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Cushing's disease is characterized by muscular weakness due to adrenal or pituitary dysfunction. To provide effective treatment, it is important to detect childhood Cushing's disease as early as possible. Age at onset of symptoms and age at diagnosis (months) for 15 children suffering from the disease were given in the article "Treatment of Cushing's Disease in Childhood and Adolescence by Transphenoidal Microadenomectomy" (New Engl. J. of Med., 1984:889). Here are the values of the differences between age at onset of symptoms and age at diagnosis: \(\begin{array}{llllllll}-24 & -12 & -55 & -15 & -30 & -60 & -14 & -21 \\ -48 & -12 & -25 & -53 & -61 & -69 & -80 & \end{array}\) a. Does the accompanying normal probability plot cast strong doubt on the approximate normality of the population distribution of differences? b. Calculate a lower \(95 \%\) confidence bound for the population mean difference, and interpret the resulting bound. c. Suppose the (age at diagnosis) - (age at onset) differences had been calculated. What would be a \(95 \%\) upper confidence bound for the corresponding population mean difference?

Step by step solution

01

Calculate Sample Mean and Standard Deviation

First, we need to calculate the sample mean and sample standard deviation of the differences.Given the differences: \([-24, -12, -55, -15, -30, -60, -14, -21, -48, -12, -25, -53, -61, -69, -80]\).Calculate the sample mean (\(\bar{x}\)):\(\bar{x} = \frac{\sum x_i}{n} = \frac{-24 + (-12) + (-55) + \ldots + (-80)}{15} = -41.13\)Next, calculate the sample standard deviation (\(s\)):\(s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\).Substituting the values, \(s \approx 20.29\).

02

Conduct Normal Probability Plot Analysis

For question (a), a normal probability plot can be visually assessed to see if the points approximate a straight line. If the plot provided seems relatively linear, we conclude there is no strong evidence against normality. Without the plot here, let's assume it's linear for theoretical analysis.

03

Calculate Lower 95% Confidence Bound

For question (b), calculate the lower 95% confidence bound for the mean.The formula for the lower bound of the confidence interval for a mean is:\[\bar{x} - t_{\alpha,n-1} \times \frac{s}{\sqrt{n}}\]Use a t-table or calculator to find \(t_{0.05,14} \approx 1.761\) (1-tailed t-value).Substitute the values:Lower bound = \(-41.13 - 1.761 \times \frac{20.29}{\sqrt{15}} \approx -48.87\).Interpreting this, we are 95% confident that the population mean difference is greater than -48.87 months.

04

Calculate Upper 95% Confidence Bound for Opposite Difference

For question (c), consider swapping the age difference to (age at diagnosis) - (age at onset).This reverses the sign of each difference, so the mean becomes positive: Mean = 41.13.The upper bound for this new situation is:\[\bar{x} + t_{\alpha,n-1} \times \frac{s}{\sqrt{n}}\]Substitute the values:Upper bound = \(41.13 + 1.761 \times \frac{20.29}{\sqrt{15}} \approx 48.39\).We are 95% confident that the population mean difference is less than 48.39 months.

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Key Concepts

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

Normal Probability Plot

A normal probability plot is a graphical tool used to assess whether a data set follows a normal distribution. The plot compares ordered data values against expected values from a normal distribution. Ideally, if the set of data is normally distributed, the points on the plot will roughly form a straight line.

This visual method is straightforward and a powerful initial check for normality. If your points stray far from the line or create a noticeable curve, this might suggest that the data does not follow a normal distribution. However, some deviation is natural, especially with small sample sizes.

In the case of Cushing's disease age differences data, the exercise asks whether the normal probability plot suggests normality. A linear pattern in such a plot would imply no strong evidence against the normality of the distribution of age differences.

Confidence Interval

The confidence interval provides a range of values that likely cover the true population parameter, such as the mean. When you create a confidence interval, you're expressing a degree of uncertainty about the parameter based on sample data.

For a 95% confidence interval, you can say that if you were to take 100 different samples and compute a confidence interval for each, about 95 of them would contain the true population parameter. This makes confidence intervals very useful in medicine and research to understand potential ranges for a mean.

In the scenario regarding the childhood onset of Cushing's disease, a 95% confidence interval was computed for the population mean difference in ages. If the exercise asks for a lower or upper confidence bound, it specifically refers to either just the lower limit or the upper limit of the interval, not both.

Sample Mean and Standard Deviation

The sample mean and standard deviation are fundamental statistics in any sort of analysis. They summarize the central tendency and dispersion of a dataset, respectively.

The sample mean, denoted as \(\bar{x}\), represents the average of all data values. It is calculated by adding up all the individual values and dividing by the number of values. In the example given, a mean difference of \(-41.13\) months was computed.

The sample standard deviation \(s\), measures how much each data point deviates from the mean on average. A smaller standard deviation means the data points are close to the mean, whereas a larger one indicates more spread out data. In the context of childhood Cushing's disease, the standard deviation of about \(20.29\) indicates varied lengths in diagnosis gap times.

These calculations are pivotal for further statistical analyses such as making confidence intervals or conducting hypothesis tests.