Question

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 0.28 Control 8 0.36 0.16 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 WAN control Round your answers to two decimal places.) ka What does your estimate suggest about true average AN IAT relative to true average control IAT? O 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 only positive numbers, it is plausible that the true average AN IAT is greater than the true average control IAT are equal. Since this confidence interval includes zero, it is plausible that the true average AN IAT and true average control IAT are equal. 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 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 Fill to answer this question. Need Help? Read It Talk to a Tutor

Answer 1

(a) The CI for true average IAT under AN protocol (for unknown population SD) is given by:

X̄ ± t(α/2,n-1)*s/√n

where, X̄ is the sample mean of IAT under AN protocol =0.53 (given)

s - is the sample SD of IAT under AN protocol=0.28 (given)

n-is the sample size of AN protocol=16 (given)

t(α/2,n-1) is the two-sided t-value for degrees of freedom (n-1) i.e 16-1 i.e 15 and level of significance α.

For a 99% CI, α=1-0.99=0.01 => α/2=0.01/2=0.005

From t table, t(α/2,n-1)=t(0.005,15)=2.947

The 99% CI for true average IAT under AN protocol is given by:

X̄ ± t(α/2,n-1)*s/√n

=0.53 ± 2.947*0.28/√16

=0.53 ± 0.2063

i.e (0.3237,0.7363) Kg

i.e (0.32,0.74) Kg

Ans : (0.32,0.74) Kg

(b) The CI for the differences between true means of  IAT under AN protocol and control (for unknown population SDs) is given by:

(X̄1-X̄2) ± t(α/2,n1+n2-2)*√(s1^2/n1+s2^2/n2)

where, X̄1 is the sample mean of IAT under AN protocol =0.53 (given)

X̄2 is the sample mean of IAT under control  =0.36 (given)

s1 - is the sample SD of IAT under AN protocol=0.28 (given)

s2 - is the sample SD of IAT under control=0.16 (given)

n1-is the sample size of AN protocol=16 (given)

n2-is the sample size of control=8(given)

t(α/2,n1+n2-2) is the two-sided t-value for degrees of freedom (n1+n2-2) i.e 16+8-2 i.e 22 and level of significance α.

For a 99% CI, α=1-0.99=0.01 => α/2=0.01/2=0.005

From t table, t(α/2,n1+n2-2)=t(0.005,22)=2.819

The 99% CI for the differences between the means of AN protocol and control is given by:

(X̄1-X̄2) ± t(α/2,n1+n2-2)*√(s1^2/n1+s2^2/n2)

=(0.53-0.36) ± 2.819 * √(0.28^2/16+0.16^2/8)

=0.17± 0.2537

i.e ( -0.0837, 0.4237)

i.e ( -0.08, 0.42) kg

Ans: ( -0.08, 0.42) kg

The above interval for the differences contains zero implies that the differences is not significant and the hypothesis for equality of means of AN and control can't be rejected.

Ans: Since the confidence interval includes zero, it is plausible that the true average AN IAT and control IAT are equal.