

Solution; (1-0)997 confidence interval for true average TAT under described & AN protocol.is, 6 (7. + y + 6 V va in 2) where, x = 0.53, D-1616 = 0.27. and 0.99 10 a 0.01 ) = 0.005 2 Za 20.005 2.58 2

X = 20 0.53 - 0:34 (2.58 Vho 0.53 0.1792 0.3558 0.36 Now, Y + 2 2 2 .2. 2 0.53+ 0.27 (2.58 Vio 3Meu 0.53 +10:1742 loq 0.70 lower limit 0.36 - cepper limiti 0.70 confidence interval for true average TAT under described AN protocal is (0.36, 0.70) b) (1-2) 991 confidence interval for difference between true average TAT and true average control IAT is 1(8-9) - Zvestor . (DY) +

where, ñ = 0.53, Y = 0.36, n = 16, 92=8 te = (0.27² = 0.0729 6(0-1732 - 00289. (1-x) = 0.99 = 220.01 Zą = soos.2.58 67 (x-7) Za ſo 0.0729 - (0.53-0.36) - 2.581 + 0.0289 8 0.17 0.2336 = -0.06. lower limit - -0.00 Now, 48-5)+72 Van (6-53-62307 2:58 Voor 10.0729 16 + 0.0289 8 0.10 upper limit 0.10. confidence interval for MAN-llcontrol is (-0.06,0:40). since estimate suggest that this conefidence interval includes zero, It is plaussible that the true average AN FAT and true average control IAT are equal.