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Although there are many treatments for bulimia nervosa, some subjects fail to benefit from treatment. In a study to determine which factors predict who will benefit from treatment, an article in the British Journal of Clinical Psychology indicates that self-esteem was one of these important predictors. The table gives the mean and standard deviation of self-esteem scores prior to treatment, at post treatment, and during a follow-up: $$ \begin{array}{lccc} & \text { Pretreatment } & \text { Posttreatment } & \text { Follow-up } \\ \hline \text { Sample Mean } \bar{x} & 20.3 & 26.6 & 27.7 \\ \text { Standard Deviation } s & 5.0 & 7.4 & 8.2 \\ \text { Sample Size } n & 21 & 21 & 20 \end{array} $$ a. Use a test of hypothesis to determine whether there is sufficient evidence to conclude that the true pretreatment mean is less than 25 . b. Construct a \(95 \%\) confidence interval for the true posttreatment mean. c. In Section 10.4 , we will introduce small-sample techniques for making inferences about the difference between two population means. Without the formality of a statistical test, what are you willing to conclude about the differences among the three sampled population means represented by the results in the table?

Step by step solution

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a. Hypothesis Test for Pretreatment Mean

To test whether the true pretreatment mean is less than 25, we'll perform a one-sample t-test. Step 1: State the null and alternative hypotheses. Null hypothesis (H0): $$\mu = 25$$ Alternative hypothesis (H1): $$\mu < 25$$ Step 2: Determine the sample mean, sample standard deviation, and sample size. $$\bar{x} = 20.3, \: s = 5.0, \: n = 21$$ Step 3: Calculate the t-statistic using the formula: $$t = \frac{\bar{x} - \mu}{s / \sqrt{n}}$$ Applying the values, $$t = \frac{20.3 - 25}{5.0 / \sqrt{21}} \approx -3.93$$ Step 4: Find the critical value and p-value. For a one-tailed test with 20 degrees of freedom (n-1) and significance level $$\alpha = 0.05$$, using a t-distribution table or calculator, the critical value is $$t_{\text{crit}} \approx -1.72$$. P-value can be found using a t-distribution table or calculator, which is approximately 0.0005. Step 5: Compare t-statistic to critical value and make a decision. Since the $$t-statistic < t_{\text{crit}} (-3.93 < -1.72)$$ and p-value < $$\alpha$$ (0.0005 < 0.05), we reject the null hypothesis in favor of the alternative hypothesis. Conclusion: There is sufficient evidence to conclude that the true pretreatment mean is less than 25.

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b. Confidence Interval for Posttreatment Mean

To construct a 95% confidence interval for the true posttreatment mean, we will use the following formula: $$CI = (\bar{x} \pm t_{\text{crit}} \times \frac{s}{\sqrt{n}})$$ where $$\bar{x} = 26.6$$, s = 7.4, and n = 21. From the t-distribution table for 20 degrees of freedom (n-1) and the $$95\%$$ confidence level ($$\alpha = 0.05$$), the critical value $$t_{\text{crit}} \approx 2.086$$. Applying the values, $$CI = (26.6 \pm 2.086 \times \frac{7.4}{\sqrt{21}}) \approx (23.945, 29.255)$$ Conclusion: The 95% confidence interval for the true posttreatment mean is approximately (23.945, 29.255).

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c. Inference About the Differences Among Samples

Without performing a formal statistical test, we can make an observation about the differences among the three sampled population means. Generally, the sample means increase from pretreatment ($$\bar{x} = 20.3$$) through posttreatment ($$\bar{x} = 26.6$$) to follow-up ($$\bar{x} = 27.7$$). This suggests that the self-esteem scores improve over time as the individuals undergo treatment and follow-up.

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Key Concepts

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

Hypothesis Testing

Hypothesis testing is a fundamental aspect of statistical inference. It allows us to make decisions about a population based on sample data. Here, the goal is to determine if the true pretreatment mean for self-esteem scores is less than 25.

• First, we set up two opposing hypotheses: the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_1\)). For this exercise, the null hypothesis is that the mean \(\mu\) equals 25, while the alternative hypothesis is that \(\mu\) is less than 25.
• We then compute the test statistic using the formula for the t-statistic: \(t = \frac{\bar{x} - \mu}{s / \sqrt{n}}\).
• The next step involves identifying the critical value and calculating the p-value. Both are used to determine if the observed data falls into the "rejection region" of the hypothesis test.
• In our example, with a calculated t-statistic of \,\-3.93, which is less than the critical value of \,\-1.72, and a p-value of 0.0005, we confidently reject the null hypothesis.
• This result suggests that there is enough evidence to support the claim that the true pretreatment mean is less than 25.

Understanding hypothesis testing is crucial as it helps make data-driven decisions without making assumptions based solely on sample statistics.

Confidence Interval

A confidence interval provides a range of values that is likely to contain the population parameter. This range gives us insights into the precision of our sample measurement.

In our exercise, we're constructing a 95% confidence interval for the posttreatment mean of self-esteem scores.

• We calculate the confidence interval using the formula: \(CI = (\bar{x} \pm t_{\text{crit}} \times \frac{s}{\sqrt{n}})\).
• The sample mean \(\bar{x}\) is 26.6, the standard deviation \(s\) is 7.4, and the sample size \(n\) is 21.
• To find \(t_{\text{crit}}\), we look for this value corresponding to 20 degrees of freedom (since \(n - 1 = 20\)) and a 95% confidence level in the t-distribution table, resulting in approximately 2.086.
• Substituting the numbers, we find the confidence interval to be approximately (23.945, 29.255).
• This interval portrays where we anticipate the true mean self-esteem score lies after treatment, with a 95% confidence level.

Confidence intervals are invaluable as they provide more information than a single point estimate, helping assess the reliability of the estimate.

t-Distribution

The t-distribution is a probability distribution commonly used in statistics for small sample sizes or when the population standard deviation is unknown. It is similar in shape to the normal distribution but has heavier tails.

• The t-distribution is particularly useful for constructing confidence intervals and conducting hypothesis tests when the sample size is small, as in our exercise with sample sizes of around 20.
• The concept of degrees of freedom is crucial here. It typically equals the sample size minus one ( $n - 1$). Degrees of freedom influence the exact shape of the t-distribution.
• In our exercise, with 20 degrees of freedom, we use the t-distribution to find critical values needed for both hypothesis testing and confidence intervals.
• The heavier tails in the t-distribution allow for greater variability that might occur by chance, making it apt for smaller samples.
• This adaptability ensures more reliable and accurate inferential results, even with less data.

The t-distribution is a flexible tool that ensures the validity of statistical testing and interval estimation when normal distribution assumptions aren't perfectly met.