378+567-142÷5=?456(325-541)÷32=?
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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;
The pooled standard deviation, is therefore;
Therefore;
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)}
Therefore, we get;
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