Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Survival Status and Heart Rate in the ICU The dataset ICUAdmissions contains information on a sample of 200 patients being admitted to the Intensive Care Unit (ICU) at a hospital. One of the variables is HeartRate and another is Status which indicates whether the patient lived (Status \(=0)\) or died (Status \(=1\) ). Use the computer output to give the details of a test to determine whether mean heart rate is different between patients who lived and died. (You should not do any computations. State the hypotheses based on the output, read the p-value off the output, and state the conclusion in context.) Two-sample \(\mathrm{T}\) for HeartRate \(\begin{array}{lrrrr}\text { Status } & \text { N } & \text { Mean } & \text { StDev } & \text { SE Mean } \\ 0 & 160 & 98.5 & 27.0 & 2.1 \\ 1 & 40 & 100.6 & 26.5 & 4.2 \\ \text { Difference } & = & m u(0) & -\operatorname{mu}(1) & \end{array}\) Estimate for difference: -2.13 \(95 \% \mathrm{Cl}\) for difference: (-11.53,7.28) T-Test of difference \(=0\) (vs not \(=\) ): T-Value \(=-0.45\) P-Value \(=0.653\) DF \(=60\)

Short Answer

Expert verified
Based on the p-value of 0.653 being greater than the significance level (like 0.05), there's not enough statistical evidence to suggest that there is a difference in the mean heart rates between the patients who survived and who died.

Step by step solution

01

Understand the provided data

The output given is from a two-sample t-test comparing the mean heart rates of two groups of patients: those who survived (Status = 0) and those who died (Status = 1). The mean heart rate for survivors was 98.5, and it was 100.6 for patients who died.
02

State the hypotheses

The null hypothesis \(H_0\) is: there is no difference in the mean heart rates between both groups. So, the mean heart rate for group 0 (survived) is equal to the mean heart rate for group 1 (died). The alternate hypothesis \(H_a\) would therefore be: there is a difference in the mean heart rates between those who survived and who died.
03

Read the p-value from the output

The p-value from the output is 0.653. This p-value is so high that it significantly exceeds common-significance level thresholds like 0.1, 0.05, or 0.01.
04

Conclude based on the p-value

Since the p-value is much greater than standard significance levels (i.e., the p-value of 0.653 exceeds 0.05), there is not enough evidence to reject the null hypothesis. In this context, this means that there's not enough statistical evidence to say that there is a difference in the mean heart rates of patients who lived and who died.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Mean Heart Rate Comparison
The mean heart rate is a crucial vital sign used by clinicians to assess a patient's health condition, especially in critical care environments like the Intensive Care Unit (ICU). When comparing mean heart rates, healthcare professionals are often interested in determining whether there is a significant difference between various patient groups. In our example, the two groups are patients who survived after being admitted to the ICU and those who did not.

Using a two-sample t-test, we can compare the mean heart rate for survivors, which in this case is 98.5, to the mean heart rate for non-survivors, which is 100.6. However, a direct comparison of these numbers is not enough. We must assess whether this difference is statistically significant, which means that it is unlikely to have occurred by chance. This is where statistical hypothesis testing becomes invaluable, as it helps us determine whether observed differences in mean heart rates can be considered a generalizable phenomenon or just a random occurrence within our sample.
ICU Patient Survival Analysis
Survival analysis in the context of ICU patients involves using statistical methods to examine the expected duration of time until one or more events of interest occur, such as death in this severe healthcare setting. In this case, the 'event' is the patient's death, and the main focus is to observe differences in heart rate that could be associated with patients' survival.

By conducting a survival analysis, we can identify not only if there are differences in heart rate between survivors and non-survivors but also how this variable might influence the prognosis of ICU patients. Heart rate could be one of many factors that, when analyzed together, might provide a predictive model for ICU patient outcomes. While our simple two-sample t-test doesn't provide the full scope of survival analysis, it offers an introduction into how one variable, mean heart rate, can be scrutinized for survival association.
Null Hypothesis in Statistics
In statistics, the null hypothesis (\(H_0\)) serves as a default statement that there is no effect or no difference. When assessing the mean heart rates between patients who lived and those who died, our null hypothesis states that the mean heart rate is the same for both groups. It's a statement of 'no difference' or 'status quo.'

