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Chapter 6: Problem 187

Survival in the ICU and Infection In the dataset ICUAdmissions, the variable Status indicates whether the ICU (Intensive Care Unit) patient lived (0) or died \((1),\) while the variable Infection indicates whether the patient had an infection ( 1 for yes, 0 for no ) at the time of admission to the ICU. Use technology to find a \(95 \%\) confidence interval for the difference in the proportion who die between those with an infection and those without.

Short Answer

Expert verified
The confidence interval will be relayed as \(d \pm (1.96 * SE)\), where \(d\) is the difference in sample proportions and SE is the calculated standard error. Actual numeric answer depends on the dataset.

Step by step solution

01

Identify the Data

First, from the ICUAdmissions dataset, two pieces of data are needed: the proportion of patients who died with an infection and the proportion of patients who died without an infection. Let's denote these proportions as \(p1\) and \(p2\) respectively.
02

Calculate the Sample Proportions

Calculate the sample proportions \(p1\) and \(p2\) by dividing the number of patients who died in each category (with infection and without infection) by the total number of patients in that category. If the dataset provides frequencies, use them to calculate proportions. If not, use descriptive statistics provided.
03

Calculate the Standard Error

The standard error (SE) for the difference in sample proportions is calculated as \(\sqrt{p1(1 - p1) / n1 + p2(1 - p2) / n2}\) Where \(n1\) and \(n2\) are the sample sizes for those with infection and without an infection respectively.
04

Find the Difference in Sample Proportions

The difference in sample proportions (\(d\)) is simply \(d = p1 - p2\)
05

Calculate the Confidence Interval

A \(95 \%\) confidence interval for the difference in proportions is calculated as \(d \pm (1.96 * SE)\), using the critical value \(1.96\) for a \(95 \%\) confidence interval in a normal distribution.

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Key Concepts

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

ICU Survival Analysis
ICU Survival Analysis is crucial for understanding patient outcomes in the Intensive Care Unit. This analysis involves examining various factors, such as the presence of infections, which might impact the survival rates of ICU patients. In this context, survival means whether patients lived after treatment in the ICU or not, represented by binary data: '0' for those who lived and '1' for those who died.
A common approach is to use statistical techniques to assess the impact of different variables on survival. One such technique is the use of confidence intervals. Here, we calculate the confidence interval for differences in survival proportions between two groups: those with infections and those without. This helps healthcare professionals identify trends and potentially harmful factors, allowing for better treatment and care strategies. Evaluating ICU survival rates helps to improve patient care and outcomes by spotting potential issues and addressing them more effectively.
Proportion Difference
Proportion Difference is an essential concept when comparing outcomes between two groups. Here, we are interested in the difference in death rates among ICU patients with and without infections. Calculating proportion difference involves determining the proportion of patients who died in each group and subtracting one from the other.
First, identify the number of ICU patients who died with an infection and without. Calculate the proportion, denoted as \( p_1 \) and \( p_2 \). This is done by dividing the number of deaths by the total number of patients in each group:
  • \( p_1 \): Proportion of deaths with infection = (Number of deaths with infection / Total patients with infection).
  • \( p_2 \): Proportion of deaths without infection = (Number of deaths without infection / Total patients without infection).
Then, find the difference between these proportions: \( d = p_1 - p_2 \). This difference gives an understanding of how much more or less prevalent the outcome is in the infection group compared to the non-infection group, which can be pivotal in clinical assessments and decision-making.
Standard Error Calculation
The concept of Standard Error (SE) Calculation is key to measuring the precision of an estimated proportion difference. In statistical analysis, the SE provides an estimate of the variability of a sample statistic. Specifically here, it helps measure how much the calculated difference between the two proportions \( d \) would vary in repeated samples.
To calculate the SE for the difference in sample proportions, use the following formula:
\[ SE = \sqrt{ \frac{p_1(1 - p_1)}{n_1} + \frac{p_2(1 - p_2)}{n_2}} \]
Where:
  • \( p_1 \) and \( p_2 \) are the sample proportions of deaths with and without infection, respectively.
  • \( n_1 \) and \( n_2 \) are the respective sample sizes.
The SE quantifies uncertainty around the difference in proportions, providing a basis for constructing confidence intervals. By calculating the SE, we can determine the confidence interval for the difference in proportions, allowing healthcare practitioners to make informed decisions based on reliable data predictions.

