/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 Average Age for ICU Patients The... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Average Age for ICU Patients The ICUAdmissions dataset includes a variable indicating the age of the patient. Find and interpret a \(95 \%\) confidence interval for mean age of ICU patients using the facts that, in the sample, the mean is 57.55 years and the standard error for such means is \(S E=1.42\). The sample size of 200 is large enough to use a normal distribution.

Short Answer

Expert verified
Therefore, we can be 95% confident that the average age of ICU patients falls between 54.71 years and 60.39 years.

Step by step solution

01

Identify the z-score

First, we need to find the correct z-score for a 95% confidence interval. This z-score comes from the z-distribution and it represents the 'cut-off' point where 2.5% of the probability is in the tails, and 95% is in between. From standard statistical tables, the z-value that corresponds to the 95% interval is approximately 1.96.
02

Apply the confidence interval formula

Next, the confidence interval formula for a population mean is applied. The formula is \( \overline{X} ± (Z * SE) \), where \( \overline{X} \) is the sample mean, Z is the z-score for the desired confidence level, and SE is the standard error. Substituting the given values into the formula, we get: \( 57.55 ± (1.96 * 1.42) \).
03

Calculate the confidence interval

Finally, to get the confidence interval, we calculate the two results from the formula: Lower bound Interval = \( 57.55 - (1.96 * 1.42) = 54.71 \)Upper bound Interval = \( 57.55 + (1.96 * 1.42) = 60.39 \)

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 91Ó°ÊÓ!

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

In Exercises 5.44 to 5.49 , find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately normally distributed. A \(95 \%\) confidence interval for a mean \(\mu\) if the sample has \(n=50\) with \(\bar{x}=72\) and \(s=12,\) and the standard error is \(S E=1.70\)

Exercises 5.21 to 5.28 ask you to convert an area from one normal distribution to an equivalent area for a different normal distribution. Draw sketches of both normal distributions, find and label the endpoints, and shade the regions on both curves. The area between 1 and 2 for a standard normal distribution converted to a \(N(100,15)\) distribution.

Boys Heights Heights of 10-year-old boys (5th graders) follow an approximate normal distribution with mean \(\mu=55.5\) inches and standard deviation \(\sigma=2.7\) inches \({ }^{6}\) (a) Draw a sketch of this normal distribution and label at least three points on the horizontal axis. (b) According to this normal distribution, what proportion of 10 -year-old boys are between \(4 \mathrm{ft} 4\) in and \(5 \mathrm{ft}\) tall (between 52 inches and 60 inches)? (c) A parent says his 10 -year-old son is in the 99 th percentile in height. How tall is this boy?

In Exercises 5.56 and \(5.57,\) find the p-value based on a standard normal distribution for each of the following standardized test statistics. (a) \(z=-1.08\) for a lower tail test for a mean (b) \(z=4.12\) for an upper tail test for a proportion (c) \(z=-1.58\) for a two-tailed test for a slope

Penalty Shots in World Cup Soccer A study \(^{13}\) of 138 penalty shots in World Cup Finals games between 1982 and 1994 found that the goalkeeper correctly guessed the direction of the kick only \(41 \%\) of the time. The article notes that this is "slightly worse than random chance." We use these data as a sample of all World Cup penalty shots ever. Test at a \(5 \%\) significance level to see whether there is evidence that the percent guessed correctly is less than \(50 \%\). The sample size is large enough to use the normal distribution. The standard error from a randomization distribution under the null hypothesis is \(S E=0.043 .\)

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.