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 1.9 for 12 sexually abused students ​(sequals2.4​) and 4.9 for 23 nonabused students ​(sequals3.3​)

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 tackle the problem step by step:

### Part 1: Calculate the Standard Error

Given dataLet's tackle the problem step by step:

### Part 1: Calculate the Standard Error

Given 
- Mean family cohesion for sexually abused students (\( \bar{x}_S \)): 1.9
- Standard deviation for sexually abused students (\( s_S \)): 2.4
- Sample size for sexually abused students (\( n_S \)): 12

- Mean family cohesion for nonabused students (\( \bar{x}_N \)): 4.9
- Standard deviation for nonabused students (\( s_N \)): 3.3
- Sample size for nonabused students (\( n_N \)): 23

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:

\[ SE = \sqrt{\left(\frac{2.4^2}{12}\right) + \left(\frac{3.3^2}{23}\right)} \]
\[ SE = \sqrt{\left(\frac{5.76}{12}\right) + \left(\frac{10.89}{23}\right)} \]
\[ SE = \sqrt{0.48 + 0.4735} \]
\[ SE = \sqrt{0.9535} \]
\[ SE \approx 0.9765 \]

So, the standard error is approximately **0.9765**.

### Part 2: Calculate the Margin of Error

For a 95% confidence interval, we use the critical value (\( t^* \)) for the appropriate degrees of freedom. Since degrees of freedom can be approximated using a conservative approach (using the smaller of \( n_S - 1 \) and \( n_N - 1 \)) or using a more precise method (like the Welch-Satterthwaite equation), we'll use a conservative approach here.

Degrees of freedom:
\[ df = \min(n_S - 1, n_N - 1) = \min(12 - 1, 23 - 1) = 11 \]

The \( t \)-value for 95% confidence and 11 degrees of freedom is approximately 2.201.

The margin of error (ME) is given by:
\[ ME = t^* \times SE \]
\[ ME = 2.201 \times 0.9765 \]
\[ ME \approx 2.1497 \]

So, the margin of error is approximately **2.1497**.

### 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:
\[ \bar{x}_S - \bar{x}_N = 1.9 - 4.9 = -3.0 \]

So, the 95% confidence interval is:
\[ -3.0 \pm 2.1497 \]

Which gives us:
\[ -3.0 - 2.1497 = -5.15 \]
\[ -3.0 + 2.1497 = -0.85 \]

Therefore, the 95% confidence interval for the difference between the mean family cohesion scores of sexually abused and nonabused students is **from -5.15 to -0.85**.

### Interpretation

This confidence interval suggests that the true difference in mean family cohesion scores between sexually abused and nonabused students is between -5.15 and -0.85. Since the interval does not include zero, we can infer that there is a significant difference between the two groups, with nonabused students having higher family cohesion scores on average.

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### Related Questions

1. What is the formula for calculating the standard error for the difference between two means?
2. How do you determine the degrees of freedom for a two-sample \( t \)-test?
3. What critical value should be used for a 95% confidence interval with 11 degrees of freedom?
4. Why is it important to round the standard error and margin of error to four decimal places?
5. How would the confidence interval change if the sample sizes were larger?
6. What assumptions are necessary for the \( t \)-test to be valid in this context?
7. How do different levels of confidence affect the margin of error?
8. What are potential sources of bias in this study?

### Tip

Always check the assumptions of normality and equal variances when conducting a two-sample \( t \)-test to ensure the results are valid.

Learn how to calculate the standard error, margin of error, and construct a 95% confidence interval for the difference between mean family cohesion scores of sexually abused and nonabused college students. Detailed step-by-step solution with formulas and interpretations provided.

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