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Most popular questions from this chapter

Standard Error from a Formula and Simulation In Exercises 6.15 to \(6.18,\) find the mean and standard error of the sample proportions two ways: (a) Use StatKey or other technology to simulate at least 1000 sample proportions. Give the mean and standard error and comment on whether the distribution appears to be normal. (b) Use the formulas in the Central Limit Theorem to compute the mean and standard error. Are the results similar to those found in part (a)? Sample proportions of sample size \(n=100\) from a population with \(p=0.4\)

In each case below, two sets of data are given for a two-tail difference in means test. In each case, which version gives a smaller \(\mathrm{p}\) -value relative to the other? (a) Both options have the same standard deviations and same sample sizes but: \(\begin{array}{lll}\text { Option 1 has: } & \bar{x}_{1}=25 & \bar{x}_{2}=23\end{array}\) Option 2 has: \(\quad \bar{x}_{1}=25 \quad \bar{x}_{2}=11\) (b) Both options have the same means \(\left(\bar{x}_{1}=22,\right.\) \(\left.\bar{x}_{2}=17\right)\) and same sample sizes but: Option 1 has: \(\quad s_{1}=15 \quad s_{2}=14\) $$ \text { Option 2 has: } \quad s_{1}=3 \quad s_{2}=4 $$ (c) Both options have the same means \(\left(\bar{x}_{1}=22,\right.\) \(\bar{x}_{2}=17\) ) and same standard deviations but: Option 1 has: \(\quad n_{1}=800 \quad n_{2}=1000\) Option 2 has: \(\quad n_{1}=25 \quad n_{2}=30\)

Random samples of the given sizes are drawn from populations with the given means and standard deviations. For each scenario: (a) Find the mean and standard error of the distribution of differences in sample means, \(\bar{x}_{1}-\bar{x}_{2}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 100 from Population 1 with mean 87 and standard deviation 12 and samples of size 80 from Population 2 with mean 81 and standard deviation 15

Data 1.3 on page 10 discusses a study designed to test whether applying a metal tag is detrimental to a penguin, as opposed to applying an electronic tag. One variable examined is the date penguins arrive at the breeding site, with later arrivals hurting breeding success. Arrival date is measured as the number of days after November 1st. Mean arrival date for the 167 times metal- tagged penguins arrived was December 7 th ( 37 days after November 1 st ) with a standard deviation of 38.77 days, while mean arrival date for the 189 times electronic-tagged penguins arrived at the breeding site was November 21 st (21 days after November 1 st ) with a standard deviation of \(27.50 .\) Do these data provide evidence that metal-tagged penguins have a later mean arrival time? Show all details of the test.

Refer to a study on hormone replacement therapy. Until 2002 , hormone replacement therapy (HRT), taking hormones to replace those the body no longer makes after menopause, was commonly prescribed to post-menopausal women. However, in 2002 the results of a large clinical trial \(^{56}\) were published, causing most doctors to stop prescribing it and most women to stop using it, impacting the health of millions of women around the world. In the experiment, 8506 women were randomized to take HRT and 8102 were randomized to take a placebo. Table 6.16 shows the observed counts for several conditions over the five years of the study. (Note: The planned duration was 8.5 years. If Exercises 6.205 through 6.208 are done correctly, you will notice that several of the p-values are just below \(0.05 .\) The study was terminated as soon as HRT was shown to significantly increase risk (using a significance level of \(\alpha=0.05)\), because at that point it was unethical to continue forcing women to take HRT). Does HRT influence the chance of a woman having a fracture? $$ \begin{array}{lcc} \hline \text { Condition } & \text { HRT Group } & \text { Placebo Group } \\ \hline \text { Cardiovascular Disease } & 164 & 122 \\ \text { Invasive Breast Cancer } & 166 & 124 \\ \text { Cancer (all) } & 502 & 458 \\ \text { Fractures } & 650 & 788 \\ \hline \end{array} $$
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