Question

The​ "fear of negative​ evaluation" (FNE) scores for 11 random female students known to have an eating disorder (1) and 14 random female students who do not have an eating disorder (2) are collected.​ The higher the​ score, the greater is the fear of a negative​ evaluation.
Bulimic Students: 16, 8, 5, 15, 20, 14, 11, 16, 19, 8, 9
Normal Students: 10, 3, 13, 10, 5, 16, 20, 15, 8, 16, 4, 7, 12, 17

Assume the scores are normally distributed. The mean for the 11 students with an eating disorder is 13.82 and standard deviation 4.92. The mean for the 14 students who do not have an eating disorder is 13.14 and standard deviation 5.29.

Required:
a. Find a 95% confidence interval for the difference between the population means of the FNE scores for bulimic and normal female students. (Round to two decimal places as needed.)
b. Interpret the result of the confidence interval.

A. We are 95% confident that the difference in mean FNE scores for bulimic and normal students is outside the confidence interval.
B. We are 95% confident that the difference in mean FNE scores for bulimic and normal students is inside the confidence interval.
C. We are 5% confident that the difference in mean FNE scores for bulimic and normal students is inside the confidence interval.
D. We are 5% confident that the difference in mean FNE scores for bulimic and normal students is outside the confidence interval.

c. What assumptions are required for the interval of part (a) to be statistically valid? Select all that apply.

Answer 1

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, $\overline x_1$ = 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, $\overline x_2$ = 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;

$\left (\bar{x}_{1}- \bar{x}_{2} \right )\pm t_{\alpha /2}\sqrt{ \hat \sigma^2 \times\left( \dfrac{1}{n_{1}}+\dfrac{1}{n_{2}} \right)}$

The pooled standard deviation, is therefore;

$\hat{\sigma} =\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}$

Therefore;

$\hat{\sigma} =\sqrt{\dfrac{\left ( 11-1 \right )\cdot4.92^{2} +\left ( 14-1 \right )\cdot 5.29^{2}}{11+14-2}} \approx 5.13241$

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)}

$C.I. = \left (13.82- 13.14 \right )\pm 2.07 \times\sqrt{5.13241^2 \times\left(\dfrac{1}{11}+\dfrac{1}{14}\right)}$

Therefore, we get;

$C.I. = 0.68\pm 4.28$

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