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Chapter 10: Problem 21

Bulimia CI A study of bulimia among college women (J. Kern and T. Hastings, Journal of Clinical Psychology, vol. \(51,1995, \mathrm{p} .499)\) 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 2.0 for 13 sexually abused students \((s=2.1)\) and 4.8 for 17 nonabused students \((s=3.2)\). a. Find the standard error for comparing the means. b. Construct a \(95 \%\) confidence interval for the difference between the mean family cohesion for sexually abused students and non-abused students. Interpret.

Short Answer

Expert verified
The standard error is approximately 1.36, and the 95% confidence interval is (-5.58, 0.02), indicating no significant difference.

Step by step solution

01

Calculate Pooled Standard Deviation

The pooled standard deviation combines the standard deviations of two independent samples. It is given by:\[S_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}\]where \(s_1 = 2.1\) for sexually abused students and \(s_2 = 3.2\) for non-abused students. The sample sizes \(n_1 = 13\) and \(n_2 = 17\). Substitute the values:\[S_p = \sqrt{\frac{(13 - 1)(2.1)^2 + (17 - 1)(3.2)^2}{13 + 17 - 2}} \approx 2.78\]
02

Calculate Standard Error for the Difference in Means

Use the pooled standard deviation to calculate the standard error (SE) for the difference in means:\[SE = S_p \times \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}\]Substitute \(S_p = 2.78\), \(n_1 = 13\), and \(n_2 = 17\) to find:\[SE = 2.78 \times \sqrt{\frac{1}{13} + \frac{1}{17}} \approx 1.36\]
03

Calculate 95% Confidence Interval

To find the 95% confidence interval, use the formula:\[CI = (\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2, df} \times SE\]The means are \(\bar{x}_1 = 2.0\) and \(\bar{x}_2 = 4.8\). Calculate the mean difference:\[\bar{x}_1 - \bar{x}_2 = 2.0 - 4.8 = -2.8\]Using a t-table, find \(t_{0.025, 28} \approx 2.048\). Substitute values:\[CI = -2.8 \pm 2.048 \times 1.36 = (-5.58, 0.02)\]
04

Interpret the Confidence Interval

The confidence interval \((-5.58, 0.02)\) means that with 95% confidence, the difference in mean family cohesion scores ranges from -5.58 for sexually abused students being lower to 0.02. Since the interval includes zero, there might not be a statistically significant difference.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Deviation
Understanding the standard deviation helps in comprehending data variability. It indicates how spread out the values in a dataset are from the mean. The formula for standard deviation is:\[SD = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \]- **Higher Values**: Suggests that data points are more spread out.- **Lower Values**: Indicates that they are closer to the mean.In this problem, the standard deviations are 2.1 for sexually abused students and 3.2 for non-abused students. These values tell us that scores for non-abused students vary more from their group mean than those of abused students.
Standard Error
The standard error (SE) measures how much the sample mean of the data is expected to fluctuate from the true population mean. It's crucial for making inferences about population parameters from sample data.For a difference in means between two samples, SE is calculated as:\[SE = S_p \times \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} \]This formula utilizes the pooled standard deviation (\(S_p\)), which accounts for both groups' variability, and the sample sizes. In this exercise, the calculated SE was approximately 1.36, providing an estimate for the variability of the mean difference.
Pooled Standard Deviation
The pooled standard deviation is derived when you have two sets of data and wish to combine their variances in a weighted manner. This approach assumes both samples have similar variance.The formula is:\[S_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}} \]Calculating it involves two steps:- Multiply each sample's variance (standard deviation squared) by its sample size minus one.- Combine these products and divide by the total degrees of freedom.For the study, this resulted in a pooled standard deviation of 2.78, blending the variabilities from the two student groups.
Mean Difference
The mean difference is simply the subtraction of one group's mean from another's. It shows the gap or difference in average scores between the two groups.In the context of our study:\[\bar{x}_1 - \bar{x}_2 = 2.0 - 4.8 = -2.8 \]This negative value indicates that non-abused students typically report higher family cohesion than abused students. When interpreting, a confidence interval is often used to understand the precision of this estimate. The negative mean difference lends insight into possible impacts of abuse on family cohesion perceptions.

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Most popular questions from this chapter

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