It is essential to define a null hypothesis clearly because it sets the stage for statistical testing. We are essentially 'putting it to the test' to see if the evidence we have collected provides strong enough data to refute it. The alternate hypothesis (\(H_a\) or \(H_1\) is the counterpart to the null, which posits that there is a difference (in this case, a difference in mean heart rates between the two groups of ICU patients). The role of a two-sample t-test is to determine which hypothesis is more likely true, based on the collected data.
P-Value Interpretation
The p-value, a fundamental concept in statistics, measures the probability of obtaining the observed results, or more extreme, when the null hypothesis is true. In layman's terms, it tells us how 'surprised' we should be by our data if we assumed the null hypothesis was true.

A low p-value (typically below 0.05) suggests that the observed data is highly unlikely under the null hypothesis, leading us to reject the null in favor of the alternative hypothesis. Conversely, a high p-value indicates that the data is consistent with the null hypothesis. Therefore, in our ICU heart rate study, with a p-value of 0.653, we find no statistically significant grounds to reject the null hypothesis. The likelihood of observing such a difference or more in mean heart rates, given that there is in truth no difference, is relatively high, at 65.3%. This high p-value tells us that the evidence does not suggest a significant difference in the mean heart rates for patients who lived compared to those who died.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the endpoints of the t-distribution with \(2.5 \%\) beyond them in each tail if the samples have sizes \(n_{1}=15\) and \(n_{2}=25\).

A data collection method is described to investigate a difference in means. In each case, determine which data analysis method is more appropriate: paired data difference in means or difference in means with two separate groups. To study the effect of women's tears on men, levels of testosterone are measured in 50 men after they sniff women's tears and after they sniff a salt solution. The order of the two treatments was randomized and the study was double-blind.

Use a t-distribution and the given matched pair sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distribution of the differences is relatively normal. Assume that differences are computed using \(d=x_{1}-x_{2}\). Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1}<\mu_{2}\) using the paired data in the following table: $$ \begin{array}{lllllllll} \hline \text { Treatment } 1 & 16 & 12 & 18 & 21 & 15 & 11 & 14 & 22 \\ \text { Treatment } 2 & 18 & 20 & 25 & 21 & 19 & 8 & 15 & 20 \\ \hline \end{array} $$

What Gives a Small P-value? 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: Option 1 has: \(\quad \bar{x}_{1}=25 \quad \bar{x}_{2}=23\) $$ \text { Option } 2 \text { has: } \quad \bar{x}_{1}=25 \quad \bar{x}_{2}=11 $$ (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: \(\quad s_{1}=15 \quad s_{2}=14\) $$ \text { Option } 2 \text { has: } \quad s_{1}=3 \quad s_{2}=4 $$ (c) Both options have the same means \(\left(\bar{x}_{1}=22,\right.\) \(\left.\bar{x}_{2}=17\right)\) and same standard deviations but: Option 1 has: \(\quad n_{1}=800 \quad n_{2}=1000\) $$ \text { Option } 2 \text { has: } \quad n_{1}=25 \quad n_{2}=30 $$

In Exercise \(6.107,\) we see that plastic microparticles are contaminating the world's shorelines and that much of the pollution appears to come from fibers from washing polyester clothes. The same study referenced in Exercise 6.107 also took samples from ocean beaches. Five samples were taken from each of 18 different shorelines worldwide, for a total of 90 samples of size \(250 \mathrm{~mL}\). The mean number of plastic microparticles found per \(250 \mathrm{~mL}\) of sediment was 18.3 with a standard deviation of 8.2 . (a) Find and interpret a \(99 \%\) confidence interval for the mean number of polyester microfibers per \(250 \mathrm{~mL}\) of beach sediment. (b) What is the margin of error? (c) If we want a margin of error of only ±1 with \(99 \%\) confidence, what sample size is needed?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free