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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 95% confidence interval for the difference in the proportion of deaths between those with an infection and those without, is given by \([d - 1.96*SE, d + 1.96*SE]\), where \(d\) is the difference in sample proportions and \(SE\) is the standard error.

Step by step solution

01

Identify Sample Sizes and Proportions

First, from the dataset 'ICUAdmissions', identify the number of people in each group (with infection and without infection) and the number of deaths in each group. These will be your sample sizes and observed deaths.
02

Calculate Sample Proportions

Next, calculate the proportion of deaths in each group by dividing the number of deaths by the total number in each group. Let's denote the proportion of deaths in infected group as \(P1\) and in the non-infected group as \(P2\). Then find the difference in sample proportions, \(d = P1 - P2\).
03

Compute Standard Error

The standard error \((SE)\) for the difference in sample proportions is calculated using the formula \(SE = \sqrt{[P1*(1 - P1) / n1] + [P2*(1 - P2) / n2]}\), where \(n1\) and \(n2\) are the sample sizes of the infected and non-infected groups respectively.
04

Calculate Confidence Interval

The formula for a 95% confidence interval for the difference in proportions is \([d - 1.96*SE, d + 1.96*SE]\). Substitute the values of \(d\) and \(SE\) in the formula to get the 95% confidence interval for the difference in the proportions of deaths between the group with infection and the group without infection.

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Key Concepts

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

Sample Proportion
The concept of sample proportion is fundamental in understanding how data from a sample can be used to make inferences about a larger population. A sample proportion is simply the fraction of the sample that has a particular attribute. In the context of our problem, the sample proportion is the number of ICU patients who died divided by the total number of patients in the group. This can be determined separately for patients with an infection and those without. For example, if out of 100 infected patients, 30 died, the sample proportion of deaths among infected patients is \( P1 = \frac{30}{100} = 0.30 \). Similarly, the proportion can be calculated for non-infected patients. The sample proportion is an estimate of the true population proportion, and thus provides an essential basis for further statistical analysis, such as determining the confidence interval.
Standard Error
Standard error is a measure that reflects the variability or dispersion of sample proportion estimates from the true population proportion. It tells us how much the sample proportion \( P1 \) might differ if different samples were taken. When dealing with proportions, the formula for standard error is \( SE = \sqrt{\frac{P(1-P)}{n}} \), where \( P \) is the sample proportion and \( n \) is the sample size. In the context of comparing two proportions, such as in the case of infected and non-infected ICU patients, the standard error of the difference in proportions accounts for the variability in both groups. Hence, it combines the variabilities into one formula: \( SE = \sqrt{\frac{P1(1-P1)}{n1} + \frac{P2(1-P2)}{n2}} \), helping us measure how much the sample proportions \( P1 \) and \( P2 \) might differ from each other.
Difference in Proportions
Difference in proportions is a statistic that measures the disparity between two sample proportions. For our ICU data, it represents the difference between the proportions of deceased patients who were infected and those who were not. Calculating this difference involves subtracting the proportion of deaths in the non-infected group from the proportion of deaths in the infected group: \( d = P1 - P2 \). This difference can tell us whether there is a substantial discrepancy between the two groups, or if any observed difference might simply be due to random sample variability. Understanding how significant this difference is often involves further statistical analysis, such as confidence interval estimation and hypothesis testing, providing evidence for any conclusions.
Statistical Hypothesis Testing
Statistical hypothesis testing is a method used to make inferences or draw conclusions about a population parameter based on sample data. When comparing the proportions of two groups, hypothesis testing allows us to determine whether any observed difference in proportions is statistically significant or simply due to chance. In the context of our ICU study, we might establish a null hypothesis such as "there is no difference in death proportions between infected and non-infected patients." Using the calculated difference \( d \) and the accompanying standard error, we can perform a test, such as a z-test, to assess the null hypothesis. The result of this test helps us decide whether to reject the null hypothesis in favor of an alternative hypothesis, indicating a statistically significant difference between the groups at a certain confidence level, such as 95%.

