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Surgery in the ICU and Gender In the dataset ICUAdmissions, the variable Service indicates whether the ICU (Intensive Care Unit) patient had surgery (1) or other medical treatment (0) and the variable Sex gives the gender of the patient \((0\) for males and 1 for females.) Use technology to test at a \(5 \%\) level whether there is a difference between males and females in the proportion of ICU patients who have surgery.

Short Answer

Expert verified
The final answer depends on the result of the hypothesis test. If the p-value is less than 0.025, there is a difference in proportions of males and females who had surgery. Otherwise, there's not enough evidence to conclude a difference.

Step by step solution

01

Formulate Hypotheses

First, formulate the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)). \(H_0\) would be that the proportions of males and females who had surgery are same and \(H_a\) would be that the proportions are not same. It can be expressed as \(H_0: p_m = p_f\) and \(H_a: p_m \neq p_f\), where \(p_m\) and \(p_f\) are the proportions of males and females who had surgery respectively.
02

Conduct Z test

Conduct a two-proportion Z-test. This involves calculating the pooled proportion and then the test statistic, which is the difference in sample proportions divided by the standard error. The level of significance is given as 0.05. After finding the z-score, refer to the standard z-table to find the p-value.
03

Decision Rule

Analyze the result. In a two-tail test, if the p-value is less than half of the level of significance (i.e. less than 0.025 in this case), we reject the null hypothesis, else we fail to reject.
04

Conclude

After conducting the test, make a conclusion. If the null hypothesis is rejected, then it can be concluded that there is a significant difference between the proportions of males and females who had surgery in the ICU. Otherwise, there is not enough evidence to conclude a difference.

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Key Concepts

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

Understanding the Two-Proportion Z-Test
The two-proportion Z-test is a statistical method used to determine if there is a significant difference between the proportions of two distinct groups. For example, in the context of ICU data analysis, we might compare the proportion of male patients who have undergone surgery to the proportion of female patients.

The calculation involves assessing the difference between the two sample proportions, considering the variation expected in sampling. The formula used is: \[Z = \frac{(p_1 - p_2)}{\text{SE}}\]where \(p_1\) and \(p_2\) are sample proportions, and SE is the standard error of the difference between proportions. The standard error incorporates the pooled sample proportion, \(p\), which is a weighted average of the two sample proportions, reflecting the combined variation of both groups.

Conducting a two-proportion Z-test involves these specific steps: calculating the test statistic (Z-score), comparing it to a critical value from the Z-table, and then using the corresponding p-value to interpret whether the results are statistically significant based on the chosen alpha level, such as 5%.
Intensive Care Unit (ICU) Data Analysis
The analysis of data from an Intensive Care Unit (ICU) involves a critical examination of variables related to patient care, demographics, and outcomes. In this instance, the goal is to examine whether the proportion of surgeries differs between male and female patients.

In handling ICU data analysis for this purpose, it is essential to:
  • Establish a clear understanding of the variables - in this case, sex of the patients and whether they had surgery.
  • Ensure that the data is accurate and represents the population in question.
  • Use appropriate statistical methods such as the two-proportion Z-test to interpret the data reliably.
One must carefully consider confounding factors and potential biases that could influence the results. This is especially pertinent in the medical field, where such data can directly impact patient care strategies and resource allocation.
Null and Alternative Hypotheses
Statistical hypothesis testing is grounded in formulating two competing statements: the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)). The null hypothesis represents a position of no effect or no difference, which in the case of ICU data would be that gender has no effect on the rate of surgery (i.e. \(p_m = p_f\)).

On the other hand, the alternative hypothesis suggests there is a significant effect or difference (i.e. \(p_m eq p_f\)). In a statistical test, evidence is collected to decide whether to support the null hypothesis or to endorse the alternative hypothesis. The outcome hinges on whether the observed data deviates sufficiently from what the null hypothesis would predict.

