Answer:
a. The 95% confidence interval for the difference in mean is C.I. = -3.6 < μ₂ - μ₁ < 4.96
B. We are 95% confident that the difference in mean FNE scores for bulimic and normal students is inside the confidence interval
c. The assumptions made are;
The variance of the two distributions are equal
Step-by-step explanation:
The given parameters are;
The mean for the 11 students with an eating disorder,
= 13.82
The standard deviation for the 11 students with an eating disorder, s₁ = 4.92
The mean for the 14 students who do not have an eating disorder,
= 13.14
The standard deviation for the 14 students with an eating disorder, s₂ = 5.29
a. The 95% confidence interval for the difference in mean is given as follows;
![\left (\bar{x}_{1}- \bar{x}_{2} \right )\pm t_{\alpha /2}\sqrt{ \hat \sigma^2 \times\left( \dfrac{1}{n_{1}}+\dfrac{1}{n_{2}} \right)}](https://tex.z-dn.net/?f=%5Cleft%20%28%5Cbar%7Bx%7D_%7B1%7D-%20%5Cbar%7Bx%7D_%7B2%7D%20%20%5Cright%20%29%5Cpm%20t_%7B%5Calpha%20%2F2%7D%5Csqrt%7B%20%5Chat%20%5Csigma%5E2%20%5Ctimes%5Cleft%28%20%5Cdfrac%7B1%7D%7Bn_%7B1%7D%7D%2B%5Cdfrac%7B1%7D%7Bn_%7B2%7D%7D%20%5Cright%29%7D)
The pooled standard deviation, is therefore;
![\hat{\sigma} =\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}](https://tex.z-dn.net/?f=%5Chat%7B%5Csigma%7D%20%3D%5Csqrt%7B%5Cdfrac%7B%5Cleft%20%28%20n_%7B1%7D-1%20%5Cright%20%29%5Ccdot%20s_%7B1%7D%5E%7B2%7D%20%2B%5Cleft%20%28%20n_%7B2%7D-1%20%5Cright%20%29%5Ccdot%20s_%7B2%7D%5E%7B2%7D%7D%7Bn_%7B1%7D%2Bn_%7B2%7D-2%7D%7D)
Therefore;
![\hat{\sigma} =\sqrt{\dfrac{\left ( 11-1 \right )\cdot4.92^{2} +\left ( 14-1 \right )\cdot 5.29^{2}}{11+14-2}} \approx 5.13241](https://tex.z-dn.net/?f=%5Chat%7B%5Csigma%7D%20%3D%5Csqrt%7B%5Cdfrac%7B%5Cleft%20%28%2011-1%20%5Cright%20%29%5Ccdot4.92%5E%7B2%7D%20%2B%5Cleft%20%28%2014-1%20%5Cright%20%29%5Ccdot%205.29%5E%7B2%7D%7D%7B11%2B14-2%7D%7D%20%5Capprox%205.13241)
Where at degrees of freedom, df = n₁ + n₂ - 2 = 25 - 2 = 23 the critical-t = 2.07
\left (13.82- 13.14 \right )\pm 2.07 \times\sqrt{5.13241^2 *(\dfrac{1}/{11}+\dfrac{1}{14}\right)}
![C.I. = \left (13.82- 13.14 \right )\pm 2.07 \times\sqrt{5.13241^2 \times\left(\dfrac{1}{11}+\dfrac{1}{14}\right)}](https://tex.z-dn.net/?f=C.I.%20%3D%20%5Cleft%20%2813.82-%2013.14%20%20%5Cright%20%29%5Cpm%202.07%20%5Ctimes%5Csqrt%7B5.13241%5E2%20%5Ctimes%5Cleft%28%5Cdfrac%7B1%7D%7B11%7D%2B%5Cdfrac%7B1%7D%7B14%7D%5Cright%29%7D)
Therefore, we get;
![C.I. = 0.68\pm 4.28](https://tex.z-dn.net/?f=C.I.%20%3D%200.68%5Cpm%204.28)
C.I. = -3.6 < μ₂ - μ₁ < 4.96
b. Therefore, given that the confidence interval extends from positive to negative, therefore, there is a possibility that there is no difference between the mean FNE scores for bulimic and normal students
B. We are 95% confident that the difference in mean FNE scores for bulimic and normal students is inside the confidence interval
c. The assumptions made are;
The variance of the two distributions are equal