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Most popular questions from this chapter

Does Red Increase Men's Attraction to Women? Exercise 1.99 on page 44 described a study \(^{46}\) which examines the impact of the color red on how attractive men perceive women to be. In the study, men were randomly divided into two groups and were asked to rate the attractiveness of women on a scale of 1 (not at all attractive) to 9 (extremely attractive). Men in one group were shown pictures of women on a white background while the men in the other group were shown the same pictures of women on a red background. The results are shown in Table 6.14 and the data for both groups are reasonably symmetric with no outliers. To determine the possible effect size of the red background over the white, find and interpret a \(90 \%\) confidence interval for the difference in mean attractiveness rating.

Can Malaria Parasites Control Mosquito Behavior? Are malaria parasites able to control mosquito behavior to their advantage? A study \(^{43}\) investigated this question by taking mosquitos and giving them the opportunity to have their first "blood meal" from a mouse. The mosquitoes were randomized to either eat from a mouse infected with malaria or an uninfected mouse. At several time points after this, mosquitoes were put into a cage with a human and it was recorded whether or not each mosquito approached the human (presumably to bite, although mosquitoes were caught before biting). Once infected, the malaria parasites in the mosquitoes go through two stages: the Oocyst stage in which the mosquito has been infected but is not yet infectious to others and then the Sporozoite stage in which the mosquito is infectious to others. Malaria parasites would benefit if mosquitoes sought risky blood meals (such as biting a human) less often in the Oocyst stage (because mosquitos are often killed while attempting a blood meal) and more often in the Sporozoite stage after becoming infectious (because this is one of the primary ways in which malaria is transmitted). Does exposing mosquitoes to malaria actually impact their behavior in this way? (a) In the Oocyst stage (after eating from mouse but before becoming infectious), 20 out of 113 mosquitoes in the group exposed to malaria approached the human and 36 out of 117 mosquitoes in the group not exposed to malaria approached the human. Calculate the Z-statistic. (b) Calculate the p-value for testing whether this provides evidence that the proportion of mosquitoes in the Oocyst stage approaching the human is lower in the group exposed to malaria. (c) In the Sporozoite stage (after becoming infectious), 37 out of 149 mosquitoes in the group exposed to malaria approached the human and 14 out of 144 mosquitoes in the group not exposed to malaria approached the human. Calculate the z-statistic. (d) Calculate the p-value for testing whether this provides evidence that the proportion of mosquitoes in the Sporozoite stage approaching the human is higher in the group exposed to malaria. (e) Based on your p-values, make conclusions about what you have learned about mosquito behavior, stage of infection, and exposure to malaria or not. (f) Can we conclude that being exposed to malaria (as opposed to not being exposed to malaria) causes these behavior changes in mosquitoes? Why or why not?

In Exercises 6.188 to 6.191 , use the t-distribution to find a confidence interval for a difference in means \(\mu_{1}-\mu_{2}\) given the relevant sample results. Give the best estimate for \(\mu_{1}-\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed. A \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the sample results \(\bar{x}_{1}=75.2, s_{1}=10.7, n_{1}=30\) and \(\bar{x}_{2}=69.0, s_{2}=8.3, n_{2}=20 .\)

When we want \(95 \%\) confidence and use the conservative estimate of \(p=0.5,\) we can use the simple formula \(n=1 /(M E)^{2}\) to estimate the sample size needed for a given margin of error ME. In Exercises 6.40 to 6.43, use this formula to determine the sample size needed for the given margin of error. A margin of error of 0.04 .

In Exercises 6.152 and \(6.153,\) find a \(95 \%\) confidence interval for the difference in proportions two ways: using StatKey or other technology and percentiles from a bootstrap distribution, and using the normal distribution and the formula for standard error. Compare the results. Difference in proportion who favor a gun control proposal, using \(\hat{p}_{f}=0.82\) for 379 out of 460 females and \(\hat{p}_{m}=0.61\) for 318 out of 520 for males. (We found a \(90 \%\) confidence interval for this difference in Exercise 6.144.)

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