A study of bulimia among college women studied the connection between childhood sexual abuse and a measure of family cohesion​ (the higher the​ score, the greater the​ cohesion). The sample mean on the family cohesion scale was 

2.02.0 

for 

1111 

sexually abused students 

​(sequals=2.42.4​) 

and 

4.54.5 

for 

1919 

nonabused students 

​(sequals=3.73.7​)
The standard error is
  

​(Round to four decimal places as​ needed.)
Part 2
The margin of error is
  

​(Round to four decimal places as​ needed.)
Part 3
b. Construct a​ 95% confidence interval for the difference between the mean family cohesion for sexually abused students ​(mu Subscript S​) and nonabused students ​(mu Subscript N​). Interpret.
The​ 95% confidence interval for ​(mu Subscript Sminusmu Subscript N​) is from
  

​(Round to two decimal places as needed. Use ascending​ order.)




Let's recalculate the values with the new data provided.

### Part 1: Calculate the Standard Error

Given dataLet's recalculate the values with the new data provided.

### Part 1: Calculate the Standard Error

Given 
- Mean family cohesion for sexually abused students (\( \bar{x}_S \)): 2.0
- Standard deviation for sexually abused students (\( s_S \)): 2.4
- Sample size for sexually abused students (\( n_S \)): 11

- Mean family cohesion for nonabused students (\( \bar{x}_N \)): 4.5
- Standard deviation for nonabused students (\( s_N \)): 3.7
- Sample size for nonabused students (\( n_N \)): 19

The standard error of the difference between two means is calculated using the formula:

\[ SE = \sqrt{\left(\frac{s_S^2}{n_S}\right) + \left(\frac{s_N^2}{n_N}\right)} \]

Plugging in the values:

\[ SE = \sqrt{\left(\frac{2.4^2}{11}\right) + \left(\frac{3.7^2}{19}\right)} \]
\[ SE = \sqrt{\left(\frac{5.76}{11}\right) + \left(\frac{13.69}{19}\right)} \]
\[ SE = \sqrt{0.5236 + 0.7205} \]
\[ SE = \sqrt{1.2441} \]
\[ SE \approx 1.1154 \]

So, the standard error is approximately **1.1154**.

### Part 2: Calculate the Margin of Error

For a 95% confidence interval, we use the critical value (\( t^* \)) for the appropriate degrees of freedom. Using a conservative approach, we'll find the degrees of freedom:

Degrees of freedom:
\[ df = \min(n_S - 1, n_N - 1) = \min(11 - 1, 19 - 1) = 10 \]

The \( t \)-value for 95% confidence and 10 degrees of freedom is approximately 2.228.

The margin of error (ME) is given by:
\[ ME = t^* \times SE \]
\[ ME = 2.228 \times 1.1154 \]
\[ ME \approx 2.4854 \]

So, the margin of error is approximately **2.4854**.

### Part 3: Construct the 95% Confidence Interval

The confidence interval for the difference between the means is given by:
\[ (\bar{x}_S - \bar{x}_N) \pm ME \]

The difference between the means:
\[ \bar{x}_S - \bar{x}_N = 2.0 - 4.5 = -2.5 \]

So, the 95% confidence interval is:
\[ -2.5 \pm 2.4854 \]

Which gives us:
\[ -2.5 - 2.4854 = -4.99 \]
\[ -2.5 + 2.4854 = -0.01 \]

Therefore, the 95% confidence interval for the difference between the mean family cohesion scores of sexually abused and nonabused students is **from -4.99 to -0.01**.

### Interpretation

This confidence interval suggests that the true difference in mean family cohesion scores between sexually abused and nonabused students is between -4.99 and -0.01. Since the interval includes zero, we cannot definitively conclude that there is a significant difference between the two groups, but there is a strong indication that nonabused students might have higher family cohesion scores on average.

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### Related Questions

1. What assumptions must be met for a two-sample \( t \)-test to be valid?
2. How does increasing the sample size affect the standard error?
3. Why is it important to round the standard error and margin of error to four decimal places?
4. How would the confidence interval change if we used a 99% confidence level instead of 95%?
5. What are the implications if the confidence interval includes zero?
6. How can we interpret the family cohesion scores in terms of their impact on the well-being of the students?
7. What other factors could influence family cohesion scores apart from childhood sexual abuse?
8. How do we calculate the critical value (\( t^* \)) for different confidence levels and degrees of freedom?

### Tip

Always verify the assumptions of normality and equality of variances before applying a two-sample \( t \)-test to ensure the results are accurate and reliable.

Learn how to calculate the standard error, margin of error, and construct a 95% confidence interval for the difference in mean family cohesion scores between sexually abused and nonabused college students. This problem involves statistical analysis with specific data points and formulas.

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