The formulation of these hypotheses is a crucial first step in any hypothesis test, including the two-proportion Z-test, as it precisely defines what is being tested and sets the stage for the investigation and interpretation of findings.
P-Value Interpretation
The p-value is a foundational concept in statistical hypothesis testing, representing the probability of observing a result, or one more extreme, assuming the null hypothesis is true. In practical terms, a low p-value indicates that the observed data is unlikely under the null hypothesis and suggests rejecting \(H_0\).

When interpreting p-values, context is everything. A generally accepted threshold for significance is 5%, but this value can vary depending on the field of study and the consequences of false findings. If a p-value is below the threshold, we say there's evidence against the null hypothesis. For instance, in the ICU data analysis example, a p-value less than 0.05 would imply a statistically significant difference in the surgical rates between males and females. It is pivotal to remember that a p-value does not measure the magnitude or practical importance of a result, but simply whether it's statistically significant.

Ultimately, p-value interpretation is more nuanced than a simple pass/fail criterion. It should be considered alongside the effect size, confidence intervals, and the broader context of the research to draw robust conclusions.

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Most popular questions from this chapter

Metal Tags on Penguins and Survival Data 1.3 on page 10 discusses a study designed to test whether applying metal tags is detrimental to penguins. One variable examined is the survival rate 10 years after tagging. The scientists observed that 10 of the 50 metal tagged penguins survived, compared to 18 of the 50 electronic tagged penguins. Construct a \(90 \%\) confidence interval for the difference in proportion surviving between the metal and electronic tagged penguins \(\left(p_{M}-p_{E}\right)\). Interpret the result.

Gender Bias In a study \(^{52}\) examining gender bias, a nationwide sample of 127 science professors evaluated the application materials of an undergraduate student who had ostensibly applied for a laboratory manager position. All participants received the same materials, which were randomly assigned either the name of a male \(\left(n_{m}=63\right)\) or the name of a female \(\left(n_{f}=64\right) .\) Participants believed that they were giving feedback to the applicant, including what salary could be expected. The average salary recommended for the male applicant was \(\$ 30,238\) with a standard deviation of \(\$ 5152\) while the average salary recommended for the (identical) female applicant was \(\$ 26,508\) with a standard deviation of \(\$ 7348\). Does this provide evidence of a gender bias, in which applicants with male names are given higher recommended salaries than applicants with female names? Show all details of the test.

Do Ovulating Women Affect Men's Speech? Studies suggest that when young men interact with a woman who is in the fertile period of her menstrual cycle, they pick up subconsciously on subtle changes in her skin tone, voice, and scent. A study introduced in Exercise \(\mathrm{B} .23\) suggests that men may even change their speech patterns around ovulating women. The men were randomly divided into two groups with one group paired with a woman in the fertile phase of her cycle and the other group with a woman in a different stage of her cycle. The same women were used in the two different stages. For the men paired with a less fertile woman, 38 of the 61 men copied their partner's sentence construction in a task to describe an object. For the men paired with a woman at peak fertility, 30 of the 62 men copied their partner's sentence construction. The experimenters hypothesized that men might be less likely to copy their partner during peak fertility in a (subconscious) attempt to attract more attention to themselves. Use the normal distribution to test at a \(5 \%\) level whether the proportion of men copying sentence structure is less when the woman is at peak fertility.

Split the Bill? Exercise 2.153 on page 105 describes a study to compare the cost of restaurant meals when people pay individually versus splitting the bill as a group. In the experiment half of the people were told they would each be responsible for individual meal costs and the other half were told the cost would be split equally among the six people at the table. The data in SplitBill includes the cost of what each person ordered (in Israeli shekels) and the payment method (Individual or Split). Some summary statistics are provided in Table 6.20 and both distributions are reasonably bell-shaped. Use this information to test (at a \(5 \%\) level ) if there is evidence that the mean cost is higher when people split the bill. You may have done this test using randomizations in Exercise 4.118 on page 302 .

In Exercises 6.150 and \(6.151,\) use StatKey or other technology to generate a bootstrap distribution of sample differences in proportions and find the standard error for that distribution. Compare the result to the value obtained using the formula for the standard error of a difference in proportions from this section. Sample A has a count of 30 successes with \(n=100\) and Sample \(\mathrm{B}\) has a count of 50 successes with \(n=250